Object x is dropped from a height of 65m at the same instant object y is thrown vertically upwards with a velocity of 5m^-1. Calculate the distance at which the two objects pass each other

Answers

Answer 1

The path travelled by a projectile is known as its trajectory. The distance at which the two objects pass each other is 16.9 m.

What is a projectile?

Any object which is thrown into space and the only force acting on it is the force of gravity. The motion of a projectile is known as the projectile motion.

The height covered by the projectile from ground is:

h = ut - 1/2 gt²

h = 5t - 1/2 × 9.8 × t²

h = 5t - 4.9t²   (1)

When projectile is thrown downward, u = 0

65 - h = 1/2 gt²

65 - (5t - 4.9t² ) = 4.9t²

65 = 5t

t = 65/5 = 13 s

h = 65t - 4.9t²

= 65 × 13 - 4.9 × 169

= 16.9 m

Thus the distance at which the two objects pass each other is  16.9 m.

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Related Questions

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100PTS
1. if a ball of 60kg falls from a height of 20m to the ground calculate the potential energy (g=10mls²)
2. A girl whose weight is 30N run up a flight of stairs 5m high in 6 seconds. what is her average power.​

Answers

1.) The potential energy of an object can be calculated using the equation: PE = mgh

Where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.

So, the potential energy of a 60 kg ball falling from a height of 20 m would be:

PE = 60 kg * 10 m/s² * 20 m = 12000 J (joules)

2.) The average power can be calculated using the equation: P = W / t

Where W is the work done and t is the time taken.

In this case, the work done can be calculated as W = mgh, where m is the mass of the girl (30 N), g is the acceleration due to gravity (10 m/s²), and h is the height of the stairs (5 m).

So, W = 30 N * 10 m/s² * 5 m = 1500 J

And the average power can be calculated as:

P = 1500 J / 6 s = 250 W (watts)

Answer:

Answer is in attached photo.

Explanation:

Solution

The solution is in the attached photo, do take note to solve these questions, multiple formulas are needed:

Q1) Potential Energy = Mass x Gravitational Accceleration x Height

= m x g x h (SI Unit: J)

Q2) Work Done = Force x Distance

= F x d (SI Unit: J)

Average Power = [tex]\frac{Total \ Work \ Done}{Total \ Time}[/tex] (SI Unit: W)


A 66 kg student traveling in a car with a constant velocity has a kinetic energy of 1.1 104 J. What is the speedometer reading of the car in km/h?

Answers

Answer:

The speedometer reading of the car in km/h is 100.1 km/h. This can be calculated by using the formula K = 0.5mv2, where K is the kinetic energy, m is the mass, and v is the velocity. Rearranging this equation to solve for v yields v = √(2K/m). In this case, m = 66 kg and K = 1.1 x 104 J, so v = √(2 x 1.1 x 104 J/ 66 kg) = 100.1 km/h.

A cart rolls with low friction on a track. A fan is mounted on the cart, and when the fan is turned on, there is a constant force acting on the cart. Three different experiments are performed: Fan off: The cart is originally at rest. You give it a brief push, and it coasts a long distance along the track in the +x direction, slowly coming to a stop. Fan forward: The fan is turned on, and you hold the cart stationary. You then take your hand away, and the cart moves forward, in the +x direction. After traveling a long distance along the track, you quickly stop and hold the cart. Fan backward: The fan is turned on facing the "wrong" way, and you hold the cart stationary. You give it a brief push, and the cart moves forward, in the +x direction, slowing down and then turning around, returning to the starting position, where you quickly stop and hold the cart. The figure displays graphs of x, position along the track, vs. time. The graphs start when the cart is at rest, and end when the cart is again at rest. Match the experiment with the correct graph.

Answers

Experiment A (Fan Off) corresponds to graph C, as it shows the cart starting from rest and slowing down until it comes to a stop. Experiment B (Fan Forward) corresponds to graph A, as it shows the cart starting from rest, accelerating, and then coming to a stop. Experiment C (Fan Backward) corresponds to graph B, as it shows the cart starting from rest, accelerating, slowing down, and then turning around and returning to the starting position before coming to a stop.

Friction is the force that resists the motion of one object against another. It is a force that acts in the opposite direction of the motion. Friction occurs when two objects rub against each other, resulting in a force that resists their relative motion. This force can be caused by a variety of factors, including the surface roughness of the objects, the material they are made of, the weight of the objects, and the environment they are in.

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What makes types of light different?

Answers

Answer:

Their wavelength

Explanation:

Light is a form of energy that travels as waves. Their length — or wavelength — determines many of light's properties. For instance, wavelength accounts for light's colour and how it interacts with matter. The range of wavelengths, from super short to very long, is known as the light spectrum.. Please mark this answer the brainliest. I need it. thanks.

Hope this helps!

Which has more kinetic energy: a 0.0011-kg bullet traveling at 389 m/s or a 7.8 107-kg ocean liner traveling at 11 m/s (21 knots)?

Answers

Answer:

the ocean liner

Explanation:

KE = 1/2mv²

bullet:  KE = 1/2(0.0011 kg)((389 m/s)² = 83.2 J

ship:  KE = 1/2(7.8x10⁷ kg)(11 m/s)² = 4.7x10⁹ J

The balls are released from rest at a height of 5.0 m
at time t=0s
. Using these numbers and basic kinematics, you can determine the amount of time it takes for the balls to reach the ground.

Answers

Time taken by the ball to reach the ground is  1.0 second.

What is acceleration?

Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).

The initial height of the  balls = 5.0 meter

Initial time = 0 second

The initial speed of the ball = 01 m/s

Acceleration due to gravity = 10.0m/s²

Hence, Time taken by the ball to reach the ground is = √(2 × 5.0/10.0) second

= 1.0 second.

Therefor it take 1.0 second to ball

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At the instant a traffic light turns green, a car that has been waiting at the intersection starts ahead with a constant acceleration of 2.00 m/s2 . At that moment a truck traveling with a constant velocity of 15.0 m/s overtakes and passes the car. (a) Calculate the time necessary for the car to reach the truck. (b) Calculate the distance beyond the traffic light that the car will pass the truck. (c) Determine the speed of the car when it passes the truck.

Answers

Answer:

Explanation:

(a) The time necessary for the car to reach the truck can be calculated using the equation for average velocity:

v_avg = v0 + at

where v0 is the initial velocity of the car (which is 0 m/s at the instant the light turns green), a is the acceleration of the car (2.00 m/s^2), and t is the time elapsed.

Since the truck is traveling with a constant velocity of 15.0 m/s, we can equate the average velocity of the car to the velocity of the truck:

v0 + at = 15.0 m/s

Solving for t, we get:

t = (15.0 m/s - v0) / a = 15.0 m/s / 2.00 m/s^2 = 7.50 s

(b) The distance beyond the traffic light that the car will pass the truck can be calculated using the equation for displacement:

d = v0t + 1/2at^2

Plugging in the values we found above, we get:

d = (0 m/s)(7.50 s) + 1/2(2.00 m/s^2)(7.50 s)^2 = 56.25 m

(c) The speed of the car when it passes the truck can be calculated using the equation for velocity:

v = v0 + at = 0 m/s + (2.00 m/s^2)(7.50 s) = 15.0 m/s

A Jogger runs 2.0km due east, then 10km at 45° north of east and finally 0.5km due north- Determine the displacement using graphical method Using Scale of 5ocm =2 km​

Answers

Answer:

Explanation:

To determine the displacement of the jogger using a graphical method, we can create a vector diagram to represent the individual components of the jogger's motion. We can then find the resulting displacement by adding the vectors end to end.

Assuming that "50cm = 2km" is the scale, we can convert the distances to represent them on the diagram:

2.0 km due east = 100 cm

10 km at 45° north of east = 70.71 cm (using Pythagorean theorem)

0.5 km due north = 25 cm

Next, we can plot the vectors on a coordinate grid, with east as the x-axis and north as the y-axis.

The jogger's first move, 2 km due east, can be represented by a vector starting from the origin and extending 100 cm to the right.

The second move, 10 km at 45° north of east, can be represented by a vector starting from the end of the first vector and extending 70.71 cm up and to the right.

The third move, 0.5 km due north, can be represented by a vector starting from the end of the second vector and extending 25 cm straight up.

Finally, to find the displacement, we add the vectors end to end. The displacement is the vector that starts at the origin and extends to the endpoint of the final vector. The magnitude of the displacement can be found using the Pythagorean theorem.

Using this method, we can determine the displacement of the jogger using a graphical method and the given scale.

a long, straight metal rod has a radius of 4.60 cm and a charge per unit length of 27.4 nc/m. find the electric field at the following distances from the axis of the rod, where distances are measured perpendicular to the rod's axis

Answers

The distances from the axis of the rod (a long, straight metal rod has a radius of 4.60 cm and a charge per unit length of 27.4 nc/m.), where distances are measured perpendicular to the rod's axis

(a) 3.20 cm = 15.39 x 10³ N/c

(b) 2.0 cm = 2.43 x 10³ N/c

(c) 200 cm = 2.4 x 10² N/c

Given a long, infinite length and a constant charge per unit, the electric field is given by:

E = λ /2πε₀r

Where,

λ = wavelength (m)

ε₀ = 8.8542 x 10⁻¹²

r = radius (m)

Hence,

The distance are measured perpendicular to the rod's axis:

a. 3.20 cm

E = (27.4 x 10⁻⁹) / 2π (8.8542 x 10⁻¹²) (0.0320)

= 15.39 x 10³ N/c

b. 20 cm

E = (27.4 x 10⁻⁹) / 2π (8.8542 x 10⁻¹²) (0.2)

= 2.43 x 10³ N/c

c. 200 cm

E = (27.4 x 10⁻⁹) / 2π (8.8542 x 10⁻¹²) (2)

= 2.4 x 10² N/c

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___ tubing is commonly used in hydraulic systems. Carbon Steel. _____ fittings are typically used in hydraulic equipment when attaching tubing to a hose.

Answers

Flared tubing is commonly used in hydraulic systems. Carbon Steel fittings are typically used in hydraulic equipment when attaching tubing to a hose.

Liquid fluid power is used by hydraulic machines to do operations. Heavy-duty construction vehicles are a typical illustration. Hydraulic fluid is pumped to numerous hydraulic motors and hydraulic cylinders located all around the machine in this sort of machine and is pressured in accordance with the resistance present. Control valves direct or automate the distribution of the fluid through hoses, tubes, or pipes.

Pascal's law, which asserts that any pressure applied to a fluid inside a closed system would transmit that pressure equally everywhere and in all directions, is the foundation of hydraulic systems, just like pneumatic systems. An incompressible liquid, as opposed to a compressible gas, serves as the fluid in a hydraulic system.

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2.) For an atom of helium, it takes 40.8 electron-volts of energy to move an electron from the first energy level to the second, it takes 48.4 electron-volts to move an electron from the first energy level to the third, and it takes 51.0 electron-volts of energy to move an electron from the first energy level to the fourth.

If I were to shine a light made of photons with 51.0 electron-volts of energy, which transition(s)

will I observe? Explain your answer.

Answers

Answer:

You will observe the transition from the first energy level to the fourth energy level. This is because the energy of the photons is equal to the amount needed to move the electron from the first energy level to the fourth energy level. Any energy higher than 51.0 electron-volts would not be observed because the electron would not be able to transition to a higher energy level.

A hot-air balloon is ascending at the rate of 11 m/s and is 59 m above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?

Answers

The speed of the package just before hitting the ground is 46.942 m/s.

How to calculate the speed?

We can solve this problem using the kinematic equations of motion. Let's assume that the positive direction is upwards and take the acceleration due to gravity to be -9.8 m/s².

(a) To find how long the package takes to reach the ground, we can use the equation:

y = y0 + v0t + (1/2)at²

Plugging in the values, we get:

0 = 59 m + (11 m/s)t + (1/2)(-9.8 m/s^2)t²

Simplifying and solving for t, we get:

t = 4.79 seconds

Therefore, the package takes approximately 4.79 seconds to reach the ground.

(b) To find the speed at which the package hits the ground, we can use the equation:

v = v0 + at

Plugging in the values, we get:

v = 11 m/s + (-9.8 m/s²)(4.79 s)

= -46.942 m/s (the negative sign indicates that the package is moving downwards)

The speed of the package just before hitting the ground is 46.942 m/s.

What really is speed in physics, and what is its unit?

The pace at which distance changes over time is referred to as speed. It has a dimension of time and distance. As a result, the fundamental unit of length as well as the basic unit if duration are combined to form the High silica content of speed. As a result, the Si derived unit for speed is the meter per second.

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after you size up the scene, the following are steps involved with performing a primary assessment for a child. place them in order in which they should occur.

Answers

The correct order of steps involved in performing a primary assessment for a child is check the patient for responsiveness, summon more advanced medical personnel, check for breathing and a pulse, and quickly scan for severe bleeding. The correct order is 2, 1, 4, and 3.

The first step in performing a primary assessment is to check the patient for responsiveness, if the child is responsive or not. This involves calling out their name, tapping their shoulder, or gently shaking them to see if they respond.

The next step is to summon more advanced medical personnel. If the child is not responsive or if you suspect a serious injury, it's important to call for more advanced medical personnel such as paramedics or emergency medical technicians (EMTs) as soon as possible.

Next, check for breathing and a pulse by checking if the child's airway, breathing, and pulse to make sure they're getting enough oxygen. If the child isn't breathing, begin cardiopulmonary resuscitation (CPR) immediately.

Then, quickly scan for severe bleeding. Look for any severe bleeding, such as from a major wound or a deep cut. If the child is bleeding severely, apply pressure to the wound to stop the bleeding.

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Your question seems incomplete, but I suppose the question was:

"After you size- up the scene, the following are steps involved with performing a primary assessment for a child. Place them in the order in which they should occur.

1. summon more advanced medical personnel

2. check the patient for responsiveness

3. quickly scan for severe bleeding

4. check for breathing and a pulse."

Determine the number of cubic feet of air per minute required to cook a room having a sensible heat gain of 4500 btu per hour to a temperature of 78 F dry bulb, if the air enters the room at a temperature of 63 F and the outside temperature is 93F

Answers

It is given that, the heat gained is 4500 btu per hour. The temperature difference here is  30 F and the specific heat of air is  0.24 btu/lb°F. Then the cubic feet of air per minute is 138.8 CFM.

What is sensible heat transfer ?

The sensible heat transfer in a system can be calculated using the equation below:

q = CFM × 1.08 ×ΔT

q = CFM x 0.075 lb/ft3 x 60 min/hour x 0.24 btu/lb°F x ∆T

where, 0.24 btu/lb°F is the specific heat of the dry air.

Given that q = 4500 btu/hour.

temperature difference = 93 F - 63 F.

Then 4500 btu/hr = CFM   × 1.08 × 30 F

CFM of air = 4500  btu/hr /(1.08 × 30 F ) = 138.8 CFM.

There for the number of cubic feat of air per minute is 138.8.

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Help me please physics worksheet

Answers

Kinetic friction has a coefficient of 0.52 while static friction has a coefficient of 0.67.

What is friction?

A force called friction prevents motion from occurring when two surfaces come into contact. Based on the characteristics of the two surfaces, their roughness, and the pressure used to press them together, it moves the objects in the opposite direction from the motion or planned motion.

How do you determine it?

a. The formula friction = force of friction / normal force is used to calculate the coefficient of static friction.

The frictional force needed to move the chair is 165 N, and the normal force equals the chair's weight (25 kg * 9.8 m/s2 = 245 N).

The static friction coefficient is therefore equal to 165 N / 245 N = 0.67.

b. The formula friction = force of friction / normal force is used to determine the coefficient of kinetic friction.

The frictional force needed to maintain the chair's motion at a constant speed is 127 N, whereas the normal force remains 245 N.

Therefore, the kinetic friction coefficient is equal to 0.52 for 127 N/245 N.

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Find the distance traveled by a boy if he walks 3 km north and then turns and walks 1 km to the west.
Include both magnitude and direction.

Answers

Answer:

Explanation:

Distance = total length traveled = 3 + 1 = 4 km

Displacement² = 3² + 1² = 10

Displacement = √10 = 3.16

direction = tan⁻¹(3/-1) = 71.6⁰ north of west

Which of the following is likely to happen when a glass of cold water is warmed to room temperature?Select the correct answer below:A. The water will absorb air as the solubility of its dissolved air increases.B. Small air bubbles will form as the solubility of its dissolved air decreases.C. The water will neither absorb nor release air because the solubility of its dissolved air does not change.D. We need more information to predict what will happen to the water.

Answers

Small air bubbles will form as the solubility of its dissolved air decreases when a glass of cold water is warmed to room temperature. Option B. is correct.

It's not as easy as it seems to pour water into a glass. Gases have been dissolved in them.

Furthermore, before we warm a glass of ice-cold water to room temperature, we must understand what we are doing. In essence, we are raising the water's temperature.

We now need to think about how rising temperatures will affect how soluble gases in the water.

By raising the temperature, we are supplying thermal energy in the form of heat to the gaseous molecules. The mobility and propensity to escape from the solution both increase as the thermal energy of the dissolved gaseous molecules rises. Because of this, the solubility of gases decreases as temperature rises.

As a result, when water is heated, gaseous molecules that have been dissolved begin to rise up from the water.

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a golf ball is suit from a ship which is at a height 46 m above water surface the horizontal distance covered by Golf is​

Answers

The horizontal distance travelled by the golf is 7.75 m.

What is the height of fall of the golf?

The time taken for the golf to fall from the given height is calculated by applying the following equation.

t = √ ( 2h / g )

where;

h is the height of fall of the golfg is acceleration due to gravity

t = √ ( 2 x 46 / 9.8 )

t = 3.1 seconds

The horizontal distance travelled by the golf is calculated as follows;

d = vt

where;

v is the horizontal velocity

d = 2.5 m/s x 3.1 s

d = 7.75 m

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The complete question is below:

a golf ball is suit from a ship which is at a height 46 m above water surface the horizontal distance covered by Golf is​ if the horizontal velocity is 2.5 m/s.

The water is flowing through a tapping pipe having diameter 400mm and 200mm at section 1 and 2 respectively. The discharge through the pipe is 0.06m³/sec. The section 1 is. 10m above datum line and section 2 is 6m above datum line. Find the intensity of pressure at section 2 to that if that of section 1 is 420 KN/m².​

Answers

The intensity of pressure at section 2 is approximately 345.64 kN/m^2.

What is pressure?

Pressure is described as the force applied perpendicular to the surface of an object per unit area over which that force is distributed.

To determine the intensity of pressure at section 2, we need to use the equation of continuity and the Bernoulli equation:

we will substitute the values from the continuity equation into the Bernoulli equation which will give us :

P_1 + 1/2 * p * (Q_1 / A_1)^2 + p * g * h_1 = P_2 + 1/2 * p * (Q_2 / A_2)^2 + p * g * h_2

Substitute all the given values into the equation

Solving for P_2 = 420,000 N/m^2 + 47.06 KN/m^2 - 71.42 KN/m^2 = 345.64 KN/m^2

In conclusion,  the intensity of pressure at section 2 is approximately 345.64 kN/m^2.

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Can someone please help me label this right

Answers

The tension at the top of the horizontal circle is T = m (v²/r - g ).

The tension at the bottom of the horizontal circle is m (v²/r + g ).

What is the tension at the bottom and top of the rope?

The tension at the bottom and top of he rope is calculated by applying the following formula as shown below;

The tension at the top of the horizontal circle is calculated as;

T = ma - mg

T = mv²/r - mg

T = m (v²/r - g )

where;

v is the speedr is the radius of the circleg is acceleration due to gravity

The tension at the bottom of the horizontal circle is calculated as;

T = ma + mg

T = mv²/r + mg

T = m (v²/r + g )

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The complete question is below:

Find the expression for the tension at the bottom and top of the circle

Determine the current at point A in this circuit.

A) 2.6 A
B) 1.6 A
C) 1.8 A
D) 2.4 A

Answers

Answer:

2.6 A

Explanation:

Firstly here we need to find out the net resistance. 30Ω and 40Ω are grouped in parallel and to these 10Ω and 15Ω are in series. So we can find the net resistance as ,

[tex]\implies R_{net}=\dfrac{R_1R_2}{R_1+R_2} \\[/tex]

[tex]\implies R_{net}=\dfrac{30\times 40}{30+70}\Omega \\[/tex]

[tex]\implies R_{net}=\dfrac{120}{7}\Omega \\[/tex]

Again,

[tex]\implies R_{net}= R_1 + R_2+R_3 \\[/tex]

[tex]\implies R_{net}=\dfrac{120}{7}+10+15\Omega \\[/tex]

[tex]\implies R_{net}= \dfrac{120+105+70}{7}\Omega\\[/tex]

[tex]\implies R_{net}= \dfrac{295}{7}\Omega \approx 42\Omega \\[/tex]

Now use Ohm's law as ,

[tex]\implies V = iR \\[/tex]

[tex]\implies 110V = i\times 42\Omega\\[/tex]

[tex]\implies i =\dfrac{110}{42} A\\[/tex]

[tex]\implies \underline{\underline{ i = 2.6\ A }} \\[/tex]

and we are done!

a block hangs in equilibrium from a vertical spring. the equilibrium position sags by 7.00 cm when a second identical block is added. what is the oscillation frequency of the two-block system?

Answers

The oscillation frequency of the two-block system is 1.6 Hz.

What is equilibrium position?

When a system is in equilibrium, neither its internal energy state nor its state of motion tend to change over time. A simple mechanical body is said to be in equilibrium if it neither experiences linear acceleration nor angular acceleration; unless it is disturbed by an external force, it will remain in that state indefinitely. If all of the forces acting on a single particle are vector summated to zero, equilibrium results. In addition to the states described above for the particle, a rigid body is said to be in equilibrium if the vector sum of all torques acting on it equals zero, keeping its state of rotational motion constant.

[tex]$$The oscillation frequency of the two-block is system is then found from Equation (3):$$\begin{aligned}f & =\frac{1}{2 \pi} \sqrt{\frac{k}{2 m}} \end{aligned}[/tex]

[tex]\begin{aligned}& =\frac{1}{2 \pi} \sqrt{\frac{196 \mathrm{~s}^{-2}}{2}} \\& =1.6 \mathrm{~Hz}\end{aligned}$$\\\\\text{where $2 m$ is the mass of the two blocks}[/tex]

Thus, the oscillation frequency of the two-block system is 1.6 Hz.

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All living things release energy from their food in a process called________, which happens inside their cells

Answers

All living things release energy from their food in a process called respiration, which happens inside their/the cells.

there is a single electron at a distance from the point charge. on which of the following quantities does the force on the electron depend?

Answers

The force acting on the electron is dependent upon the distance seen between positive charge and indeed the electron, the charges on the electron, and indeed the charge of something like the positive charge.

Why is there a positive charge?

A material that has more protons than electrons is said to have a positive charge. We are aware that electrons seem to be positively charged and protons become positively charged. Consequently, positively charged objects have more protons than electrons.

The positive charge is absent:

The positive charge on the protons and neutrons (red) results from an imbalance between protons and electrons, more particularly, when there are more protons than electrons. Protons can be added to an atom or other substance with both a neutral charge to produce a positive charge.

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The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be found by taking the vector sum of the fields created by individual point charges. Note that if a charge is placed in an electric field created by q', q will not significantly affect the electric field if it is small compared to q'.

Imagine an isolated positive point charge with a charge Q (many times larger than the charge on a single electron).

There is a single electron at a distance from the point charge. On which of the following quantities does the force on the electron depend?

Check all that apply.

A the distance between the positive charge and the electron

B the charge on the electron

C the mass of the electron

D the charge of the positive charge

E the mass of the positive charge

F the radius of the positive charge

G the radius of the electron

I got A, B, and D for this part of the problem and was correct. However,

For the same situation as in Part A, on which of the following quantities does the electric field at the electron's position depend?

Check all that apply.

A the distance between the positive charge and the electron

B the charge on the electron

C the mass of the electron

D the charge of the positive charge

E the mass of the positive charge

F the radius of the positive charge

G the radius of the electron

explain three examples of a workshop

Answers

The required three examples of a workshop are given as 1) Woodworking Workshop, 2) Photography Workshop, and 3) a Writing Workshop.

What are workshops?

Workshops are the places where production operations take place.

Woodworking Workshop: A workshop that teaches the skills and techniques involved in woodworking such as using tools, working with different types of wood, and constructing wooden objects such as furniture, toys, and crafts.

Photography Workshop: A workshop that focuses on teaching the fundamentals of photography such as composition, lighting, and editing techniques. Participants learn how to use a camera and develop their own style through practical exercises and group discussions.

Writing Workshop: A workshop that provides a supportive environment for writers to develop their craft and receive feedback from peers and a writing instructor. Topics covered in a writing workshop may include character development, plot structure, dialogue, and revision techniques.

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a new laptop was delivered to your home and left on the porch while you were at work. it is very cold outside, and you don't know how long it has been there. you unpack the laptop. you would like to know if it is working properly. which of the following is the best action to take?

Answers

The optimal course of action is to allow the laptop to warm up to room temperature before turning it on.

The room atmosphere and the outside temperature of the laptop are very different in terms of temperature.

When the user was at work and the laptop was discovered on the deck, it was delivered to his house. He is unable to recall how long it has been really chilly outside. The user opens his laptop and checks to see if everything is working correctly. Instead, turn on the machine at room temperature and let it preheat up.

When a computer is still in a frigid environment, it must warm up to room temperature for roughly 6 to 24 hours before being turned on. Otherwise, the heat the components produce might cause the laptop to become moist with water. The system's components might then be harmed by that water.

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A ball is dropped, from

rest, from the top of a building. There are two motion sensors positioned

outside of two windows in the building to record the velocities of the ball

as it passes them. Sensor A is located above sensor B. Sensor A records

the ball’s velocity to be -15.5 m

s

and sensor B records the ball’s velocity to

be -17.2 m

s

. a) how far apart are the sensors?
b) how far from sensor a is the top of the building where the ball was released
c) if the time from when the ball passes sensor b until it hits the ground is 3.5s how tall is the building

Answers

Answer:

The sensors are approximately [tex]2.83\; {\rm m}[/tex] apart.

Sensor [tex]\texttt{a}[/tex] is approximately [tex]12.2\; {\rm m}[/tex] from the top of the building.

Height of the building is approximately [tex]135\; {\rm m}[/tex].

(Assumptions: air resistance on the ball is negligible; [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)

Explanation:

Make use of the SUVAT equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex], where [tex]x[/tex] represents displacement, [tex]v[/tex] and [tex]u[/tex] are the final and initial velocity, and [tex]a[/tex] is the acceleration.

In other words, as the velocity of the object changes from [tex]u\![/tex] to [tex]v\![/tex] at a rate of [tex]a[/tex], position of the object would have changed by [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex].

a)

When the ball is at sensor [tex]\texttt{a}[/tex], velocity of the ball was [tex]u = (-15.5)\; {\rm m\cdot s^{-1}}[/tex].

Shortly after, when the ball is at sensor [tex]\texttt{b}[/tex], velocity of the ball was [tex]v = (-17.2)\; {\rm m\cdot s^{-2}}[/tex].

Under the assumptions, [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex]. Apply the SUVAT equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex] to find the change in the position of the ball between sensor [tex]\texttt{a}[/tex] ([tex]u = (-15.5)\; {\rm m\cdot s^{-1}}[/tex]) and [tex]\texttt{b}[/tex] ([tex]v = (-17.2)\; {\rm m\cdot s^{-2}}[/tex]):

[tex]\begin{aligned}x &= \frac{(-17.2)^{2} - (-15.5)^{2}}{2\, (-9.81)} \; {\rm m} \approx (-2.83)\; {\rm m}\end{aligned}[/tex].

Hence, the distance between sensor [tex]\texttt{a}[/tex] and [tex]\texttt{b}[/tex] is approximately [tex]2.83\; {\rm m}[/tex].

b)

Since the ball was released from rest, the initial velocity of the ball would be [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex].

Let [tex]v[/tex] denote the velocity of the ball at sensor [tex]\texttt{a}[/tex]: [tex]v = (-15.5)\; {\rm m\cdot s^{-1}}[/tex].

Apply the SUVAT equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex] to find the change in the position of the ball between the top of the roof ([tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]) and sensor [tex]\texttt{a}[/tex] ([tex]v = (-15.5)\; {\rm m\cdot s^{-1}}[/tex]):

[tex]\begin{aligned}x &= \frac{(-15.5)^{2} - (0)^{2}}{2\, (-9.81)} \; {\rm m} \approx (-12.2)\; {\rm m}\end{aligned}[/tex].

In other words, the distance between sensor [tex]\texttt{a}[/tex] and the top of the roof would be approximately [tex]12.2\; {\rm m}[/tex].

c)

Another SUVAT equation, [tex]v = u + a\, t[/tex], gives the velocity of an object after accelerating for a duration of [tex]t[/tex] (at a rate of [tex]a[/tex], starting from an initial velocity of [tex]u[/tex].)

Make use of this SUVAT equation to find the velocity of the ball right before hitting the ground. Specifically, velocity of the ball was [tex]u = (-17.2)\; {\rm m\cdot s^{-1}}[/tex] at sensor [tex]\texttt{b}[/tex]. After accelerating at [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex] for [tex]t = 3.5\; {\rm s}[/tex], velocity of the ball would be:

[tex]\begin{aligned}v &= u + a\, t \\ &= (-17.2)\; {\rm m\cdot s^{-1}} + (-9.81)\, (3.5)\; {\rm m\cdot s^{-1}} \\ &= (-51.535)\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].

Again, the velocity of the ball was [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] at the top of the building. Apply the SUVAT equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex] to find the change in the position of the ball between the top of the building ([tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]) and right before hitting the ground ([tex]v = (-51.535)\; {\rm m\cdot s^{-1}}[/tex]):

[tex]\begin{aligned}x &= \frac{(-51.535)^{2} - (0)^{2}}{2\, (-9.81)} \; {\rm m} \approx (-135)\; {\rm m}\end{aligned}[/tex].

What is a wave function?

Answers

a function that satisfies a wave equation and describes the properties of a wave.

Four small spheres, each charged to +15 nC, form a square 2.0 cm on each side. From far away, a proton is shot toward the square along a line perpendicular to the square and passing through its center. What minimum initial speed does the proton need to pass through the square of charges

Answers

The speed of the charged particles or proton or electron will be 2.75 × 10⁵ m/s.

What is Kinetic energy?

The kinetic energy of an object is the energy which it possesses due to its motion. Kinetic energy is defined as the work needed to accelerate an object of a given mass from the state of rest to its stated velocity.

At the center of the square, all the kinetic energy of the proton is converted to potential energy of the proton. Therefore, far away the kinetic energy of proton must be equal to the potential energy of the person at the center.

Speed of the electron far away from the square, v = 2.75 × 10⁵ m/s

Given, that the charge, q = 15 × 10⁻⁹C

Each side of square, l = 0.02m

Distance of center of square from any charge r = √2l/2 = r= √2 × 0.02/ 2

r = 0.014m

q = 15nC

q = 15nC

I = 0.02m

q = 15nC

q = 15nC

Electric potential at the center of the square is given by:

Vc = 4 × (kq/r)

Electric potential at point far away is V∞ = 0

Potential Energy difference between the two point is

ΔV = Ve - V∞

ΔV = 4×(kq/r)

For the proton to pass through the center of square the proton must have kinetic energy equal to the potential energy at the center of the square.

The kinetic energy of the proton K = mv²

The potential energy of the proton U = eΔV

For the proton to pass through

K(at ∞) = U(Center)

1/2 mv² = 4kqe/ r

v = [tex]\sqrt{(8kq)/(mr)}[/tex]

v = √(8 × 9 × 10⁹ × 15 × 10⁻⁹ × 1.6 × 10⁻¹⁹)/ (1.67 × 10⁻²⁷ × 0.014) m/s

v = 2.75 × 10⁵ m/s

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The radius of the Earth is 6370 km. If a satellite orbits 150 km above the Earth's surface, what is the velocity of the orbit? The Earth's mass is 5.97 × 1024 kg. Give your answer in m-s¹, to three significant figures and do not include units.​

Answers

Answer:

Approximately [tex]7.61 \times 10^{3}\; {\rm m\cdot s^{-1}}[/tex].

Explanation:

Add the radius of the Earth to the altitude of the satellite to find the radius of the orbit:

[tex]\begin{aligned}r &= (6370 + 150)\; {\rm km} \\ &= 6880\; {\rm km} \\ &= 6.88\times 10^{6}\; {\rm m}\end{aligned}[/tex].

Look up the gravitational constant:

[tex]G \approx 6.6743\times 10^{-11}\; {\rm m^{3}\, kg^{-1}\, s^{-2}[/tex].

Let [tex]m[/tex] denote the mass of the satellite. At an orbital velocity of [tex]v[/tex], the (centripetal) net force on the satellite would be:

[tex]\displaystyle (\text{net force}) = \frac{m\, v^{2}}{r}[/tex].

Let [tex]M[/tex] denote the mass of planet Earth. At a distance of [tex]r[/tex] from the center of the Earth, the gravitational attraction on the satellite would be:

[tex]\displaystyle (\text{gravitational attraction}) = \frac{G\, M\, m}{r^{2}}[/tex].

When the satellite is at the correct orbital velocity, the net force on the satellite would be equal to the gravitational attraction from the Earth. In other words:

[tex]\displaystyle (\text{net force}) = (\text{gravitational attraction})[/tex].

[tex]\displaystyle \frac{m\, v^{2}}{r} = \frac{G\, M\, m}{r^{2}}[/tex].

Rearrange the equation and solve for orbital velocity [tex]v[/tex]:

[tex]\begin{aligned}v^{2} &= \frac{G\, M}{r}\end{aligned}[/tex].

[tex]\begin{aligned}v &= \sqrt{\frac{G\, M}{r}} \\ &\approx \sqrt{\frac{(6.6743 \times 10^{-11})\, (5.97 \times 10^{24})}{6.88 \times 10^{6}}}\; {\rm m\cdot s^{-1}} \\ &\approx 7.61 \times 10^{3}\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

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