assuming you start at the top of the ramp that is 10 m tall with velocity of 10 m/s and your car has a mass of 1000 kg, what is your pe g

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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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."

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 ).

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;

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

According to the given data the acceleration of each car is a) 3.0 m/^s b) 4.0 m/s^2 c) -3.0 m/s^2 d) 2.5 m/s^2

The rate about which velocity fluctuates is titled acceleration. Acceleration typically signals a shift in speed, but not necessarily. An item that continues a circular course while maintaining a constant speed is still moving forward because the orientation of its motion is rotating.

The following equation can be used to determine each car's acceleration: acceleration = (final velocity - beginning velocity) / time period. For each car, the computation is as follows:

a) acceleration = (11.0 - 2.0) / 3.0 = 3.0 m/s^

b) acceleration = (3.0 - (-5.0)) / 2.0 = 4.0 m/s^2

c) acceleration = (-5.0 - 1.0) / 2.0 = -3.0 m/s^2

d) acceleration = (25.0 - 0.0) / 10.0 = 2.5 m/s^2

Because the vehicle is slowing down rather than accelerating, the acceleration of automobile "c" is negative.

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Answer:

humerus and all hand bones

Explanation:

not skull bones

in axial skel.

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.

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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Holes are much lighter and more mobile than electrons in semiconductors. This difference in effective mass is a key factor in determining the electronic properties of materials such as B. Its conductivity, charge carrier mobility, and optical properties. 

The effective mass (meff) of an electron or hole in a solid is a measure of how a particle behaves as if it had a certain mass in a vacuum, even when interacting with crystal lattices and other particles in the solid. is. In general, the effective mass of an electron or hole depends on its energy and momentum.

In semiconductors with an energy bandgap, the valence band (VB) is separated from the conduction band (CB) by the bandgap energy (Eg). The energy of CB is higher than that of VB, the electrons in CB are free to move and conduct electricity, while the holes in VB are immobile and carry positive charge.

If the effective mass of electrons in CB is five times the effective mass of holes in VB, then the effective mass ratio can be expressed as

meff,electron / meff,hole = 5

Taking the reciprocals of both sides and rearranging them gives:

meff,hole / meff,electron = 1/5

This means that the effective mass of holes in VB is about one-fifth that of electrons in CB. In other words, holes are much lighter and more mobile than electrons in semiconductors. This difference in effective mass is a key factor in determining the electronic properties of materials such as: B. Its conductivity, charge carrier mobility, and optical properties. 

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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

Answer:

it is true

Explanation:

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].

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].

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].

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].

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.

When considering fictitious elements Z and M, it is likely that element Z will have a higher ionization energy, meaning that more energy will be needed to remove an electron from it.

This means that when the two elements react, element Z will be more likely to gain electrons, while element M will be more likely to lose electrons. This means that element Z will be oxidized, while element M will be reduced.

Ionization energy, also known as ionization potential, is the amount of energy required to remove an electron from an atom or a molecule.

This process is referred to as ionization, and it involves the transfer of an electron from a neutral atom or molecule to form an ion with a positive charge.

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The first hill was 6.17 meters high.

Velocity is defined as the displacement of the object in a given amount of time and is referred to as velocity.

The work done on Amy and her bicycle by gravitational force can be calculated as follows:

W = mgh

where m is the mass of Amy and her bicycle, g is the acceleration due to gravity (9.8 m/s²), and h is the change in height.

When Amy is coasting down the first hill, her potential energy is converted to kinetic energy, so the work done by gravity is equal to the initial increase in kinetic energy:

W = (1/2)mv²

where v is the velocity. Setting these two equations equal to each other, we get:

mgh = (1/2)mv²

Solving for h, we get:

h = (1/2)v²/g

Plugging in the known values, we get:

h = (1/2)(11 m/s)²/9.8 m/s² = 6.17 m

Thus, the first hill was 6.17 m high.

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Option A is Correct. Betelgeuse star has a larger surface area if Betelgeuse and Rigel both have the same brightness but Rigel's temperature is lower.

Rigel is usually always brighter than Orionis, despite the fact that the second-brightest star in each constellation is commonly referred to as "beta" (Betelgeuse). Therefore, red-glow objects are cooler than white-glow or blue-glow objects, despite what your water taps may claim.

Therefore, blue-white stars like Rigel and red stars like Betelgeuse have substantially lower surface temperatures. At 3,500 Kelvin, Betelgeuse's surface is warm. The brightness of Betelgeuse is extremely intense (40,000 times that of our Sun), but its surface is cold (less than 4000 K).

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

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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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

The estimated change in water volume is -0.00056 [tex]m^3[/tex], which represents the volume reduction due to the applied pressure. 

The formula for volume change is:

ΔV = -V * ΔP/B

where:

ΔV = volume change

V = initial volume

ΔP = pressure change

B = bulk modulus of water

After plugging in the values ​​it looks like this:

ΔV = -1 [tex]m^3[/tex] * (35 x 10^6 Pa) / (2.2 x [tex]10^9[/tex] Pa)

= -0.00056[tex]m^3[/tex]

Therefore, the estimated change in volume of water is -0.00056 [tex]m^3[/tex], which represents the decrease in volume due to applied pressure. 

The volume is determined by multiplying the dimensions of length, width, and height. Good news for cubes comes from the fact that each of these dimensions weighs exactly the same. As a result, we can add three to the length of any side. The formula that results is as follows: Front * Back Volume

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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!

Earth constantly moves. The Earth rotates around its axis once per day. The axis is the fictitious line passing through the middle of the planet, from the North to the South Poles.

It appears that the sun is moving across the sky as Earth spins, the sun is actually whirling. There are 24 hours in a day because it takes 24 hours to complete one rotation.

In other words, if the sun is visible in the morning beginning around 6:00 AM, the Earth will have spun around by 6:00 AM the following morning, and the sun will be seen in roughly the same location.

The Earth circles the sun and rotates on its axis. The Earth completes one full rotation of the sun in a little over 365 days.

Therefore, Earth constantly moves. The Earth rotates around its axis once per day. The axis is the fictitious line passing through the middle of the planet, from the North to the South Poles.

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

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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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

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

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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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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The interior volume of the tank is 2.70 ft³ and its length is 34.0 inches.

The force delivered perpendicularly to an object's surface per unit area across which this force is dispersed is known as pressure.

To calculate the interior volume of the tank, we can use the ideal gas law:

PV = nRT

At standard temperature and pressure (STP), one mole of an ideal gas occupies 22.4 liters (0.79 cubic feet). Therefore, 50 SCF of air is equivalent to:

n = (50 SCF) / (0.79 SCF/mol) = 63.3 mol

The temperature and pressure of the tank are not at STP, so we need to use the ideal gas law to calculate the volume. Rearranging the equation to solve for V, we get:

V = nRT/P

We can assume that the temperature remains constant at 80 F (which is equivalent to 26.7 C), and the pressure is 3000 psi. The gas constant R is 8.314 J/(mol K), or 10.73 psi·ft³/(mol K).

[tex]$\mathrm{V = \frac{ (63.3 mol) \times (10.73 psi\cdot ft^3 / (mol K)) \times (26.7 C + 273.15)}{(3000 psi)} }[/tex]

= 2.70 cubic feet

To find the length of the tank, we can use the formula for the volume of a cylinder:

[tex]\mathbf{V = \pi r^2h}[/tex]

where r is the radius (which is half the diameter) and h is the height (which is the length of the tank). We know that the diameter is 6 inches, so the radius is 3 inches (0.25 feet).

Substituting the known values into the equation, we get:

2.70 cubic feet = π (0.25 ft)² h

Solving for h, we get:

h = 34.0 inches

Therefore, the interior volume of the tank is 2.70 cubic feet and its length is 34.0 inches

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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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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].

Answer:

Explanation:

Potential energy means the energy of a material when it is not moving.

Ep= mgh

That's going to depend on the height of the ball above ground, and the amount of charge on the ball.

Sadly, you haven't shared either quantity with us.

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