Answer:
Explanation:
graphing is helpful
helps visualize the line
of your equation
An airplane, starting at rest, takes off on a 600. m long runway accelerating at a rate of 12 m/s/s. How many seconds does it take to reach the end of the runway?
x = 1/2 at²
where x = length of runway, a = acceleration, and t = time.
600 m = 1/2 (12 m/s²) t²
t² = (1200 m) / (12 m/s²)
t² = 100 s²
t = 10 s
A 849-kg car starts from rest on a horizontal roadway and accelerates eastward for 5.00 s when it reaches a speed of 35.0 m/s. What is the average force exerted on the car during this time
Answer:
The average force exerted on the car during this time is 5,943 N
Explanation:
Given;
mass of the car, m = 849 kg
initial velocity of the car, u = 0
time of motion of the car, t = 5.00 s
final velocity of the car, v = 35 m/s
The average force exerted on the car during this time is given by;
[tex]F = ma \\\\F = \frac{m(v-u)}{t}\\\\F = \frac{849(35-0)}{5}\\\\F = \frac{849(35)}{5}\\\\ F = 849*7\\\\F = 5,943 \ N[/tex]
Therefore, the average force exerted on the car during this time is 5,943 N
Answer:
5943N
Let's say (+x) = eastward
Average horizontal acceleration
ax = vx -v0x/5.00s
= 35.0m/s-0/5.00s
= +7.09m/s
From here we apply the second law of newton
During this period average horizontal force acting on car
Summation x = max = (849kg)(+7.09m/s²)
= 5943N
+5.943x10³N
= 5.94kN east ward.
If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be Imax
The complete question is;
A person with body resistance between his hands of 10 kΩ accidentally grasps the terminals of a 16-kV power supply. What is the power dissipated in his body?
A) If the internal resistance of the power supply is 1600 Ω , what is the current through the person's body?
B) What is the power dissipated in his body?
C) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be I_max = 1.00mA or less?
Answer:
A) I = 1.379 A
B) P = 19016.41 W
C) r = 15990000 Ω
Explanation:
A) We are given;
Internal resistance of the power supply; r = 1600 Ω
Body resistance between hands; R = 10kΩ = 10000 Ω
Power supply voltage; E =16 kV = 16000 V
Formula for the current through the person's body with internal resistance is given by;
I = E/(R + r)
Thus;
I = 16000/(10000 + 1600)
I = 1.379 A
B) Formula for power dissipated is;
P = I²R
P = 1.379² × 10000
P = 19016.41 W
C) Now, we are told that the maximum current should be I_max = 1.00mA or less. So, I_max = 0.001 A
Thus, from I = E/(R + r) and making r the subject, we have;
r = (E/I) - R
r = (16000/0.001) - 10000
r = 15990000 Ω
A tower crane has a hoist motor rated at 159 hp. If the crane is limited to using 72.0 % of its maximum hoisting power for safety reasons, what is the shortest time in which the crane can lift a 5550 kg load over a distance of 89.0 m
Answer:
The value is [tex]t = 56.68 \ s [/tex]
Explanation:
From the question we are told that
The rating of the hoist motor is [tex]k = 159hp = 159 *746 =118614 \ W[/tex]
The percentage of it power used is [tex]z = 0.72 * 118614=85402.08 \ W[/tex]
The mass of the load is m = 5550 kg
The distance is h = 89.0 m
The potential energy required to lift the load through that distance is
[tex]E = m * g * h[/tex]
=> [tex]E = 5550 * 9.8 * 89.0[/tex]
=> [tex]E = 4840710 \ J[/tex]
Generally the time taken is mathematically represented as
[tex]t = \frac{E}{ z}[/tex]
=> [tex]t = \frac{4840710}{ 85402.08}[/tex]
=> [tex]t = 56.68 \ s [/tex]
A solid nonconducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a certain distance rl (r
(A) E/8
(B) E 78.
(C) E/2
(D) 2E
(E) 8E
Answer:
A ) E/8
Explanation:
If the sphere of radius R carries charge Q, then the volumetric charge density is:
ρ₁ = [Q/ (4/3)*π*R³]
Therefore the net charge inside r ( r < R ) is:
q₁ = ρ * (4/3)*π*r³
And E = K * q₁/r K = 9,98 *10⁹ [N*m²/C²]
E = K * ρ * (4/3)*π*r³/r
E = K * ρ * (4/3)*π*r²
If now the charge is distributed over a sphere of radius 2R
ρ₂ = [Q/ (4/3)*π*(2R)³]
ρ₂ = [Q/ (4/3)*π*8*R³]
Then ρ₂ < ρ₁ in fact ρ₂ = (1/8)*ρ₁
The electric field depends on the net charge enclosed by a gaussian surface, and the distance between the net charge and the considered point, ( considering the net charge as being at the center of the gaussian surface) In this case, there was no distance change then
E₂ = E₁/8
The right answer is lyrics A ) E/8
A car moving with an intial velocity of 60m/s is brought to rest in 30 seconds calculate the acceleration
Answer:
a = 2 [m/s^2]
Explanation:
To solve this problem we must use the expressions of kinematics, we must bear in mind that when a body is at rest its velocity is zero.
[tex]v_{f} = v_{i} - (a*t)[/tex]
where:
Vf = final velocity = 0
Vi = initial velocity = 60 [m/s]
a = desacceleration [m/s^2]
t = time = 30 [s]
Note: the negative sign of the above equation means that the car is slowing down, i.e. its speed decreases.
0 = 60 - (a*30)
a = 2 [m/s^2]
-. What is the acceleration of 4 kg trolling bag pulled by a girl with a
force of 3 N?
Answer:
Acceleration(a) = 0.75 m/s²
Explanation:
Given:
Force(F) = 3 N
Mass of thing(m) = 4 kg
Find:
Acceleration(a)
Computation:
Force(F) = ma
3 = (4)(a)
Acceleration(a) = 3/4
Acceleration(a) = 0.75 m/s²
It takes 3.8 x 10^-5 for a pulse of the radio waves from a radar to reach a plane and bounce back. How far is the plane from the radar?
Answer: 11400 m
Explanation:
Given:
t = 3.8 x 10^-5 s
v = 3 x 10^8 m/s
d = ?
Formula:
d = vt
= (3.8 x 10^-5 s)(3 x 10^8 m/s)
= 11400 m
hope this helps :)
The uniform movement allows to find the results for the distance from the radar to the plane is: 5.7 10³ m or 5.7 km
Kinematics studies the motion of objects looking for relationships between position, velocity and acceleration, in the special case that the acceleration is zero is called uniform motion and is described by the expression
[tex]v = \frac{d}{t}[/tex]
d = v t
Where v is the velocity, d the displacement and t the time.
Radar waves are electromagnetic waves with constant velocity
v = 3 10⁸ m/s
They indicate that the time of the waves to go to the plane and return is 3.8 10⁻⁵ s, therefore if the speed is constant, the time to reach the plane is half of the total time.
t = [tex]\frac{t_{total} }{ 2}[/tex]
t = [tex]\frac{3.8 \ 10^{-5}}{2}[/tex]
t = 1.9 10⁻⁵ s
Let's calculate
d = 3 10⁸ 1.9 10⁻⁵
d = 5.7 10³
In conclusion with the uniform movement we can find the results for the distance from the radar to the plane is: 5.7 10³ m or 5.7 km
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The diagram shows two forces acting on the dog. What are these two forces
Answer:
kenietic and potential i guess
Explanation:
block of mass m sits at rest on a rough inclined ramp that makes an angle with the horizontal. What must be true about normal force F on the block due to the ramp
Answer:
Explanation:
For a body on a ramp with mass m, the forces acting on the body along the vertical component are the weight and the normal reaction.
The weight of the body acts in the negative y direction while the normal reaction acts in the positive y direction
Taking the sum of forces along the y component
Sum Fy = -W+R = ma
Since acceleration is zero
-W+R = m(0)
-W+R = 0
-W = -R
W = R
Hence the Normal reaction force acting on the on the body is equal to normal force
Isaac walks 4 blocks north. Then he turns around and walks 1 block south.
Which of the following correctly describes Isaac's motion?
A. Isaac walked a distance of 5 blocks, and his displacement was 3
blocks north.
B. Isaac walked a distance of 5 blocks, and his displacement was 5
blocks.
C. Isaac walked a distance of 3 blocks, and his displacement was 3
blocks north.
D. Isaac walked a distance of 3 blocks north, and his displacement
was 5 blocks.
Answer:
Isaac walked a distance of 5 blocks, and his displacement was 3 blocks north.
Explanation:
Distance is what he covered from the beginning, while displacement was what he covered in a specific direction
SOH-CAH-TOA is used to solve for the ________
velocities in a full/angled projectile.
a. final (x and y)
b. overall
c. initial (x and y)
d. resultant
Answer:
c. initial (x and y)
Explanation:
When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.
Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"
Thus, this method resolves the initial x and y velocities.
1. What quantity of heat is required to raise?
the temperature of 450 grams of water
from 35°C to 85°C?
capacity of water is 4.18 J/g °C.
Answer: Calculate the energy required in joules to raise the temperature of 450 grams of water from 15°C to 85°C? (The specific heat capacity of water is 4.18 J/g/°C)
Explanation:
The quantity of heat is required to raise the temperature of of water is 94050 joule.
What is law of conservation of energy?Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.
The law of energy conservation is adhered to by all energy forms. The law of conservation of energy essentially says that the total energy of the system is conserved in a closed system, also known as an isolated system.
Mass of water: m = 450 grams
Initial temperature of water = 35° C
Final temperature of water = 85° C
Capacity of water is 4.18 J/g °C.
Hence, the quantity of heat is required to raise = mass × Capacity × raise in temperature
= 450 × 4.18 × (85 - 35) joule
= 94050 joule.
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A car at the top of a ramp starts from rest and rolls to the bottom of the ramp, achieving a certain final speed. If you instead wanted the car to achieve twice as much speed at the bottom of the ramp, how high should the ramp be compared to the first case
Answer:
It must be 4 times high.
Explanation:
Assuming that the car can be treated as a point mass, and that the ramp is frictionless, the total mechanical energy must be conserved.This means, that at any time, the following must be true:ΔK (change in kinetic energy) = ΔU (change in gravitational potential energy)⇒ [tex]m*g*h = \frac{1}{2} * m*v^{2}[/tex]
Let's call v₁, to the final speed of the car, and h₁ to the height of the ramp.So, at the bottom of the ramp, all the gravitational potential energy
must be equal to the kinetic energy of the car (Defining the bottom of
the ramp as our zero reference for the gravitational potential energy):
[tex]m*g*h_{1} = \frac{1}{2} * m*v_{1} ^{2}[/tex] (1)
Now, let's do v₂ = 2* v₁Replacing in (1) we get:[tex]m*g*h_{2} = \frac{1}{2} * m*(2*v_{1}) ^{2}[/tex] (2)
Dividing (2) by (1), and rearranging terms, we get:h₂ = 4* h₁Which exerts more force, the Earth pulling on the moon or the moon pulling on the Earth? Explain.
Answer: the earth
Explanation: Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth. Over a very long time, the moon's rotations created fiction with the Earth's tugging back, until the moon's orbit and rotational locked with Earth.
and that's why the earth pulls the moon
A plastic block of dimensions 2.00 cm x 3.00 cm x 4.00 cm has a mass of 30.0 g. What is its density?
Answer:
1.25 g/cm^3
Explanation:
mass-30.0g
volume- 4cm×2cm×3cm=24cm^3
density?
*to find density
Density=Mass/Volume
=30÷24
=1.25g/cm^3
what is the volume of an object that has a density of 65g/cm3 and a mass of 130g.
Density ρ is mass m per unit volume v, or
ρ = m / v
Solving for v gives
v = m / ρ
So the given object has a volume of
v = (130 g) / (65 g/cm³) = 2 cm³
My parrot has a mass of 1.33kg, what is it's weight here on earth
Answer:
Your parrot, from earth, that weighs 1.33kg is 1.33kg on earth. as far as i'm aware there is only one earth and everything always weighs the same on one planet as it did on that same planet.
Explanation:
What is the correct answer?
Answer:
2156 N
Explanation:
Data obtained from the question include:
Mass of satellite (m) = 220 Kg
Force (F) of gravity =?
The force of gravity exerted on the satellite on the surface of the earth can be obtained by using the following formula:
Force (F) of gravity = mass (m) × acceleration due to gravity (g)
F = mg
Mass of satellite (m) = 220 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) of gravity =?
F = mg
F = 220 × 9.8
F = 2156 N
Thus, the force of gravity exerted on the satellite on the surface of the earth is 2156 N
How long is a day in Neptune
Answer: the long day in neptune would be .18383562 years!
Explanation:also for every day is 16 hours
Can anybody tell me what I'm suppose to do. I click start the numbers comes up to the right
The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm
Answer:
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
Explanation:
The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:
[tex]U_{k,1}+K_{1} = U_{k,2}+K_{2}[/tex]
[tex]K_{2} = K_{1}+(U_{k,1}-U_{k,2})[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies of the block, measured in joules.
[tex]U_{k,1}[/tex], [tex]U_{k,2}[/tex] - Initial and final elastic potential energy, measured in joules.
And we expand the equation above by definitions of elastic potential energy and kinetic energy:
[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})[/tex]
[tex]v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the block, measured in kilograms.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final velocities of the block, measured in meters per second.
[tex]x_{1}[/tex], [tex]x_{2}[/tex] - Initial and final positions of spring, measured in meters.
If we know that [tex]v_{1} = 6\,\frac{m}{s}[/tex], [tex]k = 2000\,\frac{N}{m}[/tex], [tex]m = 2\,kg[/tex], [tex]x_{1} = 0\,m[/tex] and [tex]x_{2} = 0.15\,m[/tex], the final speed of the block moving at the instant the spring has been compressed is:
[tex]v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}[/tex]
[tex]v_{2}\approx 3.674\,\frac{m}{s}[/tex]
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east. what is the distance and displacement
Explanation:
Let east = E, and, west = opposite to east = - E.
Here, displacement:
=> 2m east + 4m west + 1m east
=> 2E + 4(-E) + 1E
=> 2E - 4E + 1E
=> - 1E
=> 1(-E)
=> 1m west
And, distance,
=> 2m + 4m + 1m = 7m
The distance of a person is 7 m and the displacement of the person is 1m west.
To find the distance and displacement, the given values are,
A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east.
What is the distance and the displacement?Displacement:
The displacement is shortest distance between initial and final position or we can say it is the straight line distance between initial and final position.If object moves in a straight line path without any turn then the path length and the displacement is always same.Distance:
The distance is the total path length of the object while it will move from initial to final position.If the object move on curved path then displacement is smaller than the distance moved by the object.Let us consider East = E and west = opposite to east = - E.
Calculating the displacement:
= 2m east + 4m west + 1m east
= 2E + 4(-E) + 1E
= 2E - 4E + 1E
= - 1E
= 1(-E)
= 1m west.
The displacement is 1m west.
Now calculating the distance,
= 2m + 4m + 1m
= 7m
The distance is 7m.
Thus, the displacement and the distance is found as 1 m west and 7m.
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100 POINTS.
Please provide explanation.
Thank you
Answer:
(a) 0.829 m/s
(b) 3.27 m/s
(c) 0.000153 m²
55.8%
Explanation:
(a) Flow rate equals velocity times cross-sectional area. (1 L = 0.001 m³)
Q = vA
(0.001 m³ / 2.00 s) = v (48 × π (0.002 m)²)
v = 0.829 m/s
(b) Use Bernoulli equation. Choose point 1 to be the exit of the pump, and point 2 to be exit of the shower head. Choose 0 elevation to be at point 1.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
(1.50 atm × 1.0×10⁵ Pa/atm) + ½ (1000 kg/m³) v² + 0 = (1 atm × 1.0×10⁵ Pa/atm) + ½ (1000 kg/m³) (0.829 m/s)² + (1000 kg/m³) (10 m/s²) (5.50 m)
1.50×10⁵ Pa + (500 kg/m³) v² = 1×10⁵ Pa + 414.5 Pa + 55000 Pa
v = 3.27 m/s
(c) Flow rate is constant.
Q = vA
(0.001 m³ / 2.00 s) = (3.27 m/s) A
A = 0.000153 m²
Flow rate is proportional to the pressure difference and the radius raised to the fourth power.
Q ∝ ΔP r⁴
Q₂/Q₁ = (ΔP₂/ΔP₁) (r₂/r₁)⁴
Q₂/Q₁ = (1.120) (0.840)⁴
Q₂/Q₁ = 0.558
The flow decreases to 55.8% of the original value.
Answer:
Explanation:
Regarding the point of "Flow rate is proportional to the pressure difference and the radius raised to the fourth power", flow rate depends on pressure, cross-section area and speed. As speed also depends on cross-section area, flow rate becomes dependent on pressure and cross-section area squared.
In a round pipe like blood vessel, the cross-section area is equal to pi*radius squared. So flow rate is proportional to the pressure difference and (radius squared) squared; i.e. the radius raised to the fourth power.
The new flow rate = (1.12)*(0.84)^4
=0.5576 or 55.76% of the original flow rate
A cannonball is fired horizontally from a 10 m high cliff at 20 m/s. How long will the cannonball be in the air? How far away will the cannonball strike the ground?
Answe ¡Buenos días! –
#3 ¡Buenas tardes! –
#4 ¡Bienvenid
Light is described as having a dual wave-particle nature. Which piece of evidence provides support for the model of light as a particle?
. Young’s double slit experiment showed that light waves show interference.
. Light reflects when it hits a surface.
. Light refracts when it moves from one medium to another.
. Light does not need a medium to travel.
Answer:
light reflects when it hits the surface
Explanation:
Youngs double slit is a evidence for wave nature,
The properties refraction are attributed as properties of waves. The phenomena of interference and diffraction also fall in this category.
So,
the answer must be B
Answer: Light does not need a medium to travel.
Explanation: I took the test and got it right :]
BRAINLIEST. Agraph is probelow. The graph shows the speed of a car traveling east over a 12 second period. Based on the information in the graph, it can be
that in the first second
Answer:speeding up constantly
Explanation:
The graph between the time and the speed of the car shows that the speed is increasing constantly, so, option C is correct.
What is speed?A moving object's rate of change in distance traveled is measured as speed. Speed is a scalar, which implies it is a measurement with a magnitude but no direction.
A thing that moves quickly and with high speed, covering a lot of ground in a short time. On the other hand, a slow-moving object traveling at a low speed covers a comparatively small distance in the same amount of time. An object with zero speed does not move at all.
Given:
The graph shows the speed of a car traveling east over a 12-second period,
As you can see from the graph, at time t = 0 sec the speed is 10 m/s,
At t = 3 sec, the speed = 15.3 m/s
At t = 6 sec, the speed = 20.3 m/s
Thus, speed is increasing constantly.
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Light of wavelength 580 nm falls on a slit that is 3.70×10−3mm wide. Part A Estimate how far the first brightest diffraction fringe is from the strong central maximum if the screen is 10.0 m away.
Answer:
Explanation:
wavelength λ = 580 x 10⁻⁹ m
slit width d = 3.7 x 10⁻⁶ m
distance of screen D = 10 m
distance of first bright fringe = 1.5 x λ D / d
= 1.5 x 580 x 10⁻⁹ x 10 / 3.7 x 10⁻⁶
= 2351.34 x 10⁻³ m
= 2351.34 mm .
Calculate the effective value of g, the acceleration of gravity, at 6700 m , above the Earth's surface. g
Answer:
The effective value of g at 6700 m above the Earth's surface is 9.79 m/s².
Explanation:
The value of g can be found using the following equation:
[tex] F = \frac{GmM}{r^{2}} [/tex]
[tex] ma = \frac{GmM}{r^{2}} [/tex]
[tex] a = \frac{GM}{r^{2}} [/tex]
Where:
a is the acceleration of gravity = g
G: is the gravitational constant = 6.67x10⁻¹¹ m³/(kg.s²)
M: is the Earth's mass = 5.97x10²⁴ kg
r: is the Earth's radius = 6371 km
Since we need to find g at 6700 m, the total distance is:
[tex] r_{T} = 6371000 m + 6700 m = 6377700 m [/tex]
Now, the value of g is:
[tex] a = \frac{GM}{r_{T}^{2}} = \frac{6.67\cdot 10^{-11} m^{3}/(kg*s^{2})*5.97 \cdot 10^{24} kg}{(6377700 m)^{2}} = 9.79 m/s^{2} [/tex]
Therefore, the effective value of g at 6700 m above the Earth's surface is 9.79 m/s².
I hope it helps you!
Two protons are a distance 3 10-9 m apart. What is the electric potential energy of the system consisting of the two protons
Answer:
The electric potential energy of the system is 7.87x10⁻²⁰ J.
Explanation:
The electric potential energy is given by:
[tex]E = \int{Fdr} = \frac{Kq_{1}q_{2}}{r}[/tex]
Where:
q₁ = q₂ is the charge of the protons = 1.62x10⁻¹⁹ C
r is the distance = 3x10⁻⁹ m
K: is the electrostatic constant = 9x10⁹ Nm²/C²
[tex] E = \frac{Kq_{1}q_{2}}{r} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*(1.62 \cdot 10^{-19} C)^{2}}{3\cdot 10^{-9} m} = 7.87 \cdot 10^{-20} J [/tex]
Therefore, the electric potential energy of the system is 7.87x10⁻²⁰ J.
I hope it helps you!
The electric potential energy of the system should be 7.87x10⁻²⁰ J.
Calculation of the electric potential energy:SInce We know that
fdr = kq1q2/r
Here
q₁ = q₂ i.e. is the charge of the protons = 1.62x10⁻¹⁹ C
r should be the distance = 3x10⁻⁹ m
K should be the electrostatic constant = 9x10⁹ Nm²/C²
Now electric potential energy should be
= (9x10⁹ Nm²/C² * 1.62x10⁻¹⁹ C) / 3x10⁻⁹ m
= 7.87x10⁻²⁰ J.
hence, The electric potential energy of the system should be 7.87x10⁻²⁰ J.
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