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marissa [1.9K]
3 years ago
14

A car moves at a constant velocity of 30 m/s and has 3.6 × 105 J of kinetic energy. The driver applies the brakes and the car st

ops in 95 m.
Calculate the force needed to stop the vehicle.
Physics
1 answer:
LekaFEV [45]3 years ago
5 0

The force needed to the stop the car is -3.79 N.

Explanation:

The force required to stop the car should have equal magnitude as the force required to move the car but in opposite direction. This is in accordance with the Newton's third law of motion. Since, in the present problem, we know the kinetic energy and velocity of the moving car, we can determine the mass of the car from these two parameters.

So, here v = 30 m/s and k.E. = 3.6 × 10⁵ J, then mass will be

K.E = \frac{1}{2} * m*v^{2}  \\\\m = \frac{2*KE}{v^{2} } = \frac{2*3.6*10^{5} }{30*30}=800 kg

Now, we know that the work done by the brake to stop the car will be equal to the product of force to stop the car with the distance travelled by the car on applying the brake.Here it is said that the car travels 95 m after the brake has been applied. So with the help of work energy theorem,

Work done = Final kinetic energy - Initial kinetic energy

Work done = Force × Displacement

So, Force × Displacement = Final kinetic energy - Initial Kinetic energy.

Force * 95 = 0-3.6*10^{5}\\ \\Force =\frac{-3.6*10^{5} }{95}=-3.79 N

Thus, the force needed to the stop the car is -3.79 N.

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soldier1979 [14.2K]

Answer:

Hello There!!

Explanation:

Electrons get their energy by absorbing light.

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6 0
3 years ago
An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e
timama [110]

Answer:

K = 80.75 MeV    

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}

<em>where E_{ph}: is the photon energy, E_{0p} and E_{0ap}: are the rest energies of the proton and the antiproton, respectively, equals to m₀c², K_{p} and K_{ap}: are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.</em>        

Therefore the kinetic energy of the antiproton is:    

K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}

<u>The proton mass is equal to the antiproton mass, so</u>:

K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}  

K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV

K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV

K_{ap} = 80.75 MeV              

Hence, the kinetic energy of the antiproton is 80.75 MeV.

I hope it helps you!

3 0
3 years ago
Calculate the gravitational attraction between two objects which masses are: mass A: 2.5kg, mass B: 5kg. The distance between th
jonny [76]

From the calculation, the gravitational force of attraction is  1.33 * 10^-14 N.

<h3>What is the gravitational force?</h3>

The gravitational force is an attractive force that acts between any two masses.

It is given by;

F = Gm1m2/r^2

F = 6.67 * × 10−11 * 2.5 * 5/(250)^2

F  = 83.4  × 10−11 /62500

F= 1.33 * 10^-14 N

Learn more about gravitational force:brainly.com/question/12528243

#SPJ1

4 0
2 years ago
The formula v = √2.5r models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radiu
mel-nik [20]

Answer:

The maximum safe speed of the car is 30.82 m/s.

Explanation:

It is given that,

The formula that models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r r, is in feet is given by :

v=\sqrt{2.5\ r}.........(1)

A highway crew measures the radius of curvature at an exit ramp on a highway as 380 feet, r = 380 feet

Put the value of r in equation (1) as :

v=\sqrt{2.5\ \times 380}

v = 30.82 m/s

So, the maximum safe speed of the car is 30.82 m/s. Hence, this is the required solution.

4 0
3 years ago
A box has a 20 N force applied to it to move it 5 m. What is the work done on the box? 4 J 4 N 25 J 100 J
Sunny_sXe [5.5K]
100 J

Explanation:
multiply the force by the distance
20 N x 5 meters = 100 J
please mark brainliest
7 0
3 years ago
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