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Law Incorporation [45]
3 years ago
12

In a bumper car arena, two cars of equal mass are heading straight toward each other. The orange one is traveling at a speed of

5 meters per second. The green one is traveling at a speed of 2 meters per second. Which of the forces most affects the motion of the bumper cars after they collide?
Physics
1 answer:
marshall27 [118]3 years ago
4 0

Answer: D.

Explanation:   Orange, at 3 meters per second if you calculate the net force being applied to the system.

hope this helps! ✌

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How much work is done by you when you push car a distance of 10 m while exerting a horizontal
Igoryamba

W = F x d

F: 450 N

d = 10 m

plug your variables into the equation to get W = 450N x 10m

7 0
3 years ago
9. Batman (mass=91 kg) jumps straight down from a bridge into a boat (mass=510 kg) in which a criminal is fleeing. The velocity
Ierofanga [76]

Answer:

V = 9.33 m/s

Explanation:

Given that,

Mass of the batsman, m_1=91\ kg

Mass of the boat, m_2=510\ kg                            

Initial speed of the boat, v = 11 m/s

Let V is the velocity of the boat after Batman lands in it. The net momentum of the system remains constant. Using the conservation of linear momentum to find it as :

510\times 11=(91+510)V

V=\dfrac{510\times11}{(91+510)}

V = 9.33 m/s

So, the velocity of the boat after Batman lands in it 9.33 m/s. Hence, this is the required solution.                                                          

5 0
3 years ago
What is the term for the depth of the water needed to float a boat?
mina [271]
<span> The term for the depth of the water needed to make a boat afloat is called the draft of the boat. It can be measured as the distance from the water surface down the lowest point of the vessel. It can be imagined as the submerged portion of the boat during navigation.</span>
8 0
3 years ago
A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a
vredina [299]

(a) 4.0 s

The acceleration of the car is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

a=-5.0 m/s^2 is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s

(b) 40 m

The stopping distance of the car can be calculated by using the equation

v^2 - u^2 = 2ad

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m

(c) -5.0 m/s^2

The deceleration is given by the problem, and its value is -5.0 m/s^2.

(d) 5000 N

The net force applied on the car is given by

F=ma

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

a=5.0 m/s^2 is the magnitude of the acceleration

Solving the formula, we find

F=(1000 kg)(5.0 m/s^2)=5000 N

(e) 2.0\cdot 10^5 J

The work done by the force applied by the car is

W=Fd

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

W=(5000 N)(40 m)=2.0\cdot 10^5 J

(f) The kinetic energy is converted into thermal energy

Explanation:

when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

8 0
4 years ago
A greyhound's velocity changes from 10 meters per second to 15 meters per second in 0.5 second. What is the greyhound's average
Elden [556K]

The answer would be 10ms 2

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4 years ago
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