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tamaranim1 [39]
3 years ago
10

a 2000 kg car moving down the road runs into a 5000 kg stationary suv. The car applies a force of 1400 n on the suv what is the

magnitude of force applied by the suv on the car
Physics
1 answer:
masha68 [24]3 years ago
3 0

Answer:

F_suv= 49050 N

Explanation:

We are told that a 2000 kg car moving down the road runs into a 5000 kg stationary suv. The car applies a force of 1400 N on the suv.

Now, according Newton's first law of motion, an object will continue in it's present state of rest except it is acted upon by an external body.

This means the force acting on the stationary Suv is force of gravity.

Thus; F_suv = 5000 × 9.81

F_suv= 49050 N

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On a sunny summer day, why does a white car with a light-colored interior stay cooler than a black car with a dark-colored inter
Oduvanchick [21]

Answer: Solar radiation reflects off the lighter colours, away from the car, thus keeping the car cool

Explanation: This is because lighter colors reflect a good amount of radiation while darker colors absorb it. Just like, Antarctica hasn't completely melted because it reflects a lot of the heat that is acting upon it. Or, you notice that you get hotter when you wear a black shirt opposed to a white one.

8 0
3 years ago
A car is traveling at 15 m/sm/s . Part A How fast would the car need to go to double its kinetic energy
GREYUIT [131]

Answer:

21.21 m/s

Explanation:

Let KE₁ represent the initial kinetic energy.

Let v₁ represent the initial velocity.

Let KE₂ represent the final kinetic energy.

Let v₂ represent the final velocity.

Next, the data obtained from the question:

Initial velocity (v₁) = 15 m/s

Initial kinetic Energy (KE₁) = E

Final final energy (KE₂) = double the initial kinetic energy = 2E

Final velocity (v₂) =?

Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:

KE = ½mv²

NOTE: Mass (m) = constant (since we are considering the same car)

KE₁/v₁² = KE₂/v₂²

E /15² = 2E/v₂²

E/225 = 2E/v₂²

Cross multiply

E × v₂² = 225 × 2E

E × v₂² = 450E

Divide both side by E

v₂² = 450E /E

v₂² = 450

Take the square root of both side.

v₂ = √450

v₂ = 21.21 m/s

Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.

8 0
3 years ago
Calculate the kinetic energy in joules of a 1200 kg automobile moving at 18 m/s .
vodka [1.7K]

Answer:

194,400 joules of kinetic energy.

Explanation:

Remember that to calculate the Kinetic energy you need to use the next formula:

Ke=\frac{1}{2}Mass*Velocity^2

We know that Mass= 1200 kg and velocity is 18m/s, so we insert those values into the formula:

Ke=\frac{1}{2}Mass*Velocity^2\\Ke=\frac{1}{2}1200kg*(18m/s)^2\\Ke=194,400 joules

So the kinetic energy of a car moving at 18m/s with a mass of 1200 kg would be 194,400 joules.

7 0
3 years ago
Read 2 more answers
Who does more work - a man who lifts a large box from the ground up into the back of a truck, or a man who puts the same box on
Viefleur [7K]
Answer :
The man who uses the trolley
4 0
3 years ago
A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th
Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

E=-\frac{d}{dt}(BACOS\alpha )\\

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using \pi r^{2} \\ where r is the radius of 5cm or 0.05m

A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\

since we no that the angle is at 0^{0}

we determine the magnitude of the magnetic filed

B=0.5e^{-t} \\t=2s

E=-(0.5e^{-2} * 0.00785)

E=-0.000532v\\

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A

I=2.6mA

6 0
3 years ago
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