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madam [21]
3 years ago
9

A 1000-kilogram car traveling with a velocity of 20. meters per second decelerates uniformly at -5.0 meters per second2 until it

comes to rest. What is the magnitude of the impulse applied to the car to bring it to rest?
Physics
1 answer:
Elis [28]3 years ago
3 0

Answer:

-20000 kgm/s

Explanation:

Impulse: This can be defined as the product of the mass of a body and its change in velocity. The S.I unit of impulse is kgm/s.

Mathematically, impulse can be expressed as

I = m(v-u).............. Equation 1.

Where I = impulse applied to the car to bring it to rest, m = mass of the car, u = initial velocity of the car, v = final velocity of the car.

Given: m = 1000 kg, u = 20 m/s, v = 0 m/s ( to rest)

Substitute into equation 1

I = 100(0-20)

I = 1000(-20)

I = -20000 kgm/s

Hence the impulse applied to the car to bring it to rest = -20000 kgm/s

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Taya2010 [7]

Answer:

I) You walk barefoot on the hot street and it burns your toes.

II) When you get into a car with hot black leather in the middle of the summer and your skin starts to get burned.

Explanation:

In conduction mode of heat transfer we know that the energy is transferred from one system to other system due to direct contact of two bodies

Here due to this direct contact the energy is transferred via a given solid or liquid medium

In this type of heat transfer medium particles will remain in its own position only the energy is transferred.

So here we can say the correct answer will be

I) You walk barefoot on the hot street and it burns your toes.

II) When you get into a car with hot black leather in the middle of the summer and your skin starts to get burned.

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4 years ago
What is the force that opposes motion and works against the downward pull? A) friction B) gravity C) weight D) acceleration
charle [14.2K]
I'm not entirely sure, but I believe it is A Friction. because gravity pulls down, weight isn't a force, and acceleration doesn't oppose motion
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4 years ago
The driver of a 1750 kg car traveling on a horizontal road at 110 km/h suddenly applies the brakes. Due to a slippery pavement,
Alexeev081 [22]

Answer: a=-2.4525 m/s^2

d=s=190.3 m

Explanation:The only force that is stopping the car and causing deceleration is the frictional force Fr

Fr = 25% of weight

W=mg

W=1750*9.81

W=17167.5

Hence

Fr=\frac{25}{100} * -17167.5\\\\Fr=-4291.875 N

Frictional force is negative as it acts in opposite direction

According to newton second law of motion

F=ma

hence

a=Fr/m

a=-4291.875/1750\\a=-2.4525

given

u= 110 km/h

u=110*1000/3600

u=30.55 m/s

to get t we know that final velocity v=0

v^2=u^2+2as\\0=30.55^2-2*2.4525*s\\s=190.34m

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Part A)

Distance is given as sum of total path length

So here as we are given that car drives 40 miles East and 30 miles West

So here the total path length is

Distance = 40 miles + 30 miles

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Part B)

Displacement is given as straight line distance between initial and final position

So here we can see that

d_1 = 40 miles East

d_2 = 30 miles West

Now the total displacement will be

d = d_1 - d_2 = 40 - 30 = 10 miles

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