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Westkost [7]
3 years ago
10

A 2.1 times 103 - kg car starts from rest at the top of a 5.0 - m - long driveway that is inclined at 20 deg with the horizontal

. If an average friction force of 4.0 times 103 N impedes the motion, find the speed of the car at the bottom of the driveway.

Physics
1 answer:
VLD [36.1K]3 years ago
4 0

Answer:

speed of the car at the bottom of the driveway is 3.8 m/s

Explanation:

given data

mass = 2.1× 10³ kg

distance = 5 m

angle = 20 degree

average friction force = 4 × 10³ N

to find out

find the speed of the car at the bottom of the driveway

solution

we find acceleration a by  force equation that is

force = mg×sin20 - friction force

ma = mg×sin20 - friction force

put here value

2100a = 2100 ( 9.8)×sin20 - 4000

a = 1.447 m/s²

so from motion of equation

v²-u² = 2as

here u is 0 by initial speed and v is velocity and a is acceleration and s is distance

v²-0 = 2(1.447)(5)

v = 3.8

speed of the car at the bottom of the driveway is 3.8 m/s

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A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is
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Answer:

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Explanation:

u = Initial velocity

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Equation of motion

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-18.2^2}{2\times 162}\\\Rightarrow a=-1.02234\ m/s^2

The acceleration of the craft should be 1.02234 m/s²

F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N

Weight of the craft

W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N

Thrust

F_t=F+W\\\Rightarrow F_t=8526.3156+13344\\\Rightarrow F_t=21870.3156\ N

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

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A container with rigid walls is filled with 4.0 mol of air with Cv=2.5R Then the temperature is increased from 17 degrees C to 3
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Since there's no change in volume (rigid walls), W = 0.

U = Q

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U = (4.0 mol) (2.5 × 8.314 J/mol/K) (354 C − 17 C)

U = 28,000 J

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3 years ago
I need help with #10
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