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zmey [24]
3 years ago
6

the engine of a car of amass of 2000 kg produced a force of 15000N find the acceleration of the car​

Physics
1 answer:
weqwewe [10]3 years ago
3 0
I believe the answer would be 7.5 m/s^2
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A projectile is fired at v 0 = 381.0 m/s at an angle of θ = 73.5 ∘ , with respect to the horizontal. Assume that air friction wi
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Answer:

Explanation:

velocity of projection, vo = 381 m/s

angle of projection, θ = 73.5°

The formula for the range is

R=\frac{u^{2}Sin2\theta }{g}

R=\frac{381^{2}Sin147 }{9.8}

R = 8067.4 m

Range in shorten by 34.1 %

So, the new range is

R' = 8067.4 - 34.1 x 8067.4/100

R' = 5316.4 m

5 0
3 years ago
A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between
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Answer:

-589.05 J

Explanation:

Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner

So, W = ΔK

W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)

So, substituting the values of the variables into the equation, we have

W = 1/2m(v₁² - v₀²)

W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)

W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)

W = 1/2 × 72.9 kg(-16.1604 m²/s²)

W = 1/2 × (-1178.09316 kgm²/s²)

W = -589.04658 kgm²/s²

W = -589.047 J

W ≅ -589.05 J

4 0
2 years ago
Select the correct answer.
mamaluj [8]

Answer:

amount of charge on the source charge

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2 years ago
A force of 16 N is applied to a 2 kg mass over 4 m. The surface is horizontal and frictionless. If the object is initially at re
KATRIN_1 [288]
C) Calculate the final speed.
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2 years ago
What is the speed of a bobsled who's distance-time graph indicates that it traveled 106m in 26s?
swat32

To find the speed of the bobsled, you divide the distance by time. So, when you divide 106 by 26, you get 4.07. This means the speed of the bobsled is about 4 miles per hour.

3 0
3 years ago
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