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kow [346]
3 years ago
13

In coming to a stop, a car leaves skid marks 85m long on the highway. Assuming a deceleration of 3.0m/s/s , estimate the speed o

f the car just before braking.
Physics
1 answer:
muminat3 years ago
4 0
Based on your question where a car leaves skid marks 85m long on the highway nad getting to stop. The deceleration of the car is 3m/s^2 to estimate the speed of the car just before braking first is to analyze the problem then apply the necessary formulas. the possible answer is  510m/s
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The net horizontal force on a car is 981 N. The car has a mass of 1550 kg and the force is applied when the car has a speed of 2
viktelen [127]

Answer:

Distance, d = 778.05 m                          

Explanation:

Given that,

Force acting on the car, F = 981 N

Mass of the car, m = 1550 kg

Initial speed of the car, v = 25 mi/h = 11.17 m/s

We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

F=ma\\\\a=\dfrac{F}{m}\\\\a=\dfrac{981}{1550}\\\\a=0.632\ m/s^2

Let d is the distance covered by car. Using second equation of motion as :

d=ut+\dfrac{1}{2}at^2\\\\d=11.17\times 35+\dfrac{1}{2}\times 0.632\times (35)^2\\\\d=778.05\ m

So, the car will cover a distance of 778.05 meters.

5 0
4 years ago
A football player kicks the ball at a 45 degree angle. Without an effect from the wind, the ball would travel 60.0m horizontally
Ronch [10]
B when the ball is at its maxium height 
5 0
3 years ago
Why is polaris used as a celestial reference point?
ser-zykov [4K]

At the equator, the North Celestial and South Celestial Poles would lie on the horizon where the meridian intersects the horizon. Polaris is called the "North Star." ... Polaris is special because the Earth's North Pole points almost exactly towards it.

8 0
3 years ago
How do you calculate the gravitational force on a rubber ball which has a mass of 50g
AfilCa [17]
P (gravitational force) = m (mass) x g
<=> P = 0.05 x 10
<=> P = 0.5N
6 0
3 years ago
It's nighttime, and you ve dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 0.90 m above the ed
yanalaym [24]

Answer:

The distance of the goggle from the edge is 5.30 m

Explanation:

Given:

The depth of pool (d) = 3.2 m

let 'i' be the angle of incidence

thus,

i = tan^{-1}(\frac{2.2}{0.90})

i = 67.75°

Now, Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

where,

r is the angle of refraction

n₁ is the refractive index of medium 1 = 1 for air

n₂ is the refractive index of medium 1 = 1.33 for water

now,

1 × sin 67.75° = 1.33 × sin(r)

or

r = 44.09°

Now,  

the distance of googles = 2.2 + d×tan(r)  = 2.2 + (3.2 × tan(44.09°) = 5.30 m

Hence, <u>the distance of the goggle from the edge is 5.30 m</u>

5 0
3 years ago
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