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zalisa [80]
2 years ago
10

How much force (in Newtons) acts on a 1700 kg car going around a

Physics
1 answer:
Daniel [21]2 years ago
5 0

Answer:

b) 4781 N

Explanation:

Because there is a redius do this question is talking about the acceleration force which= mv^2/r

so a=15^2/80=2.8125 m^2/s

so the force will be = m.a

F =1700×2.8125=4781.25 N

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dusya [7]

Answer:

So 55 Newtons are needed.

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What is the magnitude of the force needed to keep a 60 newton rubber block moving across level,dry asphalt in a straight line at
pantera1 [17]

let Coefficients of Friction of Rubber on asphalt (dry) =0.7

F= Coefficients of Friction * normal force = 0.7 * 60 =42 N

so the net force of the rubber is zero, meaning it will travel at a constant speed.

When the rubber is travel at 2m/s, 42N is required to keep moving at constant speed

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Explain the movement in the sky​
Montano1993 [528]

The motion of planets is separate to the motion of stars. Like everything in the sky, they rise in the east, and set in the west, because of the earth's rotation. But night by night, their position at a given time changes because of their orbit around the sun.

7 0
2 years ago
ou are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As soo
madam [21]

Till the time car is just adjacent to the bicycle we can say

distance moved by cycle = distance moved by car

Time taken by car to accelerate from rest

t = \frac{v_f - v_i}{a}

t = \frac{49 - 0}{7} = 7 s

Time taken by cycle to accelerate

t = \frac{23 - 0}{15} = 1.53 s

now the distance moved by cycle in time "t"

d = \frac{23 + 0}{2}*1.53 + 23(t - 1.53)

distance moved by car in same time

d = \frac{7t + 0}{2}(t)

now make them equal

3.5t^2 = 17.595 - 35.19 + 23t

3.5 t^2 - 23t + 17.595 = 0

t = 5.68 s

so cycle will move ahead of car for t = 5.68 s

8 0
2 years ago
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a foul ball is hit into the stands at a baseball game. the ball rises to a height of 38 meters and is caught on its way down by
lisov135 [29]

The velocity of the ball when it was caught is 12.52 m/s.

<em>"Your question is not complete it seems to be missing the following, information"</em>,

find the velocity of the ball when it was caught.

The given parameters;

maximum height above the ground reached by the ball, H = 38 m

height above the ground where the ball was caught, h = 30 m

The height traveled by the ball when it was caught is calculated as follows;

y = H - h

y = 38 - 30 = 8 m

The velocity of the ball when it was caught is calculated as;

v_f^2 = v_0 + 2gh\\\\v_f^2 = 0 + (2\times 9.8 \times 8)\\\\v_f^2 = 156.8\\\\v_f = \sqrt{156.8} \\\\v_f = 12.52 \ m/s

Thus, the velocity of the ball when it was caught is 12.52 m/s.

Learn more here: brainly.com/question/14582703

4 0
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