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Nataly [62]
3 years ago
10

7. A bicycle accelerates at 2.0 m/s2. If the mass of the bicycle and rider together is

Physics
1 answer:
kkurt [141]3 years ago
4 0

Answer:

170 N

Explanation:

Since Force F = ma were m = mass = 85 kg and a = acceleration = 2.0 m/s².

So the net force on the bicycle is

F = ma = 85 kg × 2.0 m/s² = 170 N

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The car will gain new momentum if it's velocity is doubled or tripled.
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3 years ago
A car traveling at 20 m/s when the driver sees a child standing in the road. He takes 0.80 s to react, then steps on the brakes
mr Goodwill [35]

When driver see the child standing on road his speed is 20 m/s

So here at that instant his reaction time is 0.80 s

He will cover a total distance given by product of speed and time

d_1 = v* t

d_1 = 20 * 0.8

d_1 = 16 m

now after this he will apply brakes with acceleration a = 7 m/s^2

so the distance covered before it stop is given by

v_f^2 - v_i^2 = 2 a d

0 - 20^2 = 2*(-7)*d_2

d_2 = 28.6 m

so the total distance covered by it

d = d_1 + d_2

d = 16 + 28.6 = 44.6 m

<em>so it will cover a total distance of 44.6 m</em>

3 0
3 years ago
Scientists can use spectral analysis of stars to determine which of the following?
RoseWind [281]
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3 years ago
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which type of lightening device would you use for each of the following needs: an economical light source in a manufacturing pla
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Btw only someone who is nice will answer tour question. You can't expect for explanition when the question is only worth 5 points. Not trying to be mean sorry if i am being mean
7 0
3 years ago
A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of
Fudgin [204]

Answer:

Part a)

v_f = v_x = 32.77 m/s

Part b)

T = 4.68 s

Explanation:

Part a)

Shell is fired at speed of 40 m/s at angle of 35 degree

so here we have

v_x = 40 cos35 = 32.77 m/s

v_y = 40 sin35 = 22.94 m/s

since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero

so at the highest point the speed is given

v_f = 32.77 m/s

Part b)

After completing the motion we know that the displacement of the object will be zero in Y direction

so we have

\Delta y = 0

0 = v_y t - \frac{1}{2}gt^2

T = \frac{2v_y}{g}

T = \frac{2(22.94)}{9.81} = 4.68 s

7 0
3 years ago
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