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bija089 [108]
3 years ago
6

A ball of putty with a mass of 0.250 kg is thrown with a velocity of +7.50 m/s directly toward a nearby wall. The putty sticks t

o the wall. What is the impulse on the putty as it comes to a stop
Physics
1 answer:
nlexa [21]3 years ago
8 0
The impulse experienced by an object is the force. • time.
The momentum change of an object is the mass. • velocity change.
The impulse equals the momentum change.

Your impulse is 1.875 I hope this helps
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3.5 Newton of force is needed to accelerate 140g of ball at 25 m/s^2 using the formula f=ma
Change 140g to kg
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If a car increases its velocity from 1m/s to 3.6km/hr in 5 seconds what’s its acceleration
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In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 78.5 m/s. Th
Andrews [41]

Answer:

FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

Explanation:

First you have to consider that the Ford Thunderbird (FT) follows a rectilinear motion with varying acceleration, while Mercedez Benz (MB) has a constant velocity (no acceleration). So if you finde the time spent by FT in each section, and the distance, then you will find the distance for MB.

1) Vf² = Vi² + 2ad, where Vf: final velocity, Vi: ionitial velocity, a: acceleration and d: distance.

For the first portion  (0 m/s)² = (78.5 m/s)² + 2a(250 m) ⇒

-(78.5 m/s)² / 2(250m) = a ⇒ a = -12.3 m/s².

Now, you can find the corresponding time for this section with the following formule: Vf = Vi + at ⇒ 0 m/s = 78.5 m/s + (-12.3 m/s²) t

⇒ t= (-78.5 m/s)/ (-12.3 m/s²) ⇒ t= 6.4 seconds.

2) Then FT spent 5 seconds in the pit.

3) The the FT accelerates until reach 78.5 m/s again in a distance of 370 m.

Vf² = Vi² + 2ad ⇒ (78.5 m/s)² = (0 m/s)² + 2a(370 m)

⇒ (78.5 m/s)²/ 2(370 m) = a ⇒ a = 8.3 m/s²

Then, Vf = Vi + at ⇒ 78.5 m/s = 0 m/2 + (8.3 m/s²) t

⇒ (78.5 m/s)/(8.3 m/s²) = t ⇒ t = 9.5 seconds.

4) Summarizing, the FT moves 620 meters (250 + 370 mts) in 20.9 seconds ( 6.4 s + 5 s + 9.5 s).

5) During this time, MB moves

Velocity = distance/ time ⇒ Velocity x time = Distance

⇒ Distance = (78.5 m/s) x  (20.9 seconds) ⇒ Distance = 1640.6 meters

6) Finally, the FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

3 0
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I need the answer to (c): Describe how the student accurately traces the path of the light emerging from the block through side
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Explanation:

i guess using a ruler.so first on the opposite side use your eye to alighn the ruler to the light from Ray box then draw a line or you can use pins and place them alighned to the Ray box an draw a line connecting the pins

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