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Klio2033 [76]
2 years ago
5

A friend throws a heavy ball toward you while you are standing on smooth ice. You can either catch the ball or deflect it back t

oward your friend. Which of the following options will maximize your speed right after your interaction with the ball? You should catch the ball. You should deflect the ball back toward your friend at the same speed with which it hit your hand. You should let the ball go past you without touching it. It doesn’t matter—your speed is the same regardless of what you do. You should deflect the ball back toward your friend at half the speed with which it hit your hand.
Physics
1 answer:
baherus [9]2 years ago
6 0

Answer:

Explanation:

My speed after the interaction will depend upon the impulse the ball will make on me . Now impulse can be expressed as follows

Impulse = change in momentum

change in momentum in the ball will be maximum when the ball bounces back with the same velocity which can be shown as follows

change in momentum = mv - ( - mv ) = 2mv

So when ball is bounced back with same velocity , it suffers greatest impulse from my hand . In return ,  it reacts with the same impulse on my hand pushing me with greatest impulse according to third law of motion. this maximizes my speed after the interaction.

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jarptica [38.1K]
Calculate the weight of the table through the equation,

   W = mg

where W is the weight, m is the mass, and g is the acceleration due to gravity. Substituting the known values,
   W = (0.44 kg)(9.8 m/s²) 
   <em>W = 4.312 N</em>

The components of this weight can be calculated through the equation,

   Wx = W(sin θ) 

and Wy = W(cos θ)

x - component:
   Wx = W(sin θ)
Substituting,
  Wx = (4.312 N)(sin 150°) = <em>2.156 N</em>

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6 0
2 years ago
A particle is in uniform circular motion. Assume a standard rtz coordinate system. If you deconstruct the net force acting on th
Kitty [74]

Answer:

a) One

Explanation:

In a uniform circular motion there must be a force acting to keep it in the circular track. This force can either be centripetal or a centrifugal force.

According to the Newton's first law of motion a particle continues to be in state of rest or in uniform motion until acted upon by an external force.

Here the term uniform motion need to be understood which refers to the uniform velocity of the particle in accordance to the vector laws.

3 0
2 years ago
A drag racer accelerates from rest at an average rate of +13.2 mls for a distance of 100. m. The driver coasts for 0.5 then uses
gtnhenbr [62]

Complete Question:

A drag racer accelerates from rest at an average rate of +13.2 m/s² for a distance of 100. m. The driver coasts for 0.5 s then uses the brakes and parachute to decelerate until the end of the track. If the total length of the track is 180 m, what minimum deceleration rate must the racer have in order to stop prior to the the end of the track?

Answer:

-31.92 m/s²

Explanation:

The drag races do a retiling uniform variated movement. There are 3 steps in the movement, first, it accelerates from rest until 100 m, second it coasts to 0.5 s, and then it decelerates. So, let's analyze each one of the steps:

Step 1

The initial velocity is v0 = 0 (because it was at rest), the acceleration is +13.2 m/s², and the distance ΔS = 100.0 m, so the final velocity, v, is:

v² = v0² + 2aΔS

v² = 2*13.2*100

v² = 2640

v = √2640

v = 51.38 m/s

Step 2

Know it's initial velocity is 51.38 m/s, it take 0.5s, and has the same acceleration, so, after 0.5 s, the velocity will be:

v = v0 + at

v = 51.38 + 13.2*0.5

v = 57.98 m/s

Thus, the distance it travels is:

v² = v0² + 2aΔS

57.98² = 51.38² + 2*13.2*ΔS

3361.6804 = 2639.9044 + 26.4ΔS

26.4ΔS = 721.776

ΔS = 27.34 m

Step 3

The initial velocity of the drag racer is 57.98 m/s, and it travels the final distance of the track: 180 - 100 - 27.34 = 52.66 m. So, when it stops, its final velocity will be 0. The minimum deceleration must be the one that it would stop at the end of the track (less than that it would cross the final track):

v² = v0² + 2aΔS

0 = 57.98² + 2a*52.66

-105.32a = 3361.6804

a = - 31.92 m/s²

4 0
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Answer:

Fnet=7200 N

Explanation:

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8 0
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Answer:

D:

Explanation:

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