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3 years ago

When a bat hits a ball, the impulse is what?

Physics

2 answers:

Rama09 [41]3 years ago

7 0

The answer is c. The force multiplied by the time the objects are in contact. Hope this helps! :)

german3 years ago

4 0

I believe the answer is C

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In tae-kwon-do, a hand is slammed down onto a target at a speed of 14 m/s and comes to a stop during the 3.0 ms collision. Assum

GrogVix [38]

Answer:

A) The impulse is 11,7 kg.m/s . B) The average force on the hand is 3900N

Explanation:

Givens:

Hand's mass (m)= 0,90kg

Hand's initial velocity (vi)= 14m/s

Time of collision (t)= 3ms

A) Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time. We use the formula:

/F/ = m. /Δv/

Where:

F= Force of Impulse

m= Mass

Δv= change in speed ( vf-vi)

Using that formula we get:

/F/= 0,90kg . / (0m/s-14m/s) /

/F/=0,90kg . 14m/s

/F/ = 11,7 kg. m/s

*note that (vf) is 0 because the hand stops in the action, so it's final

velocity = 0

B)The average force is equal to the change in the momentum over the change in time. We use the formula:

/F/= m. (Δv) : Δt

Where:

F= the average force from the target

m= the mass of the hand

Δv= hange in speed (vf/vi)

Δt= change in time

*if we look closely, we find the we alredy have half of the equation because we have alredy calculated m.(Δv) in point A.

We boil the equation down to:

/F/= Impulse : Δt

/F/= 11,7kg.m/s : 0,003s

/F/= 3900N

*we use 0,003s as our time because the given time was 3ms.

*the final result is expressed in Newtons because our final result ends up beeing kg.m/s² = N

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4 years ago

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nata0808 [166]

Find the given attachment

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FromTheMoon [43]

Given: Please see the attached image below.

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So in this problem, the two vectors that will have the largest magnitude are A & F when subtracted (i.e., when one vector is subtracted from the other).

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4 years ago

The block accelerates down the 30 degree incline at 3m/s2. What is the coefficient of friction of these surfaces

dolphi86 [110]

Fnet = Fg sin 30 - Ff
ma = mg sin 30 - mew Fg cos 30
ma = mg sin 30 - mew mg cos 30
a = g sin 30 - mew gcos30
a - g sin 30 = - mew g cos 30
mew = -(a - g sin30)/(g cos 30)
mew = -(3m/s2 - 9.81sin30)/(9.81 cos 30)
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