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Mama L [17]
3 years ago
14

A catcher "gives" with a baseball when catching it. If the baseball exerts a force of 475 N on the glove such that the glove is

displaced 8.97 cm, how much work is done by the ball? Answer in units of J.
Physics
1 answer:
slava [35]3 years ago
3 0

Answer:

W=42.60 J

Explanation:

Given that

F = 475 N

Acceleration due to gravity ,g= 9.8 m/s²

Height ,h= 8.97 cm = 0.0897 m ( we know 1 m = 100 cm)

As we know that work done by a force given as

W = F . d

F=force

d=Displacement

W=work done by force

Now by putting the values

W=475 x 0.0897 J

W=42.60 J

Therefore the work done by the ball will be 42.60 J.

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A ball dropped from a bridge takes three seconds to reach the water below how far is the bridge above the water?
tatuchka [14]

<u>Given that:</u>

Ball dropped from a bridge at the rate of 3 seconds

Determine the height of fall (S) = ?

      As we know that, S = ut + 1/2 ×a.t²

                          u =initial velocity = 0

                          a= g =9.81 m/s  (since free fall)

                            S = 0+ 1/2 × 9.81 × 3²

                          <em> S = 44.145 m</em>

<em>44.145 m far is the bridge from water</em>

6 0
3 years ago
A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h
hram777 [196]

Answer:

v_{2}=3.5 m/s

Explanation:

Using the conservation of energy we have:

\frac{1}{2}mv^{2}=mgh

Let's solve it for v:

v=\sqrt{2gh}

So the speed at the lowest point is v=7 m/s

Now, using the conservation of momentum we have:

m_{1}v_{1}=m_{2}v_{2}

v_{2}=\frac{1*7}{2}

Therefore the speed of the block after the collision is v_{2}=3.5 m/s

I hope it helps you!

       

8 0
3 years ago
Does a stone lying on the ground have stored energy
Vesnalui [34]
Yes, It has a stored energy in that stone.
7 0
3 years ago
A large spool in an electrician's workshop has 65 m of insulation-coated wire coiled around it. When the electrician connects a
Art [367]

Answer:

40.34\ \text{m}

Explanation:

L_1 = Length of wire = 65 m

I_1 = Initial current = 1.8 A

I_2 = Final current = 2.9 A

We know

R\propto \dfrac{1}{I}

and

R\propto L

\dfrac{V}{I}\propto L\\\Rightarrow L\propto \dfrac{1}{I}

so

\dfrac{L_2}{L_1}=\dfrac{I_1}{I_2}\\\Rightarrow L_2=\dfrac{I_1}{I_2}L_1\\\Rightarrow L_2=\dfrac{1.8}{2.9}\times 65\\\Rightarrow L_2=40.34\ \text{m}

The length of the wire remaining on the spool is 40.34\ \text{m}.

8 0
3 years ago
An electron confined in a one-dimensional box is observed, at different times, to have energies of 12 ev, 27 ev, and 48 ev
erastovalidia [21]

Answer:

l=3.5 x 10^-10m

Explanation:

just toook the test got it right good luck!

3 0
2 years ago
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