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3 years ago

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With what velocity Must a 0.53 kg softball be moving to be equal to the momentum of a 0.31 kg baseball moving at 21 m/s

1 answer:

vladimir2022 [97]3 years ago

3 0

Answer:12.28m/s

Explanation:

momentum of baseball =mass of baseball x velocity of baseball

Momentum of baseball =0.31x21

Momentum of baseball =6.51kgm/s

For a softball to have same momentum with the baseball we can say :momentum of baseball =mass of softball x velocity of softball

6.51=0.53 x velocity of softball

Velocity of softball =6.51/0.53

Velocity of softball =12.28m/s

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JulijaS [17]

Answer:

4 times the mass of Earth

Explanation:

$M_1$ = Mass of Earth

$M_2$ = Mass of the other planet

r = Radius of Earth

2r = Radius of the other planet

m = Mass of object

The force of gravity on an object on Earth is

$F=\frac{GM_1m}{r^2}$

The force of gravity on an object on the other planet is

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As the forces are equal

$\frac{GM_1m}{r^2}=\frac{GM_2m}{(2r)^2}\\\Rightarrow M_1=\frac{M_2}{4}\\\Rightarrow M_2=4M_1$

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3 years ago

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 12.8 m/s2 . At t1 the rocket e

Simora [160]

Answer:4.39 s

Explanation:

Given

initial velocity $u=0$

acceleration $a=12.8 m/s^2$

velocity acquired by sled in $t_1$ time

$v=0+at$

$v=12.8t_1$

distance traveled by sled in $t_1 s$

$v^2-u^2=2as$

$(12.8t_1)^2-0=2\times 12.8\times s_1$

$s_1=6.4\cdot t_1^2$

distance traveled in $t_2$ time with velocity $v=12.8t_1$

$s_2=v\times t_2$

$s_2=12.8\times t_1\times t_2$

$s_2=12.8\cdot t_1\cdot t_2$

$s_1+s_2=5.37\times 10^3$

$6.4t_1^2+12.8t_1t_2=5370$ ----1

$t_1+t_2=97.7 s$

$t_2=97.7-t_1$

substitute the value of $t_2$ in 1

we get

$6.4t_1^2-1250.56t_1+5370=0$

thus $t_1=\frac{1250.56-1194.33}{12.8}=4.39 s$

$t_1=4.39 s$

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Answer:

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Explanation:

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2 years ago

A panda is flying at 3 m/s using a hang-glider. If the panda’s mass is 120 kg and the hang glider’s mass is 50 kg, what is the p

WINSTONCH [101]

Answer:

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(B)

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3 years ago

A 1.3 kg ball drops vertically onto a floor, hitting with a speed of 21 m/s. It rebounds with an initial speed of 9.8 m/s. (a) W

Anarel [89]

Answer:

a) Impulse(J)=40.04 kg.m/s

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Explanation:

Given that

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We know that impulse(J) impulse is a vector quantity

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We know that P=mV

So $J=m(V_2-V_1)$

Now by putting the values

J=1.3(9.8-(-21))

So impulse(J)=40.04 kg.m/s

Force is given as

$F=\dfrac{dP}{dt}$

or we can say that

$F=m\left(\dfrac{v_2-V_1}{dt}\right)$

So $F=\dfrac{40.04}{0.0156}$

F=2566.66 N

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2 years ago