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kozerog [31]
3 years ago
5

If a ball is rolling at a velocity of 1.5 m/s and has a momentum of 10.0 kg times m/s, what is the mass of the ball?

Physics
1 answer:
stiv31 [10]3 years ago
8 0

Answer:

Mass of the ball, M = 6.667 kg

Explanation:

Given data:

Momentum, Mo = 10.0 kgm/s

Velocity of the rolling ball, V = 1.5 m/s

Mass of the body, M = ?

Momentum, Mo = Mass, M x Velocity, V

10.0 kgm/s = M x 1.5 m/s

Divide each side by 1.5 m/s

M = 10.0 kgm/s / 1.5 m/s

M = (6⅔) kg

:. Mass of the ball, M = 6.667 kg

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Which of the following is closest to 2cm?
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Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to v
butalik [34]

Answer:

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From the question we are told that:

Far point is V=34 cm

Near point is u=17 cm

Therefore

Focal Length

f=-34cm

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5 0
3 years ago
A spring gun is made by compressing a spring in a tube and then latching the spring at
inn [45]

Answer:

a)v=13.2171\,m.s^{-1}

b)H=8.9605\,m

Explanation:

Given:

mass of bullet, m=4.97\times 10^{-3}\,kg

compression of the spring, \Delta x=0.0476\,m

force required for the given compression, F=9.12 \,N

(a)

We know

F=m.a

where:

a= acceleration

9.12=4.97\times 10^{-3}\times a

a\approx 1835\,m.s^{-2}\\

we have:

initial velocity,u=0\,m.s^{-1}

Using the eq. of motion:

v^2=u^2+2a.\Delta x

where:

v= final velocity after the separation of spring with the bullet.

v^2= 0^2+2\times 1835\times 0.0476

v=13.2171\,m.s^{-1}

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

u=13.2171\,m.s^{-1}

∵At maximum height the final velocity will be zero

v=0\,m.s^{-1}

Using the equation of motion:

v^2=u^2-2g.h

where:

h= height

g= acceleration due to gravity

0^2=13.2171^2-2\times 9.8\times h

h=8.9129\,m

is the height from the release position of the spring.

So, the height from the latched position be:

H=h+\Delta x

H=8.9129+0.0476

H=8.9605\,m

4 0
3 years ago
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