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Roman55 [17]
3 years ago
10

MY PAPER IS DUE IN AN HOUR!!!

Physics
2 answers:
Doss [256]3 years ago
7 0

Answer:

Your kinetic energy is always proportional to the square of your speed,

no matter whether you're on a bicycle, a roller coaster, or a baby carriage.

The faster you're moving, the more kinetic energy you have.

Your potential energy doesn't depend on speed at all, only on height.

On a roller coaster, the minimum speeds are usually at the top of a hill

or a loop, so the potential energy would generally correspond to the

slowest speeds. But again, that's not because of the speed. It's because

of the height.

Explanation:

Delvig [45]3 years ago
6 0

Answer:

njjnbhkjgjbvggvn

Explanation:

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A cannon ball launched horizontally with a speed of 20m/s and a baseball dropped off a cliff and it accelerates at a rate of 10m
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If they both start from the same height, then they both hit the ground at the
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The cannon ball still accelerates downward at the same rate as the baseball.
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A 400 gram sample of alcohol at 10o C [Ca=2.64 J/g*oC] is mixed with 400 grams of warm water [Cw=4.186 J/g*Co] at 88o C. Assume
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4 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

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