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3 years ago

To do your homework correctly you should (5 points)

Physics

2 answers:

pychu [463]3 years ago

4 0

Answer:

C according to me

Explanation:

ITS HARD

Murrr4er [49]3 years ago

3 0

I believe c but I’m not particularly sure

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What is The distance traveled by a vehicle in 12 minutes,if it’s speed is 35km/h

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Answer: 7 km

Explanation: 12 minutes is 1/5 of an hour. 35/5=7

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Which of the following is equal to the area under a velocity-time graph

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-- The area under a velocity/time graph, between two points in time, is the difference in displacement during that period of time.

-- The area under a speed/time graph, between two points in time, is the distance covered during that period of time.

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An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie

leva [86]

To solve the problem it is necessary to apply conservation of the moment and conservation of energy.

By conservation of the moment we know that

$MV=mv$

Where

M=Heavier mass

V = Velocity of heavier mass

m = lighter mass

v = velocity of lighter mass

That equation in function of the velocity of heavier mass is

$V = \frac{mv}{M}$

Also we have that $m/M = 1/7 times$

On the other hand we have from law of conservation of energy that

$W_f = KE$

Where,

W_f = Work made by friction

KE = Kinetic Force

Applying this equation in heavier object.

$F_f*S = \frac{1}{2}MV^2$

$\mu M*g*S = \frac{1}{2}MV^2$

$\mu g*S = \frac{1}{2}( \frac{mv}{M})^2$

$\mu = \frac{1}{2} (\frac{1}{7}v)^2$

$\mu = \frac{1}{98}v^2$

$\mu = \frac{1}{g(98)(5.1)}v^2$

Here we can apply the law of conservation of energy for light mass, then

$\mu mgs = \frac{1}{2} mv^2$

Replacing the value of $\mu$

$\frac{1}{g(98)(5.1)}v^2 mgs = \frac{1}{2}mv^2$

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3 years ago

An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless

Elenna [48]

Answer:

Explanation:

Potential energy lost by mass = mgh

= 10 x 9.8 x 2 = 196 J

a ) If v be velocity of mass at the bottom , its kinetic energy will be stored in spring as elastic energy

= 1/2 m v² = 1/2 k x² , k is spring constant , x is compression , m is mass falling down

.5 x 10 v² = .5 x 500 x .75²

v = 5.3 m /s

b ) kinetic energy of mass at the bottom

= /2 m v²

= .5 x 10 x 5.3²

= 140.45 J

energy lost by mass while coming down

=potential energy at the top - kinetic energy at bottom

= 196 - 140.45

= 55.55 J .

This is equal to negative work done by friction

work done by friction = - 55.55 J

c ) Since there will be no loss of energy in compression and extension of spring so , no loss of kinetic energy will take place of mass . So it wil have same velocity that is 5.3 m /s while on its return journey.

d ) kinetic energy at the bottom = 140.45

loss of energy by friction again

= 140.45 - 55.55

= 84.9 J

If h be the height attained

mgh = 84.9

10 x 9.8 x h = 84.9

h = .866 m

( We have assumed that loss of energy in return journey will be same due to friction . )

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