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Romashka [77]
2 years ago
11

In an elastic collision, _______ energy is conserved.

Physics
1 answer:
Nonamiya [84]2 years ago
4 0

Answer:

Kinetic energy and momentum are conserved.

Explanation:

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Answer:

F=G(m1m2)/Rsquare if radius is given

F=G(m1m2)/dsquare if distance is given

where,

f =gravitational force

G =gravitational constant

m1=mass of one object

m2=mass of another object

d=distance between two object from their center r=radius of earth/planet

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3 years ago
What acceleration results from exerting a 125N force on a 0.65kg ball?
labwork [276]
first you do your pyramid f is on top and ma is
on bottom were m=mass and a=acceleration
were in this case you do f÷m=force÷mass so again in this case 125÷0.65=192.3 to be more accurate 192.3076923077
5 0
3 years ago
Newtons second law states that Force=Mass × Acceleration.The acceleration of gravity in 9.8 m/s.How much force is gravity exerti
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.98 Newton’s because you convert 100 g to kg which is .1 kg them you multiply.1 kg by 9.8 and get .98 and the units of the force are in Newton’s

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The equation for density is mass divided by volume. the mass of a substance is changed so that it is made of fewer particles. If
Finger [1]
Density is directly proportional to mass. So if there's less matter inside object, its density will also reduce.
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3 years ago
Read 2 more answers
A baseball is hit that just goes over a wall that is 45.4m high. If the baseball is traveling at 46.2 m/s at an angle of 32.7° b
mario62 [17]

Answer:

54.9 m/s at 44.9 degrees

Explanation:

If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.

Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s

Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s

SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.

Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use

Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2

We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m

If we take the first derivative of the equation of position, we get the equation for speed

V(t) = Vy0 + a * t

We know that being t2 the moment the ball goes over the wall

V(t2) = -25 m/s

Y(t2) = 45.4 m

So:

45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2

-25 = Vy0 + a * t2

Then:

Vy0 = -25 - a * t2

So:

45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2

0 = -44.4 - 25 * t2 - 1/2 * a * t2^2

a = -9.81 m/s^2

0 = -44.4 - 25 * t2 + 4.9 * t2^2

Solving this quadratic equation we get:

t1 = -1.39 s

t2 = 6.5 s

Since we are looking for a positive value we disregard t1.

Now we can obtain Vy0:

Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s

Since horizontal speed is constant Vx0 = 38.9 m/s

By Pythagoras theorem we obtain the value of the initial speed:

V0 = \sqrt{Vx0^2 + Vy0^2} = \sqrt{38.9^2 + 38.76^2} = 54.9 m/s

The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90

a = atan(Vy0/Vx0) = 44.9 degrees

5 0
3 years ago
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