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kap26 [50]
3 years ago
13

A 15lb bowling ball and a 10 pound bowling ball are

Physics
1 answer:
emmasim [6.3K]3 years ago
5 0
Same time
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Dave throws a 10 kg bowling ball straight up in the air. At the very tippy-top of its path, what is it's momentum?
svetoff [14.1K]

Answer:Zero

Explanation:

Given

mass of ball m=10\ kg

If the ball is thrown upward then at maximum point velocity of ball is zero because ball is no longer able to move upward

Momentum(P) of a particle is given by

P=mass\times velocity

P=10\times 0

P=0

Therefore at the highest point momentum is zero .

8 0
4 years ago
Which remains the same as the distance of an object from Earth changes?
Gemiola [76]

Answer:

MASS

Explanation:

weight has to do with the gravitational pull, as distance from earth decreases, so does the gravitational pull. meaning, the size of the force, the pull, and the weight would all change. mass stays the same (in this sense. if you gain weight on earth you will gain mass as well, but if you leave earth your weight will lessen and your mass will stay the same.)

hope this helped.

5 0
4 years ago
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3- The period of the simple pendulum depends on:
Alika [10]

Answer:

c. Both mass of the bob and the length of the spring.

Explanation:

#CarryOnLearning

3 0
3 years ago
A scientist studies the effect of adding different amounts of salt on the boiling point of water. He places his results in the g
den301095 [7]
The amounts are the independent boiling water is dependent

5 0
3 years ago
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A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position
Nesterboy [21]

The position of the object at time t =2.0 s is <u>6.4 m.</u>

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

v_x= \frac{dx}{dt}

Write an equation for x.

dx=v_xdt\\ x=\int v_xdt

Substitute the equation for vₓ =2t² in the integral.

x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m

Substitute 2.0s for t.

x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m

The position of the particle at t =2.0 s is <u>6.4m</u>




5 0
3 years ago
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