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bearhunter [10]
2 years ago
12

Plan a controlled experiment to investigate how an object’s position is related to the amount of its potential energy by measuri

ng how high the ball bounces in the simulation. Choose at least four heights from which you will drop the ball. Describe your experiment.
Physics
1 answer:
igor_vitrenko [27]2 years ago
3 0

Answer:

Feeling like an addict that ain't had it, up and at it in a minute

If it hadn't been invented, my limit wouldn't be infinite

I'm feeling like an infant in a womb

I'ma be here 'til the tomb

Lately I've been in my room

Lookin' and lookin' at records on the wall

Hold up

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Components grouped together for a particular function form a _____ system.
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15. Using 3 – 4 sentences, explain, in your own words, two ways you could cause an electrical current using only a wire and a ho
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3 years ago
University Physics 9.68: Accelerating Compact Disc. A computer disc drive is turned on starting from rest and has constant angul
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5 0
4 years ago
Estimate the speed of the water free surface and the time required to fill with water a cone-shaped container 1.5 m high and 1.5
Zolol [24]

Answer:

speed of water is 0.0007138m/s

Explanation:

From the law of conservation of mass

Rate of mass accumulation inside vessel = mass flow in - mass flow out

so, dm/dt = mass flow in - mass flow out

taking p as density

d \frac{dQ}{dt} = pq_i_n

where,

q(in) is the volume flow rate coming in

Q = is the volume of liquid inside tank at any time

But,

dQ = Adh

where ,

A = area of liquid surface at time t

h = height from bottom at time t

A = πr²

r is the radius of liquid surface

r = (1.5/2) \div 1.5 h = \frac{h}{2}

Hence,

\pi( \frac{h}{2} )^2\frac{dh}{dt} =q_i_n

\frac{dh}{dt} = \frac{q_i_n}{\pi (\frac{h}{2})^2 } =\frac{4q_i_n}{\pi h^2}

so, the speed of water surface at height h

v = \frac{dh}{dt} =\frac{4q_i_n}{\pi h^2}

where,

q_i_n is 75.7 L/min = 0.0757m³/min

h = 1.5m

so,

v = \frac{4 \times 0.0757}{\pi \times 1.5^2} \\\\v = 0.04283m/min

v = 0.04283 /60

v = 0.0007138m/s

Hence, speed of water is 0.0007138m/s

5 0
3 years ago
A 2.9 × 103 kg car accelerates from rest under the action of two forces. One is a forward force of 1148 N provided by traction b
Lana71 [14]

Answer:

The car must travel 51.34 m for its speed to reach 2.7 m/s

Explanation:

Mass of car = 2.9 x 10³ kg

Forward force = 1148 N

Resistive force = 941 N

Total force = 1148 - 941 = 207 N

We know

            Force = Mass x Acceleration

              207 = 2.9 x 10³ x Acceleration

            Acceleration = 0.071 m/s²  

Now we have equation of motion, v² = u² + 2as

      Initial velocity, u = 0 m/s

      Final velocity, v = 2.7 m/s

      Acceleration, a = 0.071 m/s²  

Substituting

        v² = u² + 2as

       2.7² = 0² + 2 x 0.071 x s

        s = 51.34 m

The car must travel 51.34 m for its speed to reach 2.7 m/s

5 0
4 years ago
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