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3 years ago

In Newton's cannonball experiment, if the velocity is equal the orbital velocity the Cannon ball

1 answer:

5 0

In Newton's cannonball experiment, if the velocity is equal to the orbital velocity then the cannonball will stay in Orbit.

Newtons cannonball experiment stated that the distance that a cannonball will travel, before being drawn into the Earth by the forces of gravity, is dependent on the initial velocity.

Therefore, if the cannonball is launched at a velocity that matches the orbital velocity, then it will not be able to be drawn in by gravity due to the Earth moving away from the cannonball at the same speed at which the cannonball itself is falling.

This means that the cannonball will continue to fall without reaching the Earth, therefore staying in orbit, much like that of the moon or planets around the sun.

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When a pendulum is at the position all the way to the left when it is swinging (at the top of the arc), what is true of the kine

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Potential energy + kinetic energy = constant at every moment in time

At the highest point:

potential energy is at its maximum
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3 years ago

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mario62 [17]

Answer:

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3 years ago

A disc is thrown through the air for 1.5 min with a power output of 12.5 W. How much work is done when throwing the disc?

klasskru [66]

Answer:

work = 1125 [J]

Explanation:

To solve this problem we must remember the definition of power, which is defined as the relationship between work and time. The power can be calculated using the following equation:

Power = work/time

Power = 12.5 [w]

work = joules [J]

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work = 12.5*90

work = 1125 [J]

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3 years ago

A solenoid is designed to produce a magnetic field of 3.50×10^−2 T at its center. It has a radius of 1.80 cm and a length of 46.

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Answer:

a. 2143 turns/m

b. 111.5 m

Explanation:

a. The minimum number of turns per unit length (N/L) can be found using the following equation:

$B = \frac{\mu_{0}NI}{L}$

$\frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m$

Hence, the minimum number of turns per unit length is 2143 turns/m.

b. The total length of wire is the following:

$N = 2143 turns/m*L = 2143 turns/m*46.0 \cdot 10^{-2} m = 986 turns$

Since each turn has length 2πr of wire, the total length is:

$L_{T} = N*2\pi r = 986 turn*2*\pi*1.80 \cdot 10^{-2} m = 111.5 m$

Therefore, the total length of wire required is 111.5 m.

I hope it helps you!

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3 years ago

A sports car accelerates from 0 to 25 meters per second in 4 seconds. What is its acceleration?

WITCHER [35]

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6.25 ms²

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..................

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3 years ago