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Ghella [55]
3 years ago
10

Explain the ways the earth moves and how each way affects the earth

Physics
1 answer:
Drupady [299]3 years ago
5 0

Answer:

The earth moves two ways. It spins and it moves around the sun. The spinning of the earth is called rotation. It takes the earth about 24 hours, or one day, to make one complete rotation.

Explanation:

The Earth's Three Motions. The Earth turns (rotation around the polar axis), goes along on its orbit (revolution around the Sun), swings smoothly as an unbalanced spinning top (equinoctial precession).

Because the Earth rotates on its axis, circulating air is deflected toward the right in the Northern Hemisphere and toward the left in the Southern Hemisphere. This deflection is called the Coriolis effect. Click the image for a larger view. Coastal currents are affected by local winds.

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3 years ago
A ball on the end of a string is revolving at a uniform rate in a vertical circle of radius 97.7 cm. If its speed is 3.74 m/s, a
noname [10]

The tension in the string when the ball is at the bottom of the path is 2.61 Newtons.

<h3>Tension</h3>

A tension is simply referred to as a force along the length of a flexible medium such as strings, cable, ropes etc.

Tension in a string revolving can be determined using the expression;

T = mv² / r

Where m is mass of object, v is velocity and r is radius ( length of string )

Given the data in the question;

  • Mass of ball m = 0.182kg
  • Radius ( length of string ) r = 97.7cm = 0.977m
  • Velocity = 3.74m/s
  • Tension in the string; T = ?

To determine tension in the string, we substitute our given values into the expression above.

T = mv² / r

T = (0.182kg × (3.74m/s)²) / 0.977m

T = (0.182kg × 13.9876m²/s²) / 0.977m

T = (2.5457432kgm²/s²) / 0.977m

T = 2.61kgm/s²

T = 2.61N

Therefore, the tension in the string when the ball is at the bottom of the path is 2.61 Newtons.

Learn more about Tension here: brainly.com/question/14351325

7 0
2 years ago
A baton twirler has a baton of length 0.36 m with masses of 0.48 kg at each end. Assume the rod itself is massless. The rod is f
AleksAgata [21]

Answer:

Part a)

When rotated about the mid point

\tau = 0.021Nm

Part b)

When rotated about its one end

\tau = 0.042 Nm

Explanation:

As we know that the angular acceleration of the rod is rate of change in angular speed

so we will have

\alpha = \frac{\Delta \omega}{\Delta t}

\alpha = \frac{2.4 - 0}{3.6}

\alpha = 0.67 rad/s^2

Part a)

When rotated about the mid point

I = 2mr^2

I = 2(0.48)(0.18)^2

I = 0.0311 kg m^2

now torque is given as

\tau = 0.0311 (0.67)

\tau = 0.021Nm

Part b)

When rotated about its one end

I = m(2r)^2

I = (0.48)(0.36)^2

I = 0.0622 kg m^2

now torque is given as

\tau = (0.0622)(0.67)

\tau = 0.042 Nm

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Read 2 more answers
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