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butalik [34]
3 years ago
6

The earth's radius is 6.37×10^6m; it rotates once every 24 hours. with the angular speed of 7.3 x 10^-5 what is the speed of a p

oint on the earth's surface located at 2/3 of the length of the arc between the equator and the pole, measured from equator? (Hint: what is the radius of the circle in which the point moves?)
Physics
1 answer:
alexira [117]3 years ago
6 0
The speed of the earth's surface located at 2/3 of the length of the arc between the pole which measure from the equator is 232.5 m/s.

Solution:
So the givens are, earth's radius = 6.37X10^6m, and the angular distance from the pole is 90 degrees. So 60 degrees is the 2/3.

r = 6.37x10^6 * cos(60) = 3.185x10^6m
since v = wr
v = 7.3x10^-5 * 3.185x10^6

v - 232.5 m/s
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A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 4.15 m and mass 7.98 kg, find
Ludmilka [50]

Answer:

4.535 N.m

Explanation:

To solve this question, we're going to use the formula for moment of inertia

I = mL²/12

Where

I = moment of inertia

m = mass of the ladder, 7.98 kg

L = length of the ladder, 4.15 m

On solving we have

I = 7.98 * (4.15)² / 12

I = (7.98 * 17.2225) / 12

I = 137.44 / 12

I = 11.45 kg·m²

That is the moment of inertia about the center.

Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that

τ = 11.453 kg·m² * 0.395 rad/s²

τ = 4.535 N·m

8 0
3 years ago
I need help I’ll give you 15 points for both of them
Simora [160]

Answer:

Point c

Explanation:

Because of gravity

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You stand17.5 m from a wall holding a baseball. You throw the baseball at the wall at an angle of 20.5∘ from the ground with an
Lelechka [254]

Answer: 8.8 m

Explanation:

The movement of the baseball to the wall is a example of parabolic motion. While the baseball aproach the wall it is affected by gravity.

In this case, because the Initial Velocity of the ball is a vecotr, it can be defined using its two directionals compounds. One, on the Y-axis and another on the X-axis. This can be related, on how a hypotenuse is the product of two legs of the triangle. Because of this, each one of this, can be know using the following equation:

Vx = Vo * cos(∅)

Vy = Vo * sin(∅)

Where Vo is the initial velocity 27.5 m/s, and ∅ is the angle which is 20.5°. So we calculate:

Vx = 27.5m/s * cos(20.5)

Vx = 27.5m/s * 0.936

Vx = 25.74m/s

Vy = 27.5m/s  * sin(20.5)

Vy = 27.5m/s * sin(20.5)

Vy = 9.62m/s

Now the movement is divided on two parts, one under the effect of gravity and another one with a constant velocity.

To know how tall does the baseball hit the wall, we need to know first how much time it takes the ball to reach the wall on the X-axis. The wall is 17.5m away, we velocity on Vx that is constant we can calculate as it follow:

Time (T) = Distance (D) / Velocity (V)

Where Distance is 17.5m and our Velocity is the Vx calculated before.

T = 17.5 m / 25.74m/s

T = 0.68s

This is the time it takes the ball to reach the wall.

Know with the time, we can calculate the how tall it got on that time with the following equation:

x = (Vo*t) + (\frac{1}{2} *a*t^{2} )

Where Vo is the Y-Compound of the Initial Velocity.

a is the aceleration, in this case the Gravity. Which, will be negative because is oposing the movement. Gravity is equal to 9.81 m/s^{2}

And t is the time it takes the ball to get to the wall.

x = (Vy*T) + (\frac{1}{2} *g*T^{2} )

x = ((9.62m/s)*0.68s) + (\frac{1}{2} *(9.81m/s^{2})*(0.68s)^{2} )

x = 8.8 m

This is the height that the baseball touch the wall.

6 0
3 years ago
A woodworker has made four
Lilit [14]
I think the answer is b
6 0
3 years ago
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