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4 years ago

A clock has radius of 0.5m. The outermost point on its minute hand travels along the edge. What is its tangential speed?

1 answer:

6 0

The angular speed of the minute hand is the ratio between the angle covered and the time taken. The angle covered by the minute hand is $2 \pi$ , while the time needed is $t=60 min= 3600 s$ , so the angular speed is
$\omega = \frac{2 \pi}{3600 s}=0.0017 rad/s$

And the tangential speed is given by
$v=\omega r=(0.0017 rad/s)(0.5 m)=8.72 \cdot 10^{-4} m/s$

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1-Calculate Req

ArbitrLikvidat [17]

Answer:

1. 21.66 Ohms

2. 3.38 A

3. 6.7 V

Explanation:

1. Req = 6+2 = 8 Ohms (2 and 6 are in a series circuit)

Req = 1/8 +1/4 = 3/8 = 8/3 = 2.66 Ohms (8 and 4 are parallel, so we will add them using this equation)

Req = 2.66 + 1 + 9 + 3 + 6 = 21.66 Ohms

2. I = V/R = 9/2.66 = 3.38 A (In a series circuit, the current is the same across the resistors, so we will add them and divided them by 9 volts)

3. V = IR = 3.38 x 2 = 6.7 V (In a series circuit, the voltage is different, so each resistor will have a different voltage.)

I hope this helps. I am not an expert in physics but its ok :)

<u><em>Note: If the answer benefited u, mark me as the brainliest answer if u can, thx.</em></u>

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3 years ago

Radiant energy comes from the _______.

snow_lady [41]

Answer:

C. Sun

Explanation:

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3 years ago

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What unit of measure would you use for mass?

creativ13 [48]

grams, pounds, kilos, etc

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3 years ago

A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i

nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

$U=mgh$

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J$

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

$U=mgh$

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

$U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J$

3. 0 J

As before, the gravitational potential energy of the ball is given by

$U=mgh$

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J$

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3 years ago

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Eduardwww [97]

Answer:

Motion with constant velocity of magnitude 1 m/s (uniform motion) for 4 seconds in a positive direction and then for 2 seconds uniform motion with constant velocity of magnitude 3 m/s in reverse direction .

Explanation:

The graph shows a constant velocity of 1 m/s for 4 seconds in the positive direction. After that, between 4 seconds and 6 seconds, the object reverses its motion with constant velocity of magnitude 3m/s.

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3 years ago