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Rus_ich [418]
3 years ago
9

A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is -1.67 m/s^2. Determine

the time for the feather to fall to the surface of the moon?
Physics
1 answer:
zalisa [80]3 years ago
5 0
<span>We can use a simple equation to find the time for the feather to fall to the surface. y = (1/2) g t^2 y is the height g is the acceleration due to gravity t is the time t^2 = 2y / g t = sqrt{ 2y / g } t = sqrt{ (2) (1.40 m) / (1.67 m/s^2) } t = sqrt { 1.6766 s^2 } t = 1.29 seconds It takes 1.29 seconds for the feather to fall to the surface.</span>
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How long does a lunar eclipse last? A)seconds B)Minutes C)Days D) Weeks
bonufazy [111]
Well, it happens a few weeks ahead, then for a total of 3 hours and 40 minutes.
3 0
3 years ago
a slender rod of mass m and length l is released from rest in a horizontal position. what is the rod's angular velocity when it
Makovka662 [10]

Answer:

M g H / 2 = M g L / 2      initial potential energy of rod

I ω^2 / 2 = 1/3 M L^2 * ω^2 / 2   kinetic energy attained by rod

M g L / 2 = 1/3 M L^2 * ω^2 / 2

g = 3 L ω^2

ω = (g / (3 L))^1/2

8 0
2 years ago
In one of the original Doppler experiments, a tuba was played at a frequency of 64.0 Hz on a moving flat train car, and a second
wolverine [178]

Answer:

 f_{beat} = 1.64\ Hz

Explanation:

given,

frequency of tuba.f = 64 Hz

Speed of train approaching, v = 8.50 m/s

beat frequency = ?

using Doppler's effect formula

 f' = f(\dfrac{v}{v-v_s})

v_s is the velocity of the source

v is the speed of sound, v = 340 m/s

now,

 f' = 64\times (\dfrac{340}{340 - 8.50})

       f' = 65.64 Hz

now, beat frequency is equal to

 f_{beat} = f' - f

 f_{beat} = 65.64 - 64

 f_{beat} = 1.64\ Hz

hence, beat frequency is equal to 1.64 Hz

3 0
3 years ago
An elevator motor provides 45.0 kW of power while lifting an elevator 35.0 m. If the elevator contains seven passengers each wit
mylen [45]

Find how much work ∆<em>W</em> is done by the motor in lifting the elevator:

<em>P</em> = ∆<em>W</em> / ∆<em>t</em>

where

• <em>P</em> = 45.0 kW = power provided by the motor

• ∆<em>W</em> = work done

• ∆<em>t</em> = 20.0 s = duration of time

Solve for ∆<em>W</em> :

∆<em>W</em> = <em>P</em> ∆<em>t</em> = (45.0 kW) (20.0 s) = 900 kJ

In other words, it requires 900 kJ of energy to lift the elevator and its passengers. The combined mass of the system is <em>M</em> = (<em>m</em> + 490.0) kg, where <em>m</em> is the mass of the elevator alone. Then

∆<em>W</em> = <em>M</em> <em>g h</em>

where

• <em>g</em> = 9.80 m/s² = acceleration due to gravity

• <em>h</em> = 35.0 m = distance covered by the elevator

Solve for <em>M</em>, then for <em>m</em> :

<em>M</em> = ∆<em>W</em> / (<em>g h</em>) = (900 kJ) / ((9.80 m/s²) (35.0 m)) ≈ 2623.91 kg

<em>m</em> = <em>M</em> - 490.0 kg ≈ 2133.91 kg ≈ 2130 kg

4 0
3 years ago
A 244.0 N block is at rest on a flat, frictionless table. A hooked cable applies an upward force of 24.0 N on the block. What is
blagie [28]

Answer:

268N

Explanation:

The upward force acting on the block are the reaction and the hooked table..

The total normal force acting = normal reaction + 24N

Note that the normal reaction is always equal the weight of the table

Hence the normal force acting in the block is 244.0+24 = 268.0N

4 0
3 years ago
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