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vitfil [10]
3 years ago
12

What is a gear. Please give detailed explanation

Physics
2 answers:
mestny [16]3 years ago
3 0

Answer:

gear is part of vehicle.it helps.

Viefleur [7K]3 years ago
3 0

Answer:

Explanation:

A gear is a device that transfers mechanical energy from on point to another.

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It took 500 N of force to push a car 4 meters. How much work was done?
djverab [1.8K]

Answer:

200 J

Explanation:

W = F S

W= 500 x 4

W = 2000

6 0
3 years ago
The force of gravity acting<br> on a child's mass on earth is 490 newtons. what's the child's mass?
UNO [17]
F = m • a

What we know:
- Gravity: 9.8 m/s
- Force: 490 N

Equation derived:
m = F/a
m = 490/9.8
= 50 kg
6 0
3 years ago
Read 2 more answers
Determine the frequency of the 2nd harmonic of a spring that has a 3rd harmonic resonance of f3=512 Hz.
Elena L [17]

Answer:

1) 341 Hz

Explanation:

When a string vibrates, it can vibrate with different frequencies, corresponding to different modes of oscillations.

The fundamental frequency is the lowest possible frequency at which the string can vibrate: this occurs when the string oscillate in one segment only.

If the string oscillates in n segments, we say that it is the n-th mode of vibration, or n-th harmonic.

The frequency of the n-th harmonic is given by

f_n = nf_1

where

n is the number of the harmonic

f_1 is the fundamental frequency

Here we have:

f_3=512 Hz is the frequency of the 3rd harmonic

So the fundamental frequency is

f_1=\frac{f_3}{3}=\frac{512}{3}=170.7 Hz

And so, the frequency of the 2nd harmonic is:

f_2=2f_1=2(170.7)=341.3 Hz

3 0
4 years ago
How is a full moon different from a new moon?
koban [17]
<span>A full moon is at its brightest, and here is no disk to be seen. New moons are barely visable.</span>
8 0
3 years ago
Read 2 more answers
Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at
dexar [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

   The amplitude of the lateral  force is  F = 25 \  N

   The frequency is   f = 1 \  Hz

   The mass of the bridge per unit length is  \mu  =  2000 \  kg /m

    The length of the central span is  d =  144 m

     The oscillation amplitude of the section  considered at the time considered is  A = 75 \ mm =  0.075 \  m

      The time taken for the undriven oscillation to decay to \frac{1}{e}  of its original value is  t = 6T

Generally the mass of the section considered is mathematically represented as

            m =  \mu  *  d

=>        m =  2000 * 144

=>        m =  288000 \ kg

Generally the oscillation amplitude of the section after a  time period  t is mathematically represented as

                 A(t) = A_o e^{-\frac{bt}{2m} }

Here b is the damping constant and the A_o is the amplitude of the section when it was undriven

So from the question  

               \frac{A_o}{e}  = A_o e^{-\frac{b6T}{2m} }

=>            \frac{1}{e}  =e^{-\frac{b6T}{2m} }

=>          e^{-1} =e^{-\frac{b6T}{2m} }

=>           -\frac{3T b}{m}  =  -1

=>         b  = \frac{m}{3T}

Generally the amplitude of the section considered is mathematically represented as

           A =  \frac{n * F }{ b *  2 \pi }

=>       A =  \frac{n * F }{ \frac{m}{3T}  *  2 \pi }

=>       n =  A  *  \frac{m}{3}  *  \frac{2\pi}{25}

=>       n = 0.075 *  \frac{288000}{3}  *  \frac{2* 3.142 }{25}

=>       n = 1810 \ people

3 0
3 years ago
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