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3 years ago

Average speed can be represented by the mathematical expression

Physics

1 answer:

nevsk [136]3 years ago

5 0

Average speed is defined as total distance moved in total interval of time

so it is given as

$v_{avg} = \frac{distance}{time}$

now here is we show distance by "d" and time by"t"

then we will have mathematical expression as follows

$v = \frac{d}{t}$

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One disadvantage of using proprietary licensed software is that

Vaselesa [24]

Answer:

its b

Explanation:

5 0

2 years ago

Read 2 more answers

A train whistle has a sound intensity level of 70. dB, and a library has a sound intensity level of about 40. dB. How many times

kodGreya [7K]

Answer:

The sound intensity of train is 1000 times greater than that of the library.

Explanation:

We have expression for sound intensity level,

$L=10log_{10}\left ( \frac{I}{I_0}\right )$

A train whistle has a sound intensity level of 70 dB

We have

$70=10log_{10}\left ( \frac{I_1}{I_0}\right )$

A library has a sound intensity level of about 40 dB

We also have

$40=10log_{10}\left ( \frac{I_2}{I_0}\right )$

Dividing both equations

$\frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000$

The sound intensity of train is 1000 times greater than that of the library.

3 0

2 years ago

A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon

Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force $F_p$ by the worker must be equal to the friction force $F_f$ on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

$F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N$

b. The work is done on the crate by this force is the product of its force $F_p$ and the distance traveled s = 4.35

$W_p = F_ps = 79.1*4.35 = 344 J$

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

$W_f = F_fs = -79.1*4.35 = -344 J$

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

$W_p + W_f = 344 - 344 = 0 J$