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wlad13 [49]
3 years ago
8

A person is pushing two carts that are connected with a metal bar so that the carts are moving at constant acceleration 0.30m/s2

. The masses of the carts are 100 kg and 150kg, and the mass of the connecting bar is 60 kg. Assume that the friction forces are negligible. Also assume that you are pushing the lighter cart to the right. A) Determine the force that the bar exerts on the lighter cart. B) Determine the force that the heavier cart exerts on the bar.
Physics
1 answer:
madreJ [45]3 years ago
6 0

Answer:

63N, 45N

Explanation:

Given:

Tension 1 = T1

We have the following free body equations:

F - T1 = 100*0.3 -(I)

T1 - T2 = 60*0.3 -(II)

T2 = 150*0.3 -(III)

Adding the three above equations, we have

F - T1 + T1 - T2 + T2 = (100*0.3) + (60*0.3) + (150*0.3)

Hence, F = 93N

Hence, substituting into equation I we have

T1 = Force by bar on lighter cart = 93 - (100*0.3) = 63N

Hence, force that heavier cart exerts on bar

T2 = 150*0.3 = 45N

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Anuta_ua [19.1K]

<u>The number of variations the dealership have is 40 cars</u>.

Data given;

  • The number of different product = 4
  • The number of transmission = 2 (standard or automatic)
  • The numbers of colors available = 5

<h3>Dealership variation </h3>

This is the total numbers of cars available in the dealer shop at the moment.

To get that, we simply multiply the numbers of colors available, the number of transmission and the makes of car available.

This is equal to

4*2*5=40

From the above calculation, we can say that <u>the variations the dealership offers is 40</u>

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3 years ago
Two loudspeakers on a concert stage are vibrating in phase. A listener 50.5 m from the left speaker and 26.0 m from the right on
mixer [17]

To solve this problem, it is necessary to apply the concepts related to the constructive interference caused by the wavelengths of the sound traveling in the air.

From the definition of constructive interference we know that

\Delta x = n \lambda \rightarrow \lambda = \frac{\Delta x}{n}

Where

\Delta x = Distance between speakers

n = Integer which represent he number of repetition of the spectrum

\lambda = Wavelength

At the same time the frequency is subject to the form,

f = \frac{v}{\lambda}

Where

v = 343m/s \rightarrow Velocity (of the sound at this case)

From our given values we have to \Delta x is

\Delta x = 50.5m-26m

\Delta x = 24.5m

The wavelength would be subject to the sound spectrum therefore for n = 1,

\lambda = \frac{\Delta x}{n}

\lambda = \frac{24.5}{1}

\lambda = 24.5m

Then the frequency would be,

f = \frac{v}{\lambda}

f = \frac{343}{24.5}

f = 14Hz

For the value of n = 2,

\lambda = \frac{\Delta x}{n}

\lambda = \frac{24.5}{2}

\lambda = 12.25m

Then the frequency would be,

f = \frac{v}{\lambda}

f = \frac{343}{12.25}

f = 28Hz

For n = 3,

\lambda = \frac{\Delta x}{n}

\lambda = \frac{24.5}{3}

\lambda = 8.167m

Then the frequency would be,

f = \frac{v}{\lambda}

f = \frac{343}{12.25}

f = 42Hz

From the frequencies obtained we can identify that the two lowest frequencies that can be heard due to constructive interference are 28Hz and 42Hz

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3 years ago
Find the moment of force of 60 Newton about an axis of rotation at distance 20 cm from the force​
777dan777 [17]

Answer:

10 N.m

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3 years ago
Fluid originally flows through atube at a rate of 200 cm3/s. Toillustrate the sensitivity of the Poiseuille flow rate to various
Alexxx [7]

Answer:

Q_{2}=1200cm^{3}/s

Explanation:

Given data

Q₁=200cm³/s

We know that:

F=n\frac{vA}{l}\\

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And

Q=ΔP/R

As

n₂=6.0n₁

So

Q=ΔP/R

Q=\frac{nv}{lR}\\ \frac{Q_{2}}{n_{2}}= \frac{Q_{1}}{n_{1}}\\ Q_{2}=\frac{Q_{1}}{n_{1}}*(n_{2})\\Q_{2}=\frac{200}{n_{1}}*6.0n_{1}\\ Q_{2}=1200cm^{3}/s

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The actual weight of the stone is 11 N. It is based on the Archimedes principles.

<h3>What is Archimedes principle?</h3>
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