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Lynna [10]
3 years ago
12

a block with length 1.5m width 1m height 0.5m and mass 300kg lays on the table.what is the pressure at the bottom surface of the

block ?​
Physics
1 answer:
Nesterboy [21]3 years ago
6 0

Answer:

your answer will be 320kg that would be the pressure at the bottom surface of the block

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A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end
Dominik [7]

Answer:

wavelength = 0.8989 m

Explanation:

Given data:

weight of wire is 8.25 g

length of wire is 65 cm

speed of sound in room is 344 m/s

time of returning of pulse is 7.84 ms

There are three time period

Time period = \frac{7.84\times 10^{-3}}{3} = 2.6133\times 10^{-3} s

frequency = \frac{1}{T}

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3 years ago
A train traveling at 27.5 accelerates to 42.4 m s over 75.0 s What is the displacement of the train in this time period A train
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Answer:

The displacement of the train in this time period is 2,616.86 m.

Explanation:

A Uniformly Varied Rectilinear Motion is Rectilinear because the mobile moves in a straight line, Uniformly because of there is a magnitude that remains constant (in this case the acceleration) and Varied because the speed varies, the final speed being different from the initial one.

In other words, a motion is uniformly varied rectilinear when the trajectory of the mobile is a straight line and its speed varies the same amount in each unit of time (the speed is constant and the acceleration is variable).

An independent equation of useful time in this type of movement is:

vf^{2} =vi^{2} +2*a*d <em>Expression A</em>

where:

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  • vi = initial velocity
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  • d = distance

The equation of velocity as a function of time in this type of movement is:

vf=vi + a*t

So the velocity can be calculated as: a=\frac{vf-vi}{t}

In this case:

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  • vi=27.5 m/s
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Replacing in the definition of acceleration:  a=\frac{42.4 m/s-27.5 m/s}{75 s}

a=0.199 m/s²

Now, replacing in expression A:

(42.4 m/s)^{2} =(27.5 m/s)^{2} +2*(0.199 m/s^{2}) *d

Solving:

d= \frac{(42.4 m/s)^{2} - (27.5 m/s)^{2} }{2*(0.199 m/s^{2}) }

d= 2,616.86 m

<u><em>The displacement of the train in this time period is 2,616.86 m.</em></u>

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Since sphere is performing pure rolling motion

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