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3 years ago

How can we find the volume of a small pebble with the help of measuring cyilinder

1 answer:

5 0

We can find the volume of a small pebble with the help of measuring cylinder by using the water displacement method. <span>The </span>water displacement method<span> <span>is the process of measuring the volume of an irregularly shaped object by immersing it in water. </span></span>I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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Which statement best defines why sound waves are mechanical waves?

Annette [7]

The correct answer among the choices presented above is option A. Sound waves are considered as mechanical waves because they require a medium in which they can travel through. A mechanical wave is a wave that cannot transmit energy in a vacuum. It needs something to move energy.<span />

7 0

3 years ago

Read 2 more answers

At some point in an underground horizontal pipe, the speed of water is 2.5m/s and the gauge pressure is 2×104 Pa. What is the ga

lys-0071 [83]

The gauge pressure of the pipe is 2.234 ×10⁴ Pa, if the pipe is changed for another twice the area.

To find the gauge pressure the values given are,

speed of water = 2.5 m/s

Gauge pressure = 2×10⁴ Pa

<h3>What is pressure?</h3>

Definition:

Pressure can be defined as the total applied force per unit of area. The pressure depends both on externally applied force as well the area on which it is applied.

To find the gauge pressure first we have use continuity equation,

A ₁V₁ = A₂V₂

By the given condition: As the area is increased to twice the original area

A ₁V₁ = 2A₁V₂

A₁ gets cancelled on both the sides,

V₂ =V₁/2

=2.5 /2

V₂=1.25 m/s

Then, substituting the values in Bernoulli's equation,

Equation is given as,

P₁ + 1/2 ρV₁² = P₂ + 1/2 ρV₂²

Taking 1/2 ρ as common,

P₂ = 2×10⁴ + 1/2 ×1000×(2.5² -1.25²)

= 2×10⁴ + 2343.75

=22343.75 Pa

=2.234 ×10⁴ Pa

Thus, if the pipe is changed for another twice the area then the gauge pressure is 2.234 ×10⁴ Pa.

Learn more about pressure,

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6 0

1 year ago

PLEASE HELP

Elan Coil [88]

Answer:

Explanation:

$a_{x}=0$ $v_{xo}= v_{o}cos(delta)t$ $a_{y}=-g$ $v_{yo}=sin$ θ

X-direction | Y-direction

$x=x_o+v_{xo}t$ ⇒ $x=v_{xo}cos(delta)t$ |

5 0

3 years ago

A police siren of frequency fsiren is attached to a vibrating platform. The platform and siren oscillate up and down in simple h

babunello [35]

Answer:

he maximum frequency occurs when the denominator is minimum

f’= f₀ $\frac{343}{343 + v_s}$

Explanation:

This is a doppler effect exercise, where the sound source is moving

f = fo $\frac{v}{v-v)s}$ when the source moves towards the observer

f ’=f_o $\frac{v}{v+v_{sy}}$ Alexandrian source of the observer

the maximum frequency occurs when the denominator is minimum, for both it is the point of maximum approach of the two objects

f’= f₀ $\frac{343}{343 + v_s}$

8 0

3 years ago

(a) Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The ma

andrew11 [14]

Answer:

a) $F=1.044\times 10^9\ N$

b) $F'=1.044\times 10^9\ N$

c) $F_p=1.0672\times10^{-7}\ N$

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

mass of Mars, $M=6.4\times 10^{23}\ kg$

radius of the Mars, $r=3.4\times 10^{6}\ m$

mass of human, $m=80\ kg$

Gravitation force exerted by the Mars on the human body:

$F=G.\frac{M.m}{r^2}$

where:

$G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2}$ = gravitational constant

$F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}$

$F=1.044\times 10^9\ N$

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

$F'=F$

$F'=1.044\times 10^9\ N$

When a similar person of the same mass is standing at a distance of 4 meters:

$F_p=6.67\times10^{-11}\times \frac{80\times 80}{4}$

$F_p=1.0672\times10^{-7}\ N$

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.

Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.

4 0

3 years ago