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serious [3.7K]
3 years ago
13

I need to explain how objects can have the same volume but different mass. Help??

Physics
1 answer:
kaheart [24]3 years ago
6 0
No problem, and you already know all about it.

Here are a few examples of same volume / different weight:

-- A bottle full of water is heavier than the same bottle when it's full of air.
-- Stones are heavier than styrofoam chunks the same size.
-- A bowl of meat loaf is heavier than a bowl of scrambled eggs.

In each example, two things have the same volume, but one weighs more than
the other.  I didn't say anything about mass yet, but that's easy:  As long as you
keep everything on Earth, more weight means more mass.

So how come, in each example, things with the same volume have different mass ?
This was your original question.

The answer is just the simple fact that there are millions of different substances, and
each different substance packs a different amount of mass into the same volume.

The amount of mass that a substance packs into a standard volume is called
the <em>density</em> of the substance.  Meat loaf is more dense than scrambled eggs.
Stone is more dense than styrofoam.  Water is more dense than air.  And <em>gold</em>
is 19 times as dense as water.  If you have a jar that holds a pound of water, and
you pour out the water and fill the jar with gold, the same jar holds 19 pounds of gold,
because the density of gold is 19 times the density of water.

The reason you were assigned to think about this question for homework is that
next time your Physics class meets, you'll start talking about <em>Density. </em><em /> And you're
all ready for it now.
You might be interested in
A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c
Nady [450]

(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

I= \Delta p = m \Delta v = m (v-u)

where

m is the mass of the object

\Delta v is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s

(b) 155 N

The impulse can also be rewritten as

I=F \Delta t

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

\Delta t is the duration of the collision

In this situation, we have

\Delta t = 0.06 s

So we can re-arrange the equation to find the magnitude of the average force:

F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N

6 0
3 years ago
A fire helicopter carries a 700-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee
rodikova [14]

Answer:

F = 50636.873 N

Explanation:

given,

bucket of water =  700-kg

length of cable = 20 m

Speed  = 40 m/s

angle of the cable = 38.0°

let air resistance be = F

tension in rope be = T

T cos 38° = m×g..................(1)

T sin 38^0= \dfrac{mv^2}{l} + F..........(2)

equation (1)/(2)

tan 38^0 =\dfrac{\dfrac{mv^2}{l} + F}{mg}

       0.781=\dfrac{\dfrac{700\times 40^2}{20} + F}{700\times 9.8}

           F = 50636.873 N

Hence the force exerted on the bucket is equal to F = 50636.873 N

5 0
3 years ago
Estimate the final temperature of a mole of gas at 200.0 atm and 19.0°C as it is forced through a porous plug to a final pressur
Yanka [14]

Answer : The final temperature of gas is 266.12 K

Explanation :

According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.

The formula will be:

\mu_{J,T}=(\frac{dT}{dP})_H

or,

\mu_{J,T}=(\frac{dT}{dP})_H\approx \frac{\Delta T}{\Delta P}

As per question the formula will be:

\mu_{J,T}=\frac{T_2-T_1}{P_2-P_1}   .........(1)

where,

\mu_{J,T} = Joule-Thomson coefficient of the gas = 0.13K/atm

T_1 = initial temperature = 19.0^oC=273+19.0=292.0K

T_2 = final temperature = ?

P_1 = initial pressure = 200.0 atm

P_2 = final pressure = 0.95 atm

Now put all the given values in the above equation 1, we get:

0.13K/atm=\frac{T_2-292.0K}{(0.95-200.0)atm}

T_2=266.12K

Therefore, the final temperature of gas is 266.12 K

5 0
2 years ago
.2, A car starting from rest has an acceleration of
riadik2000 [5.3K]

Answer: 2.5 m/s and 6.25 m

Explanation:

u = 0

a = 0.5 m/s²

t = 5 s

v = u + at

=  0 + 0.5 × 5

= <u>2.5 m/s</u>

s = ut + 1/2 at²

= 1/2 × 2.5 × 5

=<u> 6.25 m</u>

7 0
3 years ago
A spaceship starting from a resting position accelerates at a constant rate of 9.8 m/s. How far will the spaceship travel if its
Dvinal [7]

300 000 0 squared = 2 x 9.8 distance

KINEMATICS

Uniform or constant motion in a straight line (rectilinear). Speed or velocity constant and/or acceleration constant. If motion is up and down and/or has an up and down component then acceleration omn earth will be g. g is about 10m/s/s.


speed = distance/time

velocity = displacement/time

s=distance ... u=initial speed ... v = final speed ... a = acceleration ... t = time


v=u+at

v^2=u^2+2as

s=ut+1/2at^2

3 0
3 years ago
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