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2 years ago

A block of aluminum has a volume of 15.0 ml and a mass of 50.5 g. the density is ___________. group of answer choices

Physics

1 answer:

VARVARA [1.3K]2 years ago

8 0

A block of aluminum has a volume of 15.0 ml and a mass of 50.5 g. the density is: 3.36 g/ml

To solve this problem the density formula and the procedure that we have to use is:

d = m/v

Where:

v= volume
d= density
m= mass

Information about the problem:

v= 15.0 ml
m= 50.5 g
d = ?

Applying the density formula we get:

d = m/v

d = 50.5 g/15.0 ml

d= 3.36 g/ml

<h3>What is density?</h3>

It is a physical quantity that expresses the ratio of the body mass to the volume it occupies.

What is volume?

It is the space occupied by a body, it is calculated by multiplying its dimensions, for example: length, height and width.

Learn more about density in: brainly.com/question/1354972

#SPJ4

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Two illustrations of transverse waves. The one with shorter crests is labeled A. The wave with taller crests is labeled B. Which

suter [353]

The largest distance moved by a particle of a vibrating body from its equilibrium position is called amplitude. Wave B has higher energy because wave it has a higher amplitude.

<h3>What is the amplitude of a wave?</h3>

The largest distance moved by the particles on either side of the mean position is the amplitude of a wave.

The maximum height of either crest or trough from its equilibrium position is called amplitude.

Hence Wave B has higher energy because it has a higher amplitude.

To learn more about the amplitude of a wave refer to the link;

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2 years ago

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At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the l

Alja [10]

Answer:

Counterclockwise

Explanation:

We need to calculate the clocwkise and the counterclockwise torque.

The clockwise torque is the one generated by the child sitting on the right. This torque is given by:

$M_C = F d$

where

$F=mg=(21 kg)(9.8 m/s^2)=205.8 N$ is the force exerted by the child (his weight)

d = 1.8 m is the distance from the pivot point

So, the clockwise torque is

$M_C = (205.8 N)(1.8 m)=370.4 Nm$

The counterclockwise torque is the one generated by the adult pushing on the left, and it is given by

$M_A = F d$

where

F = 151 N is the force applied

d = 3.0 m is the distance from the pivot

Substituting,

$M_A = (151 N)(3.0 m)=453 Nm$

So, the net torque is

$M_A - M_C = 453 Nm - 370.4 Nm=82.6 Nm$

And since the counterclockwise momentum is greater than the clockwise one, the teeter-totter will rotate counterclockwise.

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3 years ago

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You're reading an article about forest fires in california and it features a photograph of firefighters battling a fire on a hil

Lelechka [254]

I will say it is about 20 m/s

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3 years ago

Helium gas is in a cylinder that has rigid walls. If the pressure of the gas is 2.00 atm, then the root-mean-square speed of the

aivan3 [116]

Answer:

$p=p_{2}-p_{1}=2.9188atm$

Explanation:

First step

To calculate by how much the pressure has been increased we must find a relation between Vrms and the pressure p

$V_{rms}=\sqrt{\frac{3RT}{M} } \\$

Where M is molar mass,R is gas constant and T is temperature

As R and M is constant so we obtain

Vrams∝√T

Second step

From Ideal gas law we know that

$pV=nRT\\$

we will find that

p∝T

So from first and second step, we can obtain that

Vrms∝√p

And a relation between both of them could be given by:

$\frac{V_{rms1} }{V_{rms2}}=\frac{\sqrt{p_{1} } }{\sqrt{p_{2}} }\\ p_{2}=[\frac{(V_{rms1})^{2} }{(V_{rms2})^{2} }]p_{1}\\ p_{2}=\frac{(276m/s)^{2} }{(176m/s)^{2} } (2atm)\\p_{2}=4.9188atm$

The pressure is increased to 4.1988 atm, so the amount will be given by:

$p=p_{2}-p_{1}=4.9188atm-2atm\\p=p_{2}-p_{1}=2.9188atm$

7 0

3 years ago

Escribe verdadero o falso :

Oduvanchick [21]

By defining kinematics and derivative relations we can find which statements are true or false:

a) False. The derivative is the instantaneous velocity.

b) True. The second drift is the instantaneous acceleration.

c) False the derivative is

d) False the derivative is

a) The velocity is defined with the variation of the position with respect to time.

$v= \frac{dx}{dt}$

Wher x is the position and t the time.

In the change of the average velocity is the average value of the velocity in an interval

$v = \frac{v_f - v_o}{t}$

We can see that the derivative is the speed in a very small timet, that is, the speed instantaneous. Therefore the statement is False.

The prime derivative of the position is the instantaneous velocity, not the average velocity.

b) Acceleration is defined as the change in position with respect to time..

$a = \frac{dv}{dt}$

Let's use the chain rule.

a = $\frac{d}{dt} \frac{dx}{dt}$

a = $\frac{d^2 x}{dt^2}$

Therefore the second derivative of the position is the instantaneous acceleration.

The statement is True.

Questions c and d ask the derivative of a function

f (x) = 2 x aⁿ

c) derivative with respect of x

$\frac{df}{dx} = 2 a^n$

The answer is False.

d) derivative with respect to a.

$\frac{df}{da} = 2n \ x a^{n-1}$

Answer d is false

In conclusion using the definition of kinematics and derivative relations we can find which statements are true.

a) False. The derivative is the instantaneous velocity.

b) True. The second drift is the instantaneous acceleration.

c) False the derivative is: $\frac{df}{dx} = 2 a^n$

d) False the derivative is: $\frac{df}{da} = 2n \ x a^{n-1}$

Learn more about the relationship between derivatives and kinematics here: brainly.com/question/15344251

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3 years ago