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3 years ago

A bar of metal had a volume of 4.5 cm^3 and a mass of 33 g. what is the density of the metal?

Physics

1 answer:

Schach [20]3 years ago

7 0

Density is defined as the ratio of mass and volume of an object

$Density = \frac{mass}{volume}$

mass = 33 g

$Volume = 4.5 cm^3$

Now we can use the formula of density

$Density = \frac{33}{4.5}$

$Density = 7.33 \frac{g}{cm^3}$

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3 years ago

What is an observation in an experiment ?

Elza [17]

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3 years ago

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 8.0 s. At thi

alisha [4.7K]

Answer:

The total number of revolution is 50 rev.

Explanation:

Given that,

Angular speed = 5.0 rev/s

Time = 8.0 s

We need to calculate the angular acceleration

Using equation of angular motion

$\omega_{f}-\omega_{i}=\alpha t$

Put the value into the formula

$5.0-0=\alpha\times8.0$

$\alpha=\dfrac{5.0}{8.0}$

$\alpha=0.625\ rev/s^2$

We need to calculate the angular displacement

Using equation of angular motion

$\theta=\omega_{i}t+\dfrac{1}{2}\alpha t^2$

Put the value into the formula

$\theta=0+\dfrac{1}{2}\times0.625\times(8.0)^2$

$\theta=20\ rev$

Now, The washer coming to rest from top spin

We need to calculate the angular acceleration

Using equation of angular motion

$\omega_{f}-\omega_{i}=\alpha t$

$\alpha=\dfrac{\omega_{f}-\omega_{i}}{t}$

$\alpha=\dfrac{0-5}{12}$

$\alpha=−0.4167\ rev/s^2$

We need to calculate the angular displacement

Using formula of displacement

$\theta'=\omega_{i}t+\dfrac{1}{2}\alpha t^2$

Put the value into the formula

$\theta'=5\times12+\dfrac{1}{2}\times(-0.4167)\times12^2$

$\theta'=30\ rev$

We need to calculate the total number of revolution

$\theta''=\theta+\theta'$

$\theta''=20+30$

$\theta''=50\ rev$

Hence, The total number of revolution is 50 rev.

5 0

3 years ago

I have a question regarding friction in rolling without slipping.

Solnce55 [7]

Explanation:

They probably put "rolls without slipping" in there to indicate that there is no loss in friction; or that the friction is constant throughout the movement of the disk. So it's more of a contingency part of the explanation of the problem.

(Remember how earlier on in Physics lessons, we see "ignore friction" written into problems; it just removes the "What about [ ]?" question for anyone who might ask.)

In this case, you can't ignore friction because the disk wouldn't roll without it.

As far as friction producing a torque... I would say that friction is a result of the torque in this case. And because the point of contact is, presumably, the ground, the friction is tangential to the disk. Meaning the friction is linear and has no angular component.

(You could probably argue that by Newton's 3rd Law there should be some opposing torque, but I think that's outside of the scope of this problem.)

Hopefully this helps clear up the misunderstanding for you.

4 0

3 years ago

A child and sled with a combined mass of 48.8 kg slide down a frictionless hill that is 7.05 m high. If the sled starts from res

nexus9112 [7]

Answer:

Speed at bottom of the hill (v) = 11.74 m/s

Explanation:

Given:

Combined mass = 48.8 kg

Height h = 7.05 m

Find:

Speed at bottom of the hill (v)

Computation:

v² = 2gh

v = √2 x 9.8 x 7.05

v = √138.18

v = 11.74 m/s

Speed at bottom of the hill (v) = 11.74 m/s

7 0

3 years ago