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3 years ago

Two steel blocks rest on a frictionless, horizontal surface. Block X has a mass of

Physics

1 answer:

stepan [7]3 years ago

3 0

Before you jump into this, you have to stand back, look at it, and decide what's going to happen.

You're pulling on some things that rest on a frictionless surface, so they have to start moving. In fact, they're going to accelerate.

Now, I want you to consider for just a second: What would happen if
the two blocks were glued together ? Then, you'd be pulling on a single
18-kg block, with 36N of force, and it would accelerate to the right at
[ A = F/M ] = 36/18 = 2 m/s² .

The actual situation is no different. The string between them has no weight,
and it's tight, and there are no other forces on the blocks, so the whole combination ... both blocks ... is still going to accelerate to the right at 2 m/s² .

OK. Now, let's look at the individual blocks:

The one in the back ... the 12 kg. It's accelerating at 2 m/s² to the right,
so the force on it must be [ F = M A ] = 12 x 2 = 24 N to the right. Where
does that force come from ? It's the tension in the string between them.

The one in the front ... the 6-kg.
It has 36 N pulling it to the right, and 24 N of string tension pulling it to the left.
The net force on this block is (26 - 24) = 12N to the right.
Its acceleration is [ A = F/M ] = 12/6 = 2m/s² to the right.

Well, whaddaya know ! Both blocks accelerate at the same rate, in the
same direction, just as if they were glued together, and the string between
them remains tight, with 24N of tension in it.

With this description, you can handle the FBD on your own, I'm sure.

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A cell phone weighing 80 grams is flying through the air at 15 m/s what is the kinetic energy

hjlf

Answer:

$9\:\mathrm{J}$

Explanation:

Kinetic energy is given by the following equation:

$KE=\frac{1}{2}mv^2$ , where $m$ is mass in $\mathrm{kg}$ and $v$ is velocity in $\mathrm{m/s}$ .

Since the cell phone's mass is given in grams, we need to convert this into kilograms:

$80\:\mathrm{g}\cdot \frac{1\:\mathrm{kg}}{1000\:\mathrm{g}}=0.08\:\mathrm{kg}$ .

Therefore, the kinetic energy of the cell phone is:

$KE=\frac{1}{2}\cdot 0.08\cdot 15^2=\fbox{$9\:\mathrm{J}$}$ .

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3 years ago

Which of the following is prohibited by the 2nd law of thermodynamics?

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(A) A device that converts heat into work with 100% efficiency

It clearly violates the second law of thermodynamics because it warns that while all work can be turned into heat, not all heat can be turned into work. Therefore, despite the innumerable efforts, the efficiencies of the bodies have only been able to reach 60% at present.

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3 years ago

The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan

ss7ja [257]

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

$I = \frac{P}{A}$

The area of a sphere is given by

$A = 4\pi r^2$

So replacing we have to

$I = \frac{P}{4\pi r^2}$

Since the question tells us to find the proportion when

$r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}$

So considering the two intensities we have to

$I_1 = \frac{P_1}{4\pi r_1^2}$

$I_2 = \frac{P_2}{4\pi r_2^2}$

The ratio between the two intensities would be

$\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}$

The power does not change therefore it remains constant, which allows summarizing the expression to

$\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2$

Re-arrange to find $I_2$

$I_2 = I_1 (\frac{r_1}{r_2})^2$

$I_2 = 100*(\frac{1}{5})^2$

$I_2 = 4W/m^2$

Therefore the intensity at five times this distance from the source is $4W/m^2$

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