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2 years ago

A student lifts a physical science book off the table and above their head. Is there work being done?

1 answer:

7 0

Answer: A force must cause a displacement in order for work to be done. A book falls off a table and free falls to the ground. Yes. This is an example of work.

Explanation:

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Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and ex

Ahat [919]

The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

What is the frequency of the sound?

Solution :

Given :

The distance between the two loud speakers, $d = 1.8 \ m$

The speaker are in phase and so the path difference is zero constructive interference occurs.

At the point $D$ , the speakers are out of phase and so the path difference is $=\frac{\lambda}{2}$

Therefore,

$$AD-BD = \frac{\lambda}{2}$

$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$

$\lambda = 2 \times 0.4985$

$\lambda = 0.99714 \ m$

Thus the frequency is :

$f=\frac{v}{\lambda}$

$f=\frac{340}{0.99714}$

$f=340.9744$ Hz

3 0

3 years ago

a length of wire is cut into five equal pieces. the five pieces are then connected in parallel, with the resulting resistance be

Natali [406]

The equivalent resistance of n resistors connected in parallel is given by
$\frac{1}{R_{eq}} = \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n}$ (1)

In our problem, the resulting resistance of the 5 pieces connected in parallel is $R_{Eq}=2.00 \Omega$ , and since the 5 pieces are identical, their resistance R is identical, so we can rewrite (1) as
$\frac{1}{R_{Eq} }= \frac{1}{2 \Omega}= \frac{1}{R}+ \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{5}{R}$
From which we find $R= 5 \cdot 2 \Omega = 10 \Omega$ .

So, each piece of wire has a resistance of $10 \Omega$ . Before the wire was cut, the five pieces were connected as they were in series. The equivalent resistance of a series of n resistors is given by
$R_{Eq}=R_1 + R_2 + ...+R_n$
So if we apply it at our case, we have
$R_{eq}=R+R+R+R+R=5 R= 5\cdot 10 \Omega= 50 \Omega$

therefore, the resistance of the original wire was $50 \Omega$ .

5 0

3 years ago

Due to the tilt of the Earth's axis,

Lynna [10]

6 months with no sunrise and the other 6 without a sunset, at some places on Earth, are the result of the orientation of Earth's axis.

3 0

3 years ago

Help please, last ride guys

Ierofanga [76]

Answer:

Solution ( for fourth attachment ) : 38°C

Tip : Remember the units °C when submitting answer

Explanation:

As you mentioned, we only need the solution for the fourth attachment.

The idea here is that the heat lost by the metal will be equal to the heat gained by the water. We know that the specific heat gained or lost will always be represented by the following formula,

q = m $*$ c

Therefore if we substitute the know values and equate the two equations knowing that " q " is common among them --- ( 1 )

0.33 $*$ 448

Remember that the change in temperature of iron (ΔT) would be represented by final temperature - initial temperature, or final temperature - 693. Similarly the change in temperature of water will be final temperature - 39. Now we can pose the final temperature as a, and solve for a through substitution --- ( 2 )

0.33 $*$ 448

From here on take a look at the attachment. It represents how to receive get a through simple algebra. Here a, the final temperature, is about 38°C. In exact terms it will be $38.03617\dots$ °C.