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andrey2020 [161]

3 years ago

13

Balanced or Unbalanced?

1 answer:

Galina-37 [17]3 years ago

6 0

Unbalanced because if it is pushing then stopping, that means that it is unbalanced.

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What type of motion does the galaxy go through

Mariana [72]

Well, it depends. Your latitude on Earth--that is, how close you are to the equator--and the time of year make a difference. I'll explain why. Your motion is made up of four pieces: the rotation of the Earth on its axis, the motion of the Earth around the Sun, the Sun's orbit about the center of the galaxy, and the motion of the whole galaxy.

8 0

3 years ago

Bill is learning to play tennis. He does pretty well hitting the ball back to his opponent but, many times he misses the ball wh

KiRa [710]

Answer: Place his feet parallel to the baseline prior to tossing the ball

5 0

3 years ago

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

$A = \frac{\pi d^2}{4}$

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

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

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

$P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}J/s$

Power is defined as the quantity of Energy per second, then

$E = 6.1*10^{-9}J$

8 0

4 years ago

Calculate the average value of an AC signal with a peak amplitude of 10V and a frequency of 10 kHz.

Solnce55 [7]

Answer:

V(average)=6.37 V

Explanation:

Given Data

Peak Voltage=10V

Frequency=10 kHZ

To Find

Average Voltage

Solution

For this first we need to find Voltage peak to peak

So

Voltage (peak to peak)= 2× voltage peak

Voltage (peak to peak)= 2×10

Voltage (peak to peak)= 20 V

Now from Voltage (peak to peak) formula we can find the Average Voltage

So

Voltage (peak to peak)=π×V(average)

V(average)=Voltage (peak to peak)/π

V(average)=20/3.14

V(average)=6.37 V

3 0

3 years ago

For an extended object, the weight force can be considered to act at which point?

JulsSmile [24]

Answer:

At the center of the object

At the end of the object farthest away from the ground

At the center of gravity of the object

At end of the object closest to the ground

Explanation:

5 0

3 years ago