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IRINA_888 [86]
2 years ago
13

The ammeter displays a reading of 0.10 A. Calculate the potential difference across the 45 Ω resistor.

Physics
1 answer:
NeX [460]2 years ago
5 0

Answer:

V = 45× 0.10= 4.5 volt

...........

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The length of the side of a cube having a density of 12.6 g/ml and a mass of 7.65 g is __________ cm.
qaws [65]

Density is the characteristic property of a substance.  It is the measure of mass of the substance divided by its volume (density= mass/volume). Manipulate the given formula to come up with the formula for the volume. Therefore, volume is equals to mass of a substance divided by its density (Vol= mass/density). Given 12.6 g/ml as density and 7.65 g mass, volume is equals to 0.60714 ml, since 1 ml = 1cm^3, volume is equals to 0.60714 cm^3 then extract the cube root of the volume to get the length of the cube in cm which is equal to 0.84677 cm.




4 0
2 years ago
A sound technician is testing the sound acoustics in a theatre for an upcoming music concert. As he moves towards the speakers,
Harlamova29_29 [7]

Answer: Increase in wave frequency

Explanation:

When we talk about acoustics we are dealing with sound waves, and one of their main components along with the velocity and wavelength is the <u>frequency.</u>

In this sense, the frequency of any wave refers to how fast (or slow) a wave oscillates. For example, in the especific case of sound waves when the oscillation is faster, the frequency is higher and the pitch gets higher as well.

6 0
2 years ago
Read 2 more answers
You are standing at the midpoint between two speakers, a distance D away from each. The speakers are playing the exact same soun
Rzqust [24]

Answer:

Explanation:

wave length of sound waves = velocity / frequency

= 340 / 170

λ = 2 m.

When the position of man is exactly at the meddle point between the speakers , sound waves from the speakers reaching man are in same phases ( path difference is zero. ) so intensity of sound is maximum .

Now , the man starts moving towards one of the speakers , his distance from one speaker becomes closer than the other creating path difference for the sound waves reaching his ears.

If he walks a distance of .5 m towards one speaker , path difference created

= .5 x 2 = 1 m

So , path difference = λ /2 ,

there will be destructive interference so minimum sound will be heard there.

When he walks a distance of 1 m , path difference created = 2m

path difference =  λ

so there is constructive interference and maximum  sound will be heard there.

Again we he walks a distance of 1.5 m , path difference created = 3 m

path difference = 3 λ /2

So there will be destructive interference so minimum sound will be heard there.

In this way we see that man starts  from a point of maximum sound intensity , reaches a point of minimum sound intensity , then reaches a point of maximum sound intensity . At last he reaches a position of minimum sound intensity.

3 0
2 years ago
Determine the torque applied to the shaft of a car that transmits 225 hp
Arisa [49]

Incomplete question.The complete question is here

Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.

Answer:

Torque=0.51 Btu

Explanation:

Given Data

Power=225 hp

Revolutions =3000 rpm

To find

T( torque )=?

Solution

As

T(Torque)=\frac{W(Work)}{2\pi n(Revolutions) }

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.

So

T=\frac{225*42.207}{2\pi 3000}\\ T=0.51 Btu

8 0
2 years ago
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating
Olegator [25]

Answer:

rod end A is strongly attracted towards the balls

rod end B is weakly repelled by the ball as it is at a greater distance

Explanation:

When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.

Therefore rod end A is strongly attracted towards the balls and

rod end B is weakly repelled by the ball as it is at a greater distance

3 0
2 years ago
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