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ss7ja [257]
3 years ago
5

How many wavelengths of a wave pass a point if the frequency of the wave is 4 hertz?

Physics
1 answer:
stepan [7]3 years ago
3 0
Four of them pass a point every second.
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An ideal gas is enclosed in a cylinder that has a movable piston on top. The piston has a mass m and an area A and is free to sl
elena-14-01-66 [18.8K]

Answer:

Explanation:

Given mass of piston \left ( m\right )

no. of moles =n

Given Pressure remains same

Temperature changes from T_1 to T_2

Work done\left ( W\right ) is given by=\int_{V_1}^{V_2}PdV

W=P\left ( V_2-V_1\right )

also PV_1=nRT_1

PV_2=nRT_2

W=nR\left ( T_2-T_1\right )

4 0
3 years ago
A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,
egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

6 0
3 years ago
Include units with all of your answers. A 1.84 kg bucket full of water is attached to a 0.76 m long string.
aleksandrvk [35]
<span>c. What is the magnitude of the tension in the string at the bottom of the circle if you are swinging it at 3.37 m/s? </span>
7 0
3 years ago
A Net Force of 9.0 N acts through a distance of 3.0 m in a time of 3.0 s. The work done is?
JulijaS [17]
Work done is the distance a force acts over.

So, the work done here is 9.0N * 3.0m = 27 J
5 0
3 years ago
Read 2 more answers
A closed circuit is formed by a solid conductor in contact with two parallel conducting rails that are closed on their left end,
castortr0y [4]

this is an equation that you need to solve for motional emf. motional emf=vBL, where v is velocity in meters/second, B is magnetic field in Teslas and L is length or distance the rails are apart from each other. when we plug everything into the formula given above, we get: motional emf=5m/s*0.80T*0.20m. solving all this we get 0.8 volts. pretty sure that since they are giving you the direction of the field, they want to know which way the current will flow . since the conductor is moving from left to right the area of the field is increasing which means magnetic flux is increasing as Ф(magnetic flux)=B(magnetic field)*A(area)*cosФ(little phi is the angle to the normal. in this case little fee is 0 degrees so the cosФ doesn't matter). so ↑Ф=B↑A. if magnetic flux is increasing, the induced magnetic field is in the opposite direction as the original magnetic field meaning the induced magnetic field will be out of the page. using the right hand rule which says that if the field is in to the page, the current should go clockwise and if the field is out of the page, the current is counterclockwise so that means that the current should be going counter clockwise since the induced field is going out of the screen. the top of the conducting wire will have its current go to the left and the bottom of the conducting wire will have the current go to the right.

8 0
2 years ago
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