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Tcecarenko [31]
3 years ago
13

When there are more waves passing through the reference point in a period of time, what wave characteristic also increase?

Physics
1 answer:
lesya [120]3 years ago
3 0
Frequency increases.
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ou are unloading a refrigerator from a delivery van. The ramp on the van is 5.0 m long, and its top end is 1.4 m above the groun
tigry1 [53]

Answer:

Explanation:

component of force along the ramp

= 370 cos θ where θ is the slope of the ramp with respect to ground.

sinθ = 1.4 / 5

θ = 16 degree

370 cos16

= 355.67 N

Work done

= component of force along the ramp x length of the ramp

= 355.67 x 5

= 1778.35 J

6 0
3 years ago
An object moves in a circle of radius R at constant speed with a period T . If you want to change only the period in order to cu
GarryVolchara [31]
The answer to that is three
5 0
3 years ago
Ok plz answer and tell me how to do it
kirza4 [7]
Answer: 25N

method: total force in the right hand direction is 100N and total force in the left hand direction is 125N. To get the net force, we add forces if they are in the same direction and substract if they are in opposite directions. since 100N and 125N are in opposite directions, we substract the larger value from the smaller value. Then we get 25N in the left hand direction as the final answer.
4 0
3 years ago
Drag the correct labels to the images. Each label can be used more than once.
coldgirl [10]

Answer:

plato answer.

Explanation:

4 0
3 years ago
#1 Not sure where to start. This is for AP Physics!
yaroslaw [1]

First,

\rho=\dfrac mV

where \rho is density, m is mass, and V is volume. We can compute the volume of the roll:

2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V

\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness x. Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

V=Ax

where A is the given area, so

4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x

\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}

If we're taking significant digits into account, the volume we found would have been V=470\,\mathrm m^3, in turn making the thickness x=250\,\mathrm{mm}.

8 0
3 years ago
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