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kondor19780726 [428]
2 years ago
11

How does the magnitude of the normal force exerted by the ramp in the figure compare to the weight of the static block? The norm

al force is:______ a. greater than the weight of the block. b. possibly greater than or less than the weight of the block, depending on whether or not the ramp surface is smooth. c. equal to the weight of the block. d. possibly greater than or equal to the weight of the block, depending on whether or not the ramp surface is smooth. less than the weight of the block.
Physics
1 answer:
DIA [1.3K]2 years ago
8 0

Answer:

less than the weight of the block.

Explanation:

From the free body diagram, we get.

The normal force is N = Mg cosθ

The tension in the string is T = Mg sinθ

Wight of the block when the block is static, W = Mg

Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,

therefore, the normal force is less than the weight of the static block.

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46 points :)
IgorC [24]

Answer:

Its is dividing by 2

Explanation:

It starts with 100 them it goes to 50, 25, 12.5 so its a cycle of dividing by 2

3 0
3 years ago
Read 2 more answers
the motion of a particle along a straight line is represented by the position versus time graph above. at which of the labeled p
atroni [7]

Point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.

On the graph, A is the point where magnitude of the acceleration of the particle is greatest as compared to other positions on the graph because the height of point A is the largest as compared to other points of the graph.

The graph shows at which point acceleration of an object is higher and lower so we can conclude that point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.

Learn more about acceleration here: brainly.com/question/933224

Learn more: brainly.com/question/25887663

3 0
2 years ago
Two mechanical waves that have positive displacements from the equilibrium position meet and coincide. What kind of
erica [24]

When two mechanical waves that have positive displacements from the equilibrium position meet and coincide, a constructive interference occurs.

Option A

<h3><u>Explanation:</u></h3>

Considering the principle of superposition of waves; the resultant amplitude of an output wave due to interference of two or more waves at any point is given by individual addition of their amplitudes at that point. Two waves with positive displacements refer to the fact that crest of the both the waves are on the same side of displacement axis, either both are positive or both are negative, similarly with their troughs.

If such two waves with their crest on crest meet at any point, by superposition principle. their individual amplitude gets added up and hence the resultant wave after interference is greater in amplitude that both the individual waves. This is termed as a constructive interference. Destructive interference on the other hand is a condition when one of the two waves has a positive displacement and other has a negative displacement (a condition of one’s crest on other’s trough); resulting in amplitude subtraction.

6 0
3 years ago
A satellite is put into an orbit at a distance from the center of the Earth equal to twice the distance from the center of the E
Sveta_85 [38]

Answer:

995 N

Explanation:

Weight of surface, w= 4000N

Gravitational constant, g, is taken as 9.81 hence mass, m of surface is W/g where W is weight of surface

m= 4000/9.81= 407.7472

Using radius of orbit of 6371km

The force of gravity of satellite in its orbit, F=\frac {GMm}{(2r)^{2}}=\frac {GMm}{4(r)^{2}}

Where G=6.67*10^{-11} and M=5.94*10^{24}

F=\frac {(6.67*10^{-11}*5.94*10^{24}*407.7472)}{4*({6.371*10^{6}m)}^{2}}

F= 995.01142 then rounded off

F=995N

6 0
2 years ago
A race car moving along a circular track has a centripetal acceleration of 15.4 m/s? If the car has
Helen [10]

Answer:

r = 58.44 [m]

Explanation:

To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.

a = v²/r

where:

a = centripetal acceleration = 15.4 [m/s²]

v = tangential speed = 30 [m/s]

r = radius or distance [m]

r = v²/a

r = 30²/15.4

r = 58.44 [m]

3 0
2 years ago
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