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Igoryamba
3 years ago
11

A box slides down a frictionless plane inclined at an angle θ above the horizontal. The gravitational force on the box is direct

ed
A) parallel to the plane in the same direction as the movement of the box.
B) parallel to the plane in the opposite direction as the movement of the box.
C) perpendicular to the plane.
D) vertically.
E) at an angle θ below the inclined plane.

Physics
1 answer:
tamaranim1 [39]3 years ago
4 0

Answer:

D) Vertically.

Explanation:

A free body diagram is used to represent all the forces acting in a body. forces like, the force of gravity as a result of the gravitational interaction between the object and the Earth (W), the frictional force opposite to the movement of the object (F_{r}), the normal force due to the plane and the object (N) and the force applied to start the movement in a particular direction (F).

As is show in the free body diagram of the system, W, which is the weight of the body as a consequence of the gravitational force, is at an angle \theta below the inclined plane. that angle between the plane and the x axis is the same that the one of the inclined plane with respect to the horizontal, Since its sides are perpendicular.  

Notice how W goes always in the direction to the center of mass of Earth in a vertical path (For comparison see figure (a) and (b)).  

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6 0
3 years ago
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The period of a sound wave coming from an instrument is 0. 002 seconds. What is the frequency of the sound? Hz.
Pie

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<h2>Period:</h2>

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Where,

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brainly.com/question/842349

5 0
2 years ago
A block with mass m = 0.450 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth
svetoff [14.1K]

Answer:

k = 26.25 N/m

Explanation:

given,

mass of the block= 0.450

distance of the block = + 0.240

acceleration = a_x = -14.0 m/s²

velocity = v_x = + 4 m/s

spring force constant (k) = ?

we know,

x = A cos (ωt - ∅).....(1)

v = - ω A cos (ωt - ∅)....(2)

a = ω²A cos (ωt - ∅).........(3)

\omega = \sqrt{\dfrac{k}{m}}

now from equation (3)

a_x = \dfrac{k}{m}x

k = \dfrac{m a_x}{x}

k = \dfrac{0.45 \times (-14)}{0.24}

k = 26.25 N/m

hence, spring force constant is equal to k = 26.25 N/m

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3 years ago
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