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daser333 [38]
3 years ago
14

a 2.0 kg block on an incline at a 60.0 degree angle is held in equilibrium by a horizontal force, what is the magnitude of this

horizontal force (disregard friction)?
Physics
2 answers:
Ymorist [56]3 years ago
8 0

Answer:

Horizontal force is 16.97 N.

Explanation:

It is given that,

Mass of the block, m = 2 kg

It is on an incline at a 60 degree angle is held in equilibrium by a horizontal force. We need to find the magnitude of horizontal force. The force acting on the block is its weight mg.

The horizontal component of force is, F_x=mg\ sin\theta

The vertical component of force is, F_y=mg\ cos\theta  

So, the horizontal component of force is,F_x=2\ kg\times 9.8\ m/s^2\ sin(60)

F_x=16.97\ N

So, the horizontal component of force is 16.97 N.

fiasKO [112]3 years ago
3 0
<span>The magnitude of this horizontal force</span> can be calculated as :

F=mg
 2x9.8=19.6N
<span>19.6cos 30= 17 Newtons
</span>
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Friction, normal force, and weight

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2 years ago
S Five particles with equal negative charges -q are placed symmetrically around a circle of radius R. Calculate the electric pot
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6 0
1 year ago
A spherical gas-storage tank with an inside diameter of 9 m is being constructed to store gas under an internal pressure of 1.50
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Answer: 33 mm

Explanation:

Given

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Internal pressure of gas, P(i) = 1.5 MPa

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Allowable stress = 340 * 0.3 = 102 MPa

σ = pr / 2t, where

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3 years ago
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