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ikadub [295]
3 years ago
11

A 50-newton box is placed on an inclined plane that makes a 30° angle with the horizontal. Calculate the component of gravitatio

nal force that acts down the inclined plane. A. 25 N B. 37 N C. 43 N D. 56 N

Physics
2 answers:
dedylja [7]3 years ago
3 0

Weight of the box = 50 Newtons

In the diagram attached,

The gravitational force is broken into two components.

The component of the gravitational force that acts along the inclined plane is 50 SinФ.

Ф = 30°

Sin 30° = 0.5

Gravitational force down the inclined plane = 50 Sin 30°

= 50 × 0.5

Gravitational force down the inclined plane = 25 Newtons or 25 N

Hence, the option A is correct.

Orlov [11]3 years ago
3 0

The correct answer is A) 25 N i took the test and got it correct

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A proton and an electron are both accelerated to the same final speed. If λp s the de Broglie wavelength of the proton and e is
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The de broglie wavelength of electron is given by

\lambda _{e}=\frac{h}{m_{e}\times v} .... (2)

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5 0
3 years ago
The axle of an automobile is acted upon by the forces and couple shown. knowing that the diameter of the solid axle is 32 mm, de
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First we need to convert the mm to inches to make our computation easier.

1mm = 0.0393701

32mm * 0.0393701 = 1.25 in

 

Solution:

C = 1/2d = ½ (1.25) = 0.625 in^4

 

Tension: tension = Te/J = 2T/ piC^3

= (2)(2500)/pi (0.0625)^3 = 6.519 x 10^3 psi = 6.519 ksi

 

Bending:

I = pi/4 * c^4 = 119.842 x 10^-3 in^4

M = (5)(600) = 3600 lb in

G = My/I = (3600)(0.625)/119.842 x 10^-3 = -18.775 x 10^2 psi = -18.775ksi

 

Gx = -18.775 ksi

Gy = 0

Txy = 6.519 ksi

 

G ave – ½ (Gx + Gy) = -9.387 ksi
R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2 + (6.519)^2 = 11.429 ksi

 

1.       G1 = Gave + R = -9/387 + 11.429 = 2.04 ksi

G2 = Gave - R = -9/387 - 11.429 = -20.8

 

Tan 2ϴp = 2txy/Gx – Gy = 2(6.519)/-9.387 = -1.3889

ϴp = -27.1 degrees and 62.9 degrees

 

 

2.       Tmax = R = 11.43 ksi

R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2 + (6.519)^2 = 11.429 ksi

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3 0
3 years ago
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