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marysya [2.9K]
2 years ago
10

Determine the scalar components Ra and Rb of the force R along the nonrectangular axes a and b. Also determine the orthogonal pr

ojection Pa of R onto axi

Physics
1 answer:
Liula [17]2 years ago
6 0

Answer: Hello your question is incomplete attached below is the complete question

Ra = 1132 N

Rb = 522.6 N

Pa = 679.7 N

Explanation:

To determine the scalar components Ra

\frac{Ra}{Sin 120^o} = \frac{750}{sin 35^o}  

therefore : Ra = \frac{sin120^o * 750}{sin 35^o} = 1132 N

To determine the scalar component Rb

\frac{Rb}{sin 25^o} = \frac{750}{sin 35^o}

therefore : Rb = \frac{sin 25^o * 750}{sin 35^o}  = 522.6 N

To determine the orthogonal projection Pa of R onto

Pa = 750 cos25^o = 679.7 N

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fenix001 [56]

Answer:

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Explanation:

Given;

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To determine the peak current carried by the axon, apply the following equation;

B = \frac{\mu I}{2\pi r}

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Re-arrange the equation and solve for "I"

B = \frac{\mu I}{2\pi r} \\\\I = \frac{B*2\pi r}{\mu} \\\\I = \frac{9*10^{-12}*2*\pi *1.3*10^{-3}}{4\pi *10^{-7}} \\\\I = 5.85 *10^{-8} \ A

Therefore, the peak current carried by the axon is 5.85 x 10⁻⁸ A

7 0
2 years ago
The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati
Vedmedyk [2.9K]

Answer:

The  tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, m=33.2\ kg

Coefficient of static friction is, \mu = 0.214

Angle of inclination is, \theta = 31.5°

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are mg\cos \theta and normal N. Now, there is no motion in the direction perpendicular to the inclined plane. So,

N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N

Consider the direction along the inclined plane.

The forces acting along the plane are mg\sin \theta and frictional force, f, down the plane and tension, T, up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N

Therefore, the  tension in the rope is 229.37 N.

3 0
1 year ago
A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the fol
qaws [65]

Answer:

Part a)

v = \sqrt{v_o^2 + g^2t^2}

Part b)

d = \frac{1}{2}gt^2

Part c)

v_f = v_o - gt

Part d)

d = \frac{1}{2}gt^2

Explanation:

Part a)

As we know that speed of package is same as that of helicopter in horizontal direction

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v_y = -gt

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Part b)

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d = \frac{1}{2}gt^2

Part c)

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v_f = v_o - gt

Part d)

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8 0
2 years ago
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Answer: 2R

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t=5s

it was correct on my do-now

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