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2 years ago

2. Explain how wave speed is independent of the source

Physics

1 answer:

hram777 [196]2 years ago

3 0

Answer:

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Explanation:

cgtct tty nguhy u n I hope help this haha

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The 94-lb force P is applied to the 220-lb crate, which is stationary before the force is applied. Determine the magnitude and d

Makovka662 [10]

Answer

given,

force = 94 lb

weight of crate = 220 lb

Assuming the static friction be equal = 0.47

kinetic friction = 0.36

Maximum force applied to move the object is when object is just start to move.

F = μ N

F = 0.47 x 220

F = 103.4 lb

As the frictional force is more than applied then the object will not move.

so, the friction force will be equal to the force applied on the object that is equal to 94 lb.

hence, the direction of force will left.

8 0

2 years ago

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umka21 [38]

You would have to put the number together

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2 years ago

A scientist designed a foam container to help keep frozen foods from melting. Which best explains how the foam works? It removes

Allushta [10]

Answer:

the foam an insulator............

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2 years ago

What could be the possible answer to the question ?<br><br>thankyou ~

Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$