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Nutka1998 [239]
2 years ago
5

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

able is 90°. If the beam is inclined at an angle of θ = 31.0° with respect to horizontal.
The horizontal component of the force exerted by the hi.nge on the beam is 8.662x10^1 N.
What is the magnitude of the force that the beam exerts on the hi.nge?
Physics
1 answer:
irinina [24]2 years ago
7 0

288.51 N is  the magnitude of the force that the beam exerts on the hi.nge.

Given

Mass 0f beam = 40 Kg

The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N

Angle between the beam and cable is = 90°

Angle between beam and the horizontal component = 31°

As the system of the beam, hi_nge and cable are in equilibrium.

The magnitude of the force that the beam exerts on the hi_nge can be calculated by -

F =The  horizontal component of force + the vertical component of force  

F = 86.62 N + 40 × 9.8 × sin 31°

F =86.62 N + 201.89 N

F = 288.51 N

Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51  N.

Learn more about components of forces here brainly.com/question/26446720

#SPJ1

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Answer:

option (B) decreases

Explanation:

According to the Wein's displacement law, the minimum wavelength of the radiated emission is inversely proportional to the absolute temperature of the body which emits radiation.

\lambda_{m}\alpha \frac{1}{T}

Where, T is the absolute temperature of the body and λm is the minimum wavelength of heat radiated.

Here, as the temperature increases, the wavelength decreases.

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Explanation:

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3 years ago
Please help with this!!!!!
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3. What is the gravitational force between a 70 kg physics student and her 1 kg textbook, at a distance of 1 meter? (This number
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ANSWER

\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}

EXPLANATION

Parameters given:

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To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

F=\frac{GMm}{r^2}

where G = gravitational constant

Therefore, the gravitational force acting between the student and the textbook is:

\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\  \\ F=4.67*10^{-9}\text{ }N \end{gathered}

That is the answer.

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