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katovenus [111]
3 years ago
9

A string exerts a force of 20 N on a box at an angle of 38° from the horizontal. What is the horizontal component of the force o

n the box?
Physics
2 answers:
dem82 [27]3 years ago
7 0

Answer:

  15.8 N

Explanation:

The component in a given direction is the magnitude of the applied force, multiplied by the cosine of the angle between its application and the direction of interest.

The horizontal component is ...

  (20 N)cos(38°) ≈ 15.8 N

choli [55]3 years ago
4 0
<h3>questions:</h3>

A string exerts a force of 20 N on a box at an angle of 38° from the horizontal. What is the horizontal component of the force on the box?

<h3>answer;</h3>

15.8 N

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In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance
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Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = X_{L} = j\omega L = 2\pi fL

where

R = resistance

X_{L} = Inductive Reactance

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

C = \frac{\epsilon_{o}A}{x}

where

x = separation between the parallel plates

Thus

C ∝ \frac{1}{x}

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}

Also,

Z ∝ I

Therefore,

\frac{Z}{I} = \frac{Z'}{I'}

\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}

{R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})

{R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})

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6 0
3 years ago
During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25
Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

dQ=0

dU=-dW

We need to calculate the number of moles and specific heat

Using formula of heat

dU=nC_{v}dT

nC_{v}=\dfrac{dU}{dT}

Put the value into the formula

nC_{v}=\dfrac{-100}{5}

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We need to calculate the heat

Using formula of heat

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Put the value into the formula

dQ=-20\times5+25

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We need to calculate the heat capacity for the second process

Using formula of heat

dQ=nC_{v}(dT_{1})

Put the value into the formula

-75=nC_{v}\times(-5)

nC_{v}=\dfrac{-75}{-5}

nC_{v}=15\ J/K

Hence, The heat capacity for the second process is 15 J/K.

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3 years ago
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