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SVEN [57.7K]
3 years ago
8

If forces acting on an object are unbalanced, which factor may result from an unbalanced force? The net force is negative. There

is zero net force. The forces cancel each other out. There is no change in motion.
Can you guys help me out, please??
Physics
2 answers:
NISA [10]3 years ago
8 0

Answer:

A.  The net force is negative

Explanation:

The net force is negative is the correct answer.  Just took the test on edge.

Oksi-84 [34.3K]3 years ago
6 0

Answer:the Forces cancel out each other

Explanation:the forces cancel out each other

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Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const
kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is  L=  20cm = 0.2m

Thermal conductivity of the wall is  K = 2.79 W/m·K

Temperature at the left side surface is T₁ =  50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is  h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

Q_{conduction} = Q_{convection}

Expression for the heat conduction process is

Q_{conduction} = \frac{K(T_1 - T)}{L}

Expression for the heat convection process is

Q_{convection} = h(T_2 - T)

Substitute the expressions of conduction and convection in equation above

Q_{conduction} = Q_{convection}

\frac{K(T_1 - T_2)}{L} = h(T_2 - T)

Substitute the values in above equation

\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC

Now heat flux through the wall can be calculated as

q_{flux} = Q_{conduction} \\\\q_{flux}  = \frac{K(T_1 - T_2)}{L}\\\\q_{flux}  = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0
3 years ago
If you put two identical cars on opposite sides of a large magnet, what happens
FrozenT [24]

Answer:

Depends on what pole it is.

Explanation:

If the poles of the cars and magnets are the same they will repel, if different, attracts.

7 0
3 years ago
Read 2 more answers
Infer why the output force exerted by a rake must be less than input force?
adell [148]
<h3><u>Answer and Explanation</u>;</h3>
  • input force refers to the force exerted on a machine, also known as the effort, while the output force is the force machines produce or the Load. The ratio of output force to input force gives the mechanical advantage of a simple machine
  • <em><u>The output force exerted by the rake must be less than the input force because one has to use force while raking. The force used to move the rake is the input force. </u></em>
  • <em><u>The rake is not going to be able to convert all of the input force into output force, the force the rake applies to move the leaves, because of friction.</u></em>
5 0
3 years ago
If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an RC discharge thr
Agata [3.3K]

Answer:

Resistance in the flash tube, R=3.97\times 10^{-3}\ \Omega

Explanation:

It is given that,

Speed of the bullet, v = 500 m/s

Distance between one RC constant, d = 1 mm = 0.001 m

Capacitance, C=503\ \mu F=503\times 10^{-6}\ F

The time constant of RC circuit is given by :

\tau=RC

R is the resistance in the flash tube

R=\dfrac{\tau}{C}..........(1)

Speed of the bullet is given by total distance divided by total time taken as :

v=\dfrac{d}{\tau}

\tau=\dfrac{d}{v}

\tau=\dfrac{0.001}{500}

\tau=0.000002\ s

Equation (1) becomes :

R=\dfrac{0.000002}{503\times 10^{-6}}

R=3.97\times 10^{-3}\ \Omega

So, the resistance in the flash tube is 3.97\times 10^{-3}\ \Omega. Hence, this is the required solution.

8 0
3 years ago
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In a given circuit, the supplied voltage is 1.5 volts. One resistor is 3 ohms and the other is 2 ohms. Use Ohm's law to determin
tino4ka555 [31]
I = E / R

If the resistors are in series, the current is 0.3 Amperes.

If the resistors are in parallel, the current is 1.25 Amperes.
4 0
4 years ago
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