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nydimaria [60]
3 years ago
13

A raised plaster cover (often called a plaster ring or mud ring) is permitted to increase the maximum number of conductors permi

tted in an outlet box when it is _____.
Physics
1 answer:
andreev551 [17]3 years ago
6 0

Answer:

When it is marked with its cubic-inch volume

Explanation:

Because this allows for best and efficient identification

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If fire needs oxygen to burn, where does the sun get oxygen if there is no oxygen in space?​
elena55 [62]

Answer:

through nuclear fusion

Explanation:

The burning of the sun is not chemical combustion. It is nuclear fusion. This combustion releases energy which we experience as the heat and light given off by the flame.

5 0
2 years ago
Read 2 more answers
Suppose you give a 10 Newton push to Ryan on skis (he weighs 50 kg), how much will he accelerate?
Talja [164]

Well we can just use F=ma. The force is 10N, the mass is 50 kg, solve for a. Well since we kg and N, no conversion is necessary. So just plugging in the numbers, we get

10N = 50 kg · a

\frac{10N}{50kg}=a

A newton is just \frac{kg·m}{s^{2}}

a=\frac{\frac{10kg·m}{s^{2}}}{50kg}

The s^2 and 50 kg you multiply

a=\frac{10kg·m}{50kg·s^{2}}

The kg's cancel and 10/50 is 1/5

\frac{1}{5}·\frac{m}{s^{2}}

So the acceleration is 1/5 m/s^2


3 0
2 years ago
A rifle fires a 2.01 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the t
Zina [86]

Answer:

The value of spring constant is 266.01 \frac{N}{m}

Explanation:

Given:

Mass of pellet m = 2.01 \times 10^{-2} kg

Height difference of pellet rise h_{f} - h_{o} = 6.03 m

Spring compression x = 9.45 \times 10^{-2} m

From energy conservation law,

Spring potential energy is stored into potential energy,

  mg(h_{f} -h_{o})  = \frac{1}{2} kx^{2}

Where k = spring constant, g = 9.8 \frac{m}{s^{2} }

  k = \frac{2mg(h_{f} -h_{o} )}{x^{2} }

  k = \frac{2 \times 9.8 \times 6.03\times 2.01 \times 10^{-2} }{(9.45\times 10^{-2} )^{2} }

  k = 266.01 \frac{N}{m}

Therefore, the value of spring constant is 266.01 \frac{N}{m}

6 0
3 years ago
What happens to the density of a fluid as the substance gains thernal energy
Natalka [10]

Density reduces

Explanation:

The density of a fluid will reduce as a substance gains more thermal energy. Heat is a form of thermal energy usually noticed as temperature differences in a matter.

  • When thermal energy is gained by a body, the temperature of the body rises.
  • The particles of the medium gains more kinetic energy.
  • This causes the substance to expand and it has more volume.
  • When a substance expands, it occupies more space.
  • The more the volume a body occupies, the lesser the density.
  • Expansion does not change the mass of the fluid but the volume changes.
  • This is why density decreases.

Learn more:

Density brainly.com/question/8441651

#learnwithBrainly

8 0
3 years ago
A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.
34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated (Q), measured in joules, by the cylindrical resistor is the power multiplied by operation time (\Delta t), measured in hours. That is:

Q = \dot Q \cdot \Delta t (1)

If we know that \dot Q = 1.2\,W and \Delta t = 86400\,s, then the amount of heat dissipated by the resistor is:

Q = (1.2\,W)\cdot (86400\,s)

Q = 103680\,J

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux (Q'), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder (A), measured in square meters:

Q' = \frac{\dot Q}{A} (2)

Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h } (3)

Where:

D - Diameter, measured in meters.

h - Length, measured in meters.

If we know that \dot Q = 1.2\,W, D = 4\times 10^{-3}\,m and h = 2\times 10^{-2}\,m, the heat flux of the resistor is:

Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }

Q' \approx 4340.589\,\frac{W}{m^{2}}

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces (r), no unit, is the ratio of the top and bottom surfaces to total surface:

r = \frac{\frac{\pi}{2}\cdot D^{2}}{A} (3)

If we know that A \approx 2.765\times 10^{-4}\,m^{2} and D = 4\times 10^{-3}\,m, then the fraction is:

r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}

r = 0.045

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

7 0
3 years ago
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