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dolphi86 [110]
3 years ago
12

If the resistance doubles while the voltage stays the same, what effect will that have on the power?

Physics
1 answer:
zepelin [54]3 years ago
4 0
The best formula to use for this question is

             Power = (voltage)² / R

So      Original power = (voltage)² / (original R)
and
           New power  =  (same voltage)² / (2 x original R)

That's equal to            (1/2) (voltage)² / (original R)

which is equal to    (1/2) (original power) .

So I guess the power decreases by half, and that's choice-B .
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How will sunlight most likely affect a black shirt on a hot summer day? the temperature of the shirt will depend on how much sun
vladimir2022 [97]

The correct answer is option (C) the temperature of the shirt will increase because all wavelengths of light are absorbed by the shirt.

The relationship of heat and light

  • Heat is a measure of the movement of particles in the body, the more particles move, the warmer the body becomes.
  • When the body absorbs light radiation, its particles vibrate in accordance with the electromagnetic radiation's wavelengths, which causes an increase in the temperature with the increase in particle movement.
  • The more wavelengths of radiation absorbed by an object, produces more heat.

Learn more about the Wavelength of light with the help of the given link:

brainly.com/question/13961990

#SPJ4

5 0
1 year ago
A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the as
Taya2010 [7]

Answer:

w =  - 508.53 joules

q = - 3091.47 joules

Explanation:

Let us convert the time in hours into seconds

0.010* 3600\\= 36

Change in internal energy

\delta E = p * \delta t

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

\delta E = - 100 * 36\\

\delta E = - 3600 Joules

Amount of work done by the system

w = - P * \delta V

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

w = - 1 * ( 5.92 -0.90)\\

w = -5.02 liter-atmospheres

Work done in Joules

- 5.02 * 101.3\\= 508.53Joules

q = \delta E - w\\

Substituting the given values we get -

q = - 3600 - (-508.53)\\q = - 3091.47

Thus

w =  - 508.53 joules

q = - 3091.47 joules

7 0
3 years ago
A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/
Elena-2011 [213]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

KE_i + PE_i = KE_f + PE_f

\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0

\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2

So by solving above equation we will have

v_f = 84.3 m/s

now in order to find the momentum we can use

P = mv

P = 1.25 \times 84.3

P = 105.4 kg m/s

6 0
3 years ago
You want to place a mirror at a blind turn on the staircase in your house. Which would be best suited for this purpose?
Eduardwww [97]
You can't see beyond a blind turn, so a mirror would allow you to see around the corner.
6 0
3 years ago
A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1
balandron [24]

Answer:

The answer is "143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}"

Explanation:

For point a:

Energy balance equation:

\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\

W=0\\\\Q=0\\\\m_e=0

From the above equation:

\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\

because the rate of air entering the tank that is h_i constant.

\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\

Since the tank was initially empty and the inlet is constant hence, m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\

Interpolate the enthalpy between T = 300 \ K \ and\ T=295\ K. The surrounding air  

temperature:

T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

Substituting the value from ideal gas:

\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}

Follow the ideal gas table.

The u_2= 298.33\ \frac{kJ}{kg} and between temperature T =410 \ K \ and\  T=240\ K.

Interpolate

\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}

Substitute values from the table.

 \frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\

For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\

For point c:

 Entropy is given by the following formula:

\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}

5 0
2 years ago
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