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faltersainse [42]
3 years ago
14

On a cold day, a heat pump absorbs heat from the outside air at 14°F (−10°C) and transfers it into a home at a temperature of 86

°F (30°C). Determine the maximum κ of the heat pump.
Physics
1 answer:
Leto [7]3 years ago
7 0

Answer:

6.575

Explanation:

T1 = 30C = 30 + 273 = 303 K

T2 = - 10 C = - 10 + 273 = 263 K

The coefficient of performance of heat pump

k = T2 / (T1 - T2)

k = 263 / (303 - 263) = 6.575

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If you drag a 50kg block across the floor which has a coefficient of friction of .30, what is
torisob [31]

Answer:

750 Newton

Explanation:

force=mass(acceleration)

6 0
2 years ago
In one hour, coal supplies 500 000 J of energy. The energy amounts to 380 000 J. How much useful energy is produced in one hour?
Debora [2.8K]

Answer:

120,000J

Corrected question;

In one hour, coal supplies 500 000 J of energy. The wasted energy amounts to 380 000 J. How much useful energy is produced in one hour?

Explanation:

Given;

Total energy Et = 500,000 J

Wasted Energy Ew = 380,000J

The amount useful energy is the amount of energy that is available for supply.

This can be derived by subtracting the wasted energy from the total energy.

Useful energy = Total Energy - wasted energy

Eu = Et - Ew

Substituting the given values;

Eu = 500,000J - 380,000

Eu = 120,000 J

The amount of useful energy produced in one hour is 120,000 J

5 0
3 years ago
17,874,000 what is the value of 1
kotegsom [21]
Salutations!

17,874,000 what is the value of 1?

The value of 1 is 10 million. The place value would be 10,000,000.

Hope I helped.
3 0
3 years ago
Read 2 more answers
A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr
VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

rR:\\\\E=k\frac{Q}{r^2}

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}

with this values of a you can use the following formula:

a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0
2 years ago
A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1
tamaranim1 [39]
The acceleration would be 6m/sThis is because of the formula, "f/m=a" to find the acceleration; We would need to subtract the force of the friction which equals 1380, then divide that by the mass (which was 230) to get the answer 6
7 0
3 years ago
Read 2 more answers
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