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4 years ago

14

A 2 block of ice is pushed along the ground with a constant 10 N force. Assuming there is no friction what is the acceleration o

f the block

1 answer:

Alex_Xolod [135]4 years ago

8 0

Acceleration of block = net force/mass = 10/2 = 5 m/s²

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A heat pump is used to maintain a house at 22°C by extracting heat from the outside air on a day when the outside air temperatur

Ilia_Sergeevich [38]

Answer:

It is enough

Explanation:

To develop the problem it is necessary to take into account the concepts related to the coefficient of performance of a pump.

The two ways in which the performance coefficient can be expressed are given by:

$COP_p = \frac{T_H}{T_H-T_L}$

Where,

$T_H =$ High Temperature

$T_L =$ Low Temperature

And the other way is,

$COP_p = \frac{\dot{Q}}{W}$

Where $\dot{Q}$ is heat rate and W the power consumed.

We have all our terms in Celsius, so we calculate the temperature in Kelvin

$T_H = 22+273 = 295K$

$T_L = 2+273 = 275k$

The rate at which heat is lost is:

$\dot{Q} = 110000kJ/h$

The power consumed by the heat pump is

$\dot{W} = 5kW$

And the coefficient of performance is

$COP_p = \frac{T_H}{T_H-T_L}$

$COP_p = \frac{295}{295-275}$

$COP_p = 14.75$

With this value we can calculate the Power required,

$COP_p = \frac{\dot{Q}}{W}$

$14.75 = \frac{110000}{W}$

$W = \frac{110000}{3600*14.75}$

$W = 2.07kW$

<em>The power consumed is consumed is 5kW which is more than 2.07kW so this heat pump powerful enough.</em>

4 0

3 years ago

What is the radius of sun and earth?

dezoksy [38]

Radius of SUN = 695,700 km

Radius of EARTH = 6,371 km

6 0

3 years ago

Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration

stepan [7]

Answer:

The magnitude of the acceleration ae of the earth due to the gravitational pull of the moon is $\mathbf{3.3187\times10^{-5}}\frac{m}{s^{2}}$

Explanation:

By Newton's gravitational law, the magnitude of the gravitational force between two objects is:

$F=G\frac{Mm}{r^{2}}$ (1)

With G the gravitational constant, M the mass of earth, m the mass of the moon and r the distance between the moon and the earth, a quick search on physics books or internet websites give us the values:

$M=5.972\times10^{24}\,kg$

$m=7.34767309\times10^{22}\,kg$

$r=384400\,km$

$G=6.674\times10^{-11}\,\frac{N\,m^{3}}{kg^{2}}$

Using those values on (1)

$F=(6.674\times10^{-11})*\frac{(5.972\times10^{24})(7.34767309\times10^{22})}{(384400\times10^{3})^{2}}$

$F\approx1.98193\times10^{20}N$

Now, by Newton's second Law we can find the acceleration of earth ae due moon's pull:

$F=M*ae\Longrightarrow ae=\frac{F}{M}=\frac{1.98193\times10^{20}}{5.972\times10^{24}}\approx\mathbf{3.3187\times10^{-5}}$

6 0

4 years ago

Car A hits car B (initially at rest and of equal mass) frombehind while going 35 m/s. Immediately after the collision, car Bmove

kolezko [41]

Answer:

The fraction of kinetic energy lost in the collision in term of the initial energy is 0.49.

Explanation:

As the final and initial velocities are known it is possible then the kinetic energy is possible to calculate for each instant.

By definition, the kinetic energy is:

k = 0.5*mV^2

Expressing the initial and final kinetic energy for cars A and B:

$ki=0.5*maVa_{i}^2+0.5*mbVb_{i}^2$

$kf=0.5*maVa_{f}^2+0.5*mbVb_{f}^2$

Since the masses are equals:

$m=ma=mb$

For the known velocities, the kinetics energies result:

$ki=0.5*mVa_{i}^2$

$ki=0.5*m(35 m/s)^2=612.5m^2/s^2*m$

$kf=0.5*mbVb_{f}^2$

$kf=0.5*m(25 m/s)^2=312.5m^2/s^2*m$

The lost energy in the collision is the difference between the initial and final kinectic energies:

$kl=ki-kf$

$kl = 612.5m^2/s^2*m-312.5 m^2/s^2*m=300 m^2/s^2*m$

Finally the relation between the lost and the initial kinetic energy:

$kl/ki = 300 m^2/s^2 * m / 612.5 m^2/s^2 * m$

$kl/ki = 24/49=0.49$

7 0

4 years ago

How long does it take to travel?<br> 120 miles at 40mph

lianna [129]

Answer:

it takes 3 hours to travel

Explanation:

120 ÷ 40 = 3

7 0

2 years ago

Read 2 more answers