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2 years ago

9

A heat pump absorbs heat from the cold outdoors at 3Â°c and supplies heat to a house at 20Â°c at a rate of 30,000 kj/h. if the p

ower consumed by the heat pump is 3 kw, the coefficient of performance of the heat pump is

1 answer:

Mazyrski [523]2 years ago

6 0

A heat pump absorbs heat from the cold outdoors at 3 C and supplies heat to a
house at 20 C at a rate of 30,000 kJ/h. If the power consumed by the heat pump
<span>is 3 kW, find the coefficient of performance of the heat pump.</span>

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The newly formed xenon nucleus is left in an excited state. Thus, when it decays to a state of lower energy a gamma ray is emitt

nevsk [136]

Answer:3.87*10^-4

Explanation:

What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca

We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.

C=f*lambda

3*10^8=f*3.44*10^-12

F=0.87*10^20 hz

Then with the frequency, find the energy emitted using equation

E=hf E = freq*Plank's constant

E=.87*10^20*6.62*10^-34

E=575.94*10^(-16)

With this energy, convert into MeV from joules.

With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.

Plugging and computing all necessary numbers gives you

3.87*10^-4 u.

6 0

2 years ago

A mineral deposit along a strip of length 8 cm has density s(x) = 0.01x(8 − x) g/cm for 0 ≤ x ≤ 8. Calculate the total mass of t

vazorg [7]

Answer:

8z

Explanation:

It is 8z

3 0

2 years ago

__________ curves help lessen the effect of the force of the forward motion on your vehicle as it enters the curve.

Rainbow [258]

Answer:

Banked

Explanation:

Banked curves are formed when the inner edge is below the outer edge.

It is done in order to ensure the reliability of the frictional force as it varies when the road is wet wet or oily. Thus in order to avoid these problems the curved roads are banked.

Banking of the curve provides the necessary centripetal force, i.e., the horizontal component of the normal reaction force to keep the vehicle i motion and thus helps in reducing the effect of the forward motion force on the vehicle.

5 0

2 years ago

A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0

2 years ago

Part A At t tt = 2.0 s s , what is the particle's position? Express your answer to two significant figures and include the appro

mars1129 [50]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The particle's position is $x_{2} = 10 m$

The particle's velocity is $v = 2 \ m/s$

Explanation:

From the question we are told that

$x = 2m$ at $t_o = 0 \ sec$

and from the graph at t = 0 $v = 6 /m$

Now the acceleration which is the slope of the graph is mathematically represented as

$a = - \frac{6 - 4}{3-2}$

$a = - 2 m/s^2$

The negative sign shows that it is a negative slope

Now to obtain the velocity at t = 2 sec

We use the equation of motion as follows

$v = v_o + at$

substituting values '

$v = 6 + (-2)(2)$

$v = 2 \ m/s$

Now to obtain the position of the particle at v = 2 m/s

We use the equation of motion as follows

$v^2 = v_o ^2 + 2 ax$

So $2 ^2 = 6^2 + 2(-2)x$

$4x = 32$

$x = 8 m$

From above $x = 2m$ at $t_o = 0 \ sec$

So the position at t = 2 s

$x_{2} = x + x_o$

$x_{2} = 2 + 8$

$x_{2} = 10 m$

7 0

2 years ago