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3 years ago

What is the best explanation for why the room with the heat pump was cooler than the rest of the house?

Physics

1 answer:

Stolb23 [73]3 years ago

6 0

A Heat pump takes heat from colder object and transfers it to warmer object.

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a 12-kg block is pulled by a string at a 30 degree angle. if the string exerts a 50 n force, the block accelerates at 4.13 m/s2.

alexdok [17]

Let, frictional force is f and normal force is N.

Balancing horizontal forces :

$50cos \ 30^o - f = 12 \times 4.13\\\\f = 50cos \ 30^o - 49.56\\\\f = -6.26\ N$

Also, balancing vertical forces, we get :

$N = mg + 50sin \ 30^o\\\\N =( 12\times 9.8 ) + (50\times sin \ 30^o)\\\\N = 142.6\ N$

Therefore, the frictional force and normal force is -6.26 N and 142.6 N.

5 0

3 years ago

On a horizontal axis whose unit is the meter, a linear load ranging from 0 to 1 ma a linear load distribution = 2 nC / m.

Dimas [21]

Answer:

The correct answer is - 8.99N/C

Explanation:

$dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k$

6 0

3 years ago

Which of the following shows acceleration? *

Ronch [10]

Answer:

a ball thrown upward

Explanation:

it is the only thing that is picking up speed

3 0

3 years ago

Read 2 more answers

Calculate the speed of the ball, vo in m/s, just after the launch. A bowling ball of mass m = 1.5 kg is launched from a spring c

klemol [59]

Answer:

$v_0=17.3m/s$

Explanation:

In this problem we have three important moments; the instant in which the ball is released (1), the instant in which the ball starts to fly freely (2) and the instant in which has its maximum height (3). From the conservation of mechanical energy, the total energy in each moment has to be the same. In (1), it is only elastic potential energy; in (2) and (3) are both gravitational potential energy and kinetic energy. Writing this and substituting by known values, we obtain:

$E_1=E_2=E_3\\\\U_e_1=U_g_2+K_2=U_g_3+K_3\\\\\frac{1}{2}kd^2=mg(d\sin\theta)+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}m(v_0\cos\theta)^2$

Since we only care about the velocity $v_0$ , we can keep only the second and third parts of the equation and solve:

$mgd\sin\theta+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}mv_0^2\cos^2\theta\\\\\frac{1}{2}mv_0^2(1-\cos^2\theta)=mg(h-d\sin\theta)\\\\v_0=\sqrt{\frac{2g(h-d\sin\theta)}{1-\cos^2\theta}}\\\\v_0=\sqrt{\frac{2(9.8m/s^2)(4.4m-(0.21m)\sin32\°)}{1-\cos^232\°}}\\\\v_0=17.3m/s$

So, the speed of the ball just after the launch is 17.3m/s.

4 0

3 years ago

Explain newton's third law of motion

sergij07 [2.7K]

Answer:

for every action, there is an equal and opposite reaction.

Explanation:

the statement means that every interaction there is a pair of forces acting on the two interacting objects.

3 0

3 years ago