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3 years ago

9

If Styrofoam has a specific heat of 1131 J/kg°C, how much energy would it take to change a 0.5 kg Styrofoam cooler from 15°C to

2°C?

1 answer:

Y_Kistochka [10]3 years ago

8 0

Answer:

.

Explanation:

You might be interested in

As an airplane accelerates in level flight, what parameter must the pilot adjust to keep level and balance Lift = Weight?

Y_Kistochka [10]

Answer:

C. Angle of Attack.

Explanation:

The pilot must adjust the angle of attack parameter. The angle of attack of this plane to get to the desired lift coefficient.

And thus, we have

Lift = Weight

3 0

2 years ago

Which device provides electrical energy to run an electric circuit

REY [17]

The correct answer is

C. The battery

The battery is a device that provides a potential difference in the circuit, and so an electromotive force (e.m.f.) which pushes the electrons in the circuit from the negative pole towards the positive pole of the battery, so they move through the circuit. Therefore, it provides electrical energy.

8 0

3 years ago

Read 2 more answers

At what speed, as a fraction of c , is a particle's total energy twice its rest energy

WINSTONCH [101]

The rest energy of a particle is
$E_0=m_0 c^2$
where $m_0$ is the rest mass of the particle and c is the speed of light.

The total energy of a relativistic particle is
$E=mc^2 = \frac{m_0 c^2}{ \sqrt{1- \frac{v^2}{c^2} } }$
where v is the speed of the particle.

We want the total energy of the particle to be twice its rest energy, so that
$E=2E_0$
which means:
$\frac{m_0c^2}{ \sqrt{1- \frac{v^2}{c^2} } }=2m_0 c^2$
$\frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }=2$
From which we find the ratio between the speed of the particle v and the speed of light c:
$\frac{v}{c}= \sqrt{1- (\frac{1}{2})^2 } =0.87$
So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.

3 0

3 years ago

. Each valve A, B, and C, when open, releases water into a tank at its own constant rate. With all three valves open, the tank f

olasank [31]

Answer:

Time taken by A and B is 1.2 hr.

Explanation:

Given that

Time taken by tank when all(A+B+C) are open = 1 hr

Time taken by tank when A+C are open = 1.5 hr

Time taken by tank when B+C are open = 2 hr

If we treat as filling of tank is a work then

Work = time x rate

Lets take work is 1 unit

1 = 1(1/a+1/b+1/c) ---------1

1 = 1.5(1/a+1/c) ----------2

1 = 2(1/b+1/c) --------3

From equation 1 and 3

1=1(1/a+1/2)

a=2

Form equation 2

1 = 1.5(1/2+1/c)

c=6

From equation 3

1 = 2(1/b+1/6)

b=3

So time taken by

A is alone to fill tank is 2 hr

B is alone to fill tank is 3 hr

C is alone to fill tank is 6 hr

So $time\ taken\ by\ A\ and\ B =\dfrac{2\times 3}{2+3} hr$

Time taken by A and B is 1.2 hr.

7 0

2 years ago

Jake is helping Fin push a box at a constant velocity up an incline that makes an angle of 30.0° above the horizontal by applyin

andre [41]

Given data

The angle of inclination of the plane is theta = 30 degree

The applied force in the inclined plane is F = 94 N

The distance moved in the inclined plane is d = 2.30 m

The coefficient of kinetic friction is u_k = 0.280

The free-body diagram of the above configuration is shown below:

Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.

(a)

The expression for the work done by the pushing force is given as:

$W=Fd$

Substitute the value in the above equation.

$\begin{gathered} W=94\text{ N}\times2.30\text{ m} \\ W=216.2\text{ J} \end{gathered}$

Thus, the work done by the pushing force is 216.2 J.

(b)

The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.

$\begin{gathered} F=F_f+mg\sin \theta \\ F=\mu_kN+mg\sin \theta \end{gathered}$

The expression for the normal reaction force is given as:

$N=mg\cos \theta$

The expression for the mass of the box is given as:

$\begin{gathered} F=\mu_k\times mg\cos \theta+mg\sin \theta \\ m=\frac{F}{\mu_kg\cos \theta+g\sin \theta} \end{gathered}$

Substitute the value in the above equation.

$\begin{gathered} m=\frac{94\text{ N}}{0.28\times9.8m/s^2\times\cos 30^o+9.8m/s^2\times\sin 30^0} \\ m=12.9\text{ kg} \end{gathered}$

Thus, the mass of the box is 12.9 kg.

7 0

8 months ago