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2 years ago

10

Write a hypothesis about how the height of the cylinder affects the temperature of the water. Use the "if . . . then . . . becau

se . . .” format and be sure to answer the lesson question: "How is potential energy converted to thermal energy in a system?”

1 answer:

IgorLugansk [536]2 years ago

4 0

If the mass of the cylinder increases, the temperature of the water increases, because a greater mass means the cylinder has more potential energy that can be converted to thermal energy, increasing the temperature of the water.

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Noah drops a rock with a density of 1.73 g/cm3 into a pond. Will the rock float or sink?

bearhunter [10]

Answer:

It will sink

Explanation:

An object in the water can float only if its density is lower than the density of the water.

In fact, for an object completely immersed in water, there are two forces acting on it:

- Its weight, $W=mg=\rho_o V g$ , downward, where $\rho_o$ is the density of the object, V its volume and g the gravitational acceleration

- The buoyant force, $B=\rho_w V g$ , upwards, there $\rho_w$ is the density of the water

We see that when the density of an object is larger than the density of the water, $\rho_o > \rho_w$ , the weight is greater than the buoyant force, $W>B$ , so the object sinks.

In this case, the rock has a density of 1.73 g/cm3, while water has a density of 1.0 g/cm^3, so the rock will sink.

5 0

2 years ago

Read 2 more answers

A 18.0−μF capacitor is placed across a 22.5−V battery for a few seconds and is then connected across a 12.0−mH inductor that has

Gelneren [198K]

Answer:

Part a)

$i = 10.4 mA$

Part b)

in this case the charge on the capacitor will become zero

Part c)

$t_1 = 0.73 ms$

Part d)

$t = 2.2 ms$

Explanation:

As we know that first capacitor is charged with the battery and then it is connected to the inductor

So here we will have

$Q = CV$

$Q = (18\mu F)(22.5 V)$

$Q = 405 \mu C$

Part a)

now since the total energy of capacitor is converted into the energy of inductor

so by energy conservation we can say

$\frac{Q^2}{2C} = \frac{1}{2}Li^2$

so maximum current is given as

$i = \sqrt{\frac{L}{C}}Q$

$i = \sqrt{\frac{12\times 10^{-3}}{18\times 10^{-6}}}(405\times 10^{-6})$

$i = 10.4 mA$

Part b)

When current is maximum then whole energy of capacitor is converted into magnetic energy of inductor

So in this case the charge on the capacitor will become zero

Part c)

Time period of oscillation of charge between the plates and inductor is given as

$T = 2\pi\sqrt{LC}$

$T = 2\pi\sqrt{(18\mu F)(12 mH)}$

$T = 2.92 ms$

now capacitor gets discharged first time after 1/4 of total time period

$t_1 = 0.73 ms$

Part d)

Since time period is T and capacitor gets discharged two times in one complete time period of the motion

so first it will discharges in T/4 time

then next T/4 it will get charged again

then next T/4 time it will again discharged

so total time taken

$t = 3T/4$

$t = 2.2 ms$

4 0

2 years ago

A solenoid of length 0.700m having a circular cross-section of radius 5.00cm stores 6.00 μJ of energy when a 0.400-A current run

AURORKA [14]

Answer:

The winding density of the solenoid, n = 104 turns/m

Explanation:

Given that,

Length of the solenoid, l = 0.7 m

Radius of the circular cross section, r = 5 cm = 0.05 m

Energy stored in the solenoid, $E=6\ \mu J=6\times 10^{-6}\ J$

Current, I = 0.4 A

To find,

The winding density of the solenoid.

Solution,

The expression for the energy stored in the solenoid is given by :

$U=\dfrac{1}{2}LI^2$

Where

L is the self inductance of the solenoid

$L=\mu_on^2lA$

n is the winding density of the solenoid

$n=\sqrt{\dfrac{2U}{\mu_oI^2l\pi r^2}}$

$n=\sqrt{\dfrac{2\times 6\times 10^{-6}}{4\pi \times 10^{-7}\times 0.7\times (0.4)^2\pi (0.05)^2}}$

n = 104 turns/m

So, the winding density of the solenoid is 104 turns/m

8 0

2 years ago

Read 2 more answers

A certain slide projector has a 100 mm focal length lens. How far away is the screen, if a slide is placed 103 mm from the lens

Vlad [161]

Answer:

3.43 m

Explanation:

f = 100 mm

u = - 103 mm

Let v be the distance between the screen and the lens of the projector.

Use lens equation

1 / f = 1 / v - 1 / u

1 / 100 = 1 / v + 1 / 103

1 / v = 1 / 100 - 1 / 103

1 / v = (103 - 100) / (100 x 103)

1 / v = 3 / 10300

v = 3433.33 mm = 3.43 m

4 0

2 years ago

A giant wall clock with diameter d rests vertically on the floor. The minute hand sticks out from the face of the clock, and its

Katyanochek1 [597]

Answer:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

Explanation:

We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.

The x-component of the length of the minute hand is:

$d_{x}(t)=dcos(\theta (t))$ (1)

d is the length of the minute hand (d=D/2)
D is the diameter of the clock
t is the time (min)

Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:

$\theta (t)=\omega *t$ (2)

Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:

$\omega=\frac{\Delta \theta}{\Delta t}=\frac{\theta_{f}-\theta_{i}}{t_{f}-t{i}}=\frac{\pi/2-0}{15-0}=\frac{\pi}{30}$

Now, let's put this value on (2)

$\theta (t)=\frac{\pi}{30}*t$

Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

I hope it helps you!

6 0

2 years ago