What type of electromagnetic wave does a cloud fall under?

An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the ce

Mamont248 [21]

Answer:

Explanation:

To find the angular velocity of the tank at which the bottom of the tank is exposed

From the information given:

At rest, the initial volume of the tank is:

$V_i = \pi R^2 h_i --- (1)$

where;

height h which is the height for the free surface in a rotating tank is expressed as:

$h = \dfrac{\omega^2 r^2}{2g} + C$

at the bottom surface of the tank;

r = 0, h = 0

∴

$h = \dfrac{\omega^2 r^2}{2g} + C$

0 = 0 + C

C = 0

Thus; the free surface height in a rotating tank is:

$h=\dfrac{\omega^2 r^2}{2g} --- (2)$

Now; the volume of the water when the tank is rotating is:

dV = 2π × r × h × dr

Taking the integral on both sides;

$\int \limits ^{V_f}_{0} \ dV = \int \limits ^R_0 \times 2 \pi \times r \times h \ dr$

replacing the value of h in equation (2); we have:

$V_f} = \int \limits ^R_0 \times 2 \pi \times r \times ( \dfrac{\omega ^2 r^2}{2g} ) \ dr$

$V_f = \dfrac{ \pi \omega ^2}{g} \int \limits ^R_0 \ r^3 \ dr$

$V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{r^4}{4} \Big]^R_0$

$V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{R^4}{4} \Big] --- (3)$

Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.

Then $V_f = V_i$

Replacing equation (1) and (3)

$\dfrac{\pi \omega^2}{g}( \dfrac{R^4}{4}) = \pi R^2 h_i$

$\omega^2 = \dfrac{4g \times h_i }{R^2}$

$\omega =\sqrt{ \dfrac{4g \times h_i }{R^2}}$

$\omega = \sqrt{\dfrac{4 \times 9.81 \ m/s^2 \times 0.7 \ m}{(0.5)^2} }$

$\omega = \sqrt{109.87 }$

$\mathbf{\omega = 10.48 \ rad/s}$

Finally, the angular velocity of the tank at which the bottom of the tank is exposed = 10.48 rad/s

6 0

3 years ago

This is a velocity versus time graph of a car starting from rest. If the area under the line is 10 meters: what is the correspon

shtirl [24]

Answer: The correct answer is option (D).

Explanation:

The area under the line is 10 meters that is displacement which means that area enclosed by the right angled triangle is 10 unit squares.

So, the it can be written as:

$\text{area of triangle}=\frac{1}{2}\times velocity\times time$

$10=\frac{1}{2}\times 2.0 m/s\times time$

$time =10 seconds$

Hence the, the correct answer is option (D).

6 0

3 years ago

The acceleration due to gravity on the moon is about 5.4 ft/s2 . If your weight is 150 lbf on the earth:

Arada [10]

Answer:

4.662 slugs

Explanation:

Your mass on the moon should always be the same as any planet you are on (due to law of mass conservation), only your weight be different as gravitational acceleration is different on each planet.

If you weight 150 lbf on Earth, and gravitational acceleration on Earth is 32.174 ft/s2. The your mass on Earth is

m = W / g = 150 / 32.174 = 4.662 slugs

which is also your mass on the moon.

4 0

3 years ago

What type of energy is released when a firecracker explodes

Ludmilka [50]

-- Sound energy. That's how you hear it explode.

-- Light energy. That's how you see it in the sky.

-- Heat energy. That's why you get burned if sparks fall on you.

-- Potential energy, as all the pieces fall to the ground.

These were all stored in it as chemical energy before it exploded.

5 0

4 years ago

Read 2 more answers

You serve a tennis ball from a height of 1.80 m above the ground. The ball leaves your racket with a velocity of 18.0 m/s at an

Delicious77 [7]

Answer:

Yes, ball will clear the net

Explanation:

First we have to find the range of projectile motion.

Data given,

Ф = 7°

Initial velocity = 18 m/s

R = (V)^2.sin2Ф/g

Now by putting values

R = 7.99 m

Now for height

h = v^2.(sinФ)^2/2g

by putting values

h = 0.245 m

Since range is less than our distance (11.83 m) from net, so still it is not clear that ball will clear the net or not.

So, now from the maximum height, we have to calculate the horizontal distance of ball to net.

Now velocity in projectile motion is in two dimensions.

V(x) = 18 m/s

V(y) = 0 m/s (because at maximum height, ball will stop and then start again, so y-component of velocity will be 0 but since there will be no acceleration along x-axis, so V(x) will be 18 m/s)

Now, by formula S = V(y)t + (1/2)gt^2

we can calculate time which is required by the ball to reach net from the maximum height it has achieved.

Now, tricky part is to calculate S, because without it we can not calculate t.

So, by data given in question, we know that the ball is served at height of 1.8 m and it achieved the height of 0.245 m. But net is at height of 1.07 m.

So, the vertical distance downward, which ball will travel from maximum height to net will be

S = 1.8 + 0.245 - 1.07

S = 0.975 m

Since we know V(y) = 0 m/s

S = (1/2)gt^2

t = (2S/g)^(1/2)

t = 0.44 s

Now time for both vertical and horizontal distance are same,

So, for horizontal distance "D(x)"

D(x) = V(x) x t (Since, no acceleration along x axis, so we can use simple formula to calculate distance)

D(x) = 18 x 0.44

D(x) = 8.029 m

Now please notice that at maximum height, range was half, so at that point ball covered distance "a"

a = 3.99 m

From maximum height to net, as we calculated, ball covered

D(x) = 8.029 m

So, total distance covered by ball

a + D(x) = 3.99 + 8.029

a + D(x) = 12.024 m

which is more than your total distance from net which is 11.83 m. So, the ball will clear the net.

7 0

3 years ago