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Natali5045456 [20]

3 years ago

6

Which of the following is true regarding the Earth's mantle?

1 answer:

Brilliant_brown [7]3 years ago

6 0

B) It’s material moves due to convection currents.

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A violin string that is 50.0 cm long has a fundamental frequency of 440 Hz. What is the

amid [387]

For a standing wave on a string, the wavelength is equal to twice the length of the string:
$\lambda=2 L$
In our problem, L=50.0 cm=0.50 m, therefore the wavelength of the wave is
$\lambda = 2 \cdot 0.50 m = 1.00 m$

And the speed of the wave is given by the product between the frequency and the wavelength of the wave:
$v=\lambda f = (1.00 m)(440 Hz)=440 m/s$

5 0

4 years ago

Read 2 more answers

2. Determine the current in a 120-watt bulb plugged into a 120-volt outlet.

Sergio039 [100]

Answer:

1 ampere

Explanation:

Power = VI

V = voltage

I = current

Given

Power = 120 watts

V = 120 volts

Therefore

120 = 120 x I

Divide both sides by 120

120/120 = 120/120 x I

1 = I

I = 1 ampere

8 0

3 years ago

Estimate the radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bul

Deffense [45]

..........................................................

8 0

3 years ago

The model below shows a calcium atom. An image has a mix of red and blue balls in its center and 4 concentric black rings around

zubka84 [21]

Answer:

8 electrons in the third energy level

Explanation:

From the description,the third energy level has 8 electron (represented by the small green balls you describe)

4 0

3 years ago

A gymnast dismounts off the uneven bars in a tuck position with a radius of 0.3m (assume she is a solid sphere) and an angular v

kifflom [539]

Here we will say that there is no external torque on the system so we will have

$L_i = L_f$

here we know that

$L_i = I_1\omega_1$

where we know that

$I_1 = \frac{2}{5}mr^2$

Also we know that

$I_2 = \frac{1}{12}mL^2$

initial angular speed will be

$\omega_1 = 2\pi(2rev/s) = 4\pi rad/s$

now from above equation

$\frac{2}{5}mr^2 (4\pi) = \frac{1}{12}mL^2 \omega$

$0.4(0.3)^2(4\pi) = \frac{1}{12}(1.5)^2\omega$

$0.452 = 0.1875 \omega$

now we have

$\omega = 2.41 rad/s$

so final speed will be 2.41 rad/s

6 0

3 years ago