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USPshnik [31]
3 years ago
12

In the Earth's mantle, heat is transferred in large convection currents. Within these currents,

Physics
2 answers:
liq [111]3 years ago
8 0

A.cooler and hotter rock rise but do not sink.

B.hotter and cooler rock sink but do not rise.

C.hotter rock rises and cooler rock sinks.

D.cooler rock rises and hotter rock sinks.

hotter rock rises and cooler rock sinks.

Answer: Option C.

<u>Explanation:</u>

The process that happens inside the surface of earth makes the rocks inside the surface to either expand or shrink in size because of the presence of heat inside the surface.

Heat rises and as the warmer rock gets further away from the super heated center of the Earth it begins to cool off and starts to sink back toward the super heated center of the Earth where it is reheated and the whole process starts over again. That is what a convection current is.

Klio2033 [76]3 years ago
3 0

Answer:Hotter rock rises and cooler rock sinks

Explanation:

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4 0
3 years ago
Read 2 more answers
A "590-W" electric heater is designed to operate from 120-V lines.
lukranit [14]

Answer:

a) 24.4 Ω

b) 4.92 A

c) 495.9 W

d)

c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.

Explanation:

b)

The formula for power is:

P = IV

where,

P = Power of heater = 590 W

V = Voltage it takes = 120 V

I = Current Drawn = ?

Therefore,

590 W = (I)(120 V)

I = 590 W/120 V

<u>I = 4.92 A</u>

<u></u>

a)

From Ohm's Law:

V = IR

R = V/I

Therefore,

R = 120 V/4.92 A

<u>R = 24.4 Ω</u>

<u></u>

c)

For constant resistance and 110 V the power becomes:

P = V²/R

Therefore,

P = (110 V)²/24.4 Ω

<u>P = 495.9 W</u>

<u></u>

d)

If the resistance decreases, it will increase the current according to Ohm's Law. As a result of increase in current the power shall increase according to formula (P = VI). Therefore, correct option is:

<u>c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.</u>

7 0
2 years ago
Could someone help me with some info for an essay?I WILL REPOST FOR 100 POINTS IF YOU CAN.
polet [3.4K]

Answer:

Self-motivation is the surest way to stay focused. Nearly all the most significant tasks in life are tests for our motivation. Facing challenges and achieving success requires focus; this means we need to be responsible for our motivation. People who are motivated towards achieving their goals are focused on success. The following ideas will help you stay focused and motivated in your work and studies.

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3 0
2 years ago
Calculate the energy of the green light emitted, per photon, by a mercury lamp with a frequency of 5.49 × 1014 hz.
Tcecarenko [31]
The energy of a photon is given by
E=hf
where
h=6.6 \cdot 10^{-34} Js is the Planck constant
f is the frequency of the photon

In our problem, the frequency of the light is 
f=5.49 \cdot 10^{14}Hz
therefore we can use the previous equation to calculate the energy of each photon of the green light emitted by the lamp:
E=hf=(6.6 \cdot 10^{-34}Js)(5.49 \cdot 10^{14} Hz)=3.62 \cdot 10^{-19} J
8 0
3 years ago
A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th
svp [43]

a) x(t)=2.0 sin (10 t) [m]

The equation which gives the position of a simple harmonic oscillator is:

x(t)= A sin (\omega t)

where

A is the amplitude

\omega=\sqrt{\frac{k}{m}} is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s

The amplitude, A, can be found from the maximum velocity of the spring:

v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m

So, the equation of motion is

x(t)= 2.0 sin (10 t) [m]

b)  t=0.10 s, t=0.52 s

The potential energy is given by:

U(x)=\frac{1}{2}kx^2

While the kinetic energy is given by:

K=\frac{1}{2}mv^2

The velocity as a function of time t is:

v(t)=v_{max} cos(\omega t)

The problem asks as the time t at which U=3K, so we have:

\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}

However, \frac{m}{k}=\frac{1}{\omega^2}, so we have

(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\

with two solutions:

\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s

\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s

c) 3 seconds.

When x=0, the equation of motion is:

0=A sin (\omega t)

so, t=0.

When x=1.00 m, the equation of motion is:

1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s

The period of a pendulum is:

T=2 \pi \sqrt{\frac{L}{g}}

where L is the length of the pendulum. By using T=0.628 s, we find

L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m






5 0
3 years ago
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