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Maurinko [17]
3 years ago
7

Help pls! Explain if you think we should continue to use geothermal power in the future

Physics
1 answer:
zhuklara [117]3 years ago
4 0

Answer:

Geothermal power can provide consistent electricity throughout the day and year - continuous baseload power and flexible power to support the needs of variable renewable energy resources, such as wind and solar. Sustainable Investment.

Explanation:

THIS IS WHY WE SHOULD USE GEOTHERMAL ENERGY IN FUTURE

YOU CAN MARK ME AS BRAINIEST IF YOU WANT

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At what altitude above the earth's surface would the acceleration due to gravity be 4.9 m/s^2? Assume the mean radius of the ear
Mandarinka [93]
We apply the gravity calculation expressed in the formula: g=GM/r2 
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
(1)       Radius = √6.674e-11*5.972e24/8             = 7058 kms    Earth radius or surface of earth from center of earth= 6400 kmsSo r= 658 kms from surface of earth.
Gravity 8m/s2 will be at 658 kms from surface of earth.
(2) half gravity= 9.8/2= 4.9 m/s2     Radius=√6.674e-11*5.972e24/4.9                 = 9019 kms        Half Gravity will exist at 9019-6400= 2619 kms from surface of earth.
4 0
3 years ago
Which part of the wave has the highest frequency?
trapecia [35]

Answer:

The last part on the right side of the diagram

Explanation:

Im on plato and just got it right :)

6 0
3 years ago
Read 2 more answers
Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki
irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

1)

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every  second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

a=1 m/s^2

which is equivalent to the gradient of the line in the velocity-time graph.

2)

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

a=-1 m/s^2 is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

T=-\frac{u}{a}=-\frac{10}{-1}=10 s

3)

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

a=-1 m/s^2 is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m

7 0
3 years ago
An unknown fluid flows at a speed of 31 m/s. Suppose the fluid has a mass of 47 kg runs at this speed. What is the fluid’s kinet
Leya [2.2K]

Answer:

22583.5J

Explanation:

KE=1/2 mv^2

=1/2*47Kg*(31m/s^2)

=23.5Kg * 961m/s^2

=22583.5J

7 0
2 years ago
A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b
jonny [76]

Answer:

Part a)

I = 3.16 \times 10^{-5} W/m^2

Part b)

P_o = 0.162 Pa

Explanation:

Part a)

Level of sound = 75 dB

now we know that

L = 10 Log\frac{I}{I_0}

here we know that

I_0 = 10^{-12} W/m^2

now we have

75 = 10 Log(\frac{I}{10^{-12}})

I = 3.16 \times 10^{-5} W/m^2

Part b)

Intensity of sound wave is given as

I = \frac{1}{2}\rho A^2\omega^2 c

here we know that

A = \frac{P_o}{Bk}

so we have

I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c

I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}

now we know

\rho = 1.2 kg/m^3

c = 340 m/s

B = 1.4 \times 10^5 Pa

now we have

3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}

P_o = 0.162 Pa

5 0
3 years ago
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