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faltersainse [42]
3 years ago
12

Tidal power can be generated by man without investing a lot of energy into the process. True False

Physics
2 answers:
tiny-mole [99]3 years ago
6 0

Answer: True

Explanation: Tidal Power can be generated with with less efforts and investment. The tidal energy is generated by the help of the tidal energy generators.

These generators are fitted in underwater turbines which has high tidal movements, which are designed to capture the kinetic motion of the surging of ocean in order to produce electricity.

Dmitriy789 [7]3 years ago
4 0

If only that were true, you would see tidal power stations
all up and down every coast by now.

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A 0.73-m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the bar
Tasya [4]

Answer:

A)  B = 5.4 10⁻⁵ T, B) the positive side of the bar is to the West

Explanation:

A) For this exercise we must use the expression of Faraday's law for a moving body

            fem = -  \frac{d \phi }{dt}

            fem = - \frac{d (B l y}{dt}= - B l v- d (B l y) / dt = - B lv

            B = - \frac{fem}{l \ v}

we calculate

             B = - 7.9 10⁻⁴ /(0.73 20)

             B = 5.4 10⁻⁵ T

B) to determine which side of the bar is positive, we must use the right hand rule

the thumb points in the direction of the rod movement to the south, the magnetic field points in the horizontal direction and the rod is in the east-west direction.

Therefore the force points in the direction perpendicular to the velocity and the magnetic field is in the east direction; therefore the positive side of the bar is to the West

4 0
3 years ago
A physics book slides off a table at 1.25ms and hits the ground after 0.4s
tresset_1 [31]

Answer:

Whats the question here?

7 0
3 years ago
A balloon is filled to a volume of 7.00*10^2 mL at a temperature of 20.0°C. The balloon is then cooled at constant pressure to a
wolverine [178]

The final volume of the gas is 238.9 mL

Explanation:

We can solve this problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas (V) is proportional to its absolute temperature (T):

\frac{V}{T}=const.

Which can be also re-written as

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where

V_1, V_2 are the initial and final volumes of the gas

T_1, T_2 are the initial and final temperature of the gas

For the gas in the balloon in this problem, we have:

V_1 = 7.00\cdot 10^2 mL = 700 mL is the initial volume

T_1=20.0^{\circ}C+273=293 K is the initial absolute temperature

V_2 is the final volume

T_2 = 1.00\cdot 10^2 K = 100 K is the final temperature

Solving for V_2,

V_2 = \frac{V_1 T_2}{T_1}=\frac{(700)(100)}{293}=238.9 mL

Learn more about ideal gases:

brainly.com/question/9321544

brainly.com/question/7316997

brainly.com/question/3658563

#LearnwithBrainly

6 0
3 years ago
James accelerates his skate board uniformly along a straight road from rest to 10 m/s in 4 seconds. What is James Acceleration?
lawyer [7]

Given:

u(initial velocity)=0

v(final velocity)= 10 m/s

t= 4 sec

Now we know that

v= u + at

Where v is the final velocity

u is the initial velocity

a is the acceleration measured in m/s^2

t is the time measured in sec

10=0+ax4

a=10/4

a=2.5 m/s^2

4 0
3 years ago
The owner of a company that manufactures drinking cups decides it would be impressive to build an inground swimming pool that is
garri49 [273]

Answer:

The depth is 5.15 m.

Explanation:

Lets take the depth of the pool = h m

The atmospheric pressure ,P = 101235 N/m²

The area of the top = A m²

The area of the bottom =  a m²

Given that A= 1.5 a

The force on the top of the pool = P A

The total pressure on the bottom = P + ρ g h

ρ =Density of the  water = 1000 kg/m³

The total pressure at the bottom of the pool =  (P + ρ g h) a

The bottom and the top force is same

(P + ρ g h) a = P A

P a +ρ g h a =  P A

ρ g h a =  P A - P a

h=\dfrac{P ( A-a)}{\rho g a}

h=\dfrac{P ( 1.5 a-a)}{\rho g a}

h=\dfrac{P ( 1.5- 1)}{\rho g}

h=\dfrac{101235 ( 1.5- 1)}{1000\times 9.81}\ m

h=5.15 m

The depth is 5.15 m.

7 0
3 years ago
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