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Ganezh [65]
3 years ago
8

Hydroelectric power is energy created by _______.

Physics
2 answers:
Verdich [7]3 years ago
4 0
The answer is c. Moving Water
Lelu [443]3 years ago
3 0
C moving water is the answer  
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Electromagnetic radiation that has a short wavelength will have a ________ frequency.
ArbitrLikvidat [17]
High frequency , it is because wavelength is inversely proportional to frequency
8 0
3 years ago
If a car travels 600m west in 25 seconds, what is its velocity?
Elena L [17]

v =  \frac{ \:the covered distance}{time} \\

v =  \frac{600}{25} =  \frac{6 \times 100}{25} = 6 \times 4 = 24 \:  \\

v = 24 \:  \frac{m}{s} \\

_________________________________

If west means the west of the axis x the velocity equal :

- 24 \:  \frac{m}{s} \\

4 0
3 years ago
A Cessna aircraft has a lift off speed of 147 km/h. What minimum constant acceleration does this require if the aircraft is to b
netineya [11]

Answer:

The acceleration is  a =51945 \ km/h^2

Explanation:

From the question we are told that

   The lift up speed is  v  = 147 \  km/h

    The distance covered for the take off run is s =  208 m = 0.208 \ km

Generally from kinematic equation we have that

      v^2 = u^2 + 2as

Here u is the initial  speed of the aircraft with value 0 m/ s give that the aircraft started from rest

So  

    147^2 = 0^2 + 2* a* 0.208

=>  a =51945 \ km/h^2

6 0
2 years ago
I need help with this answer
7nadin3 [17]

Answer:

double replacement

Explanation:

sorry if im wrong

8 0
3 years ago
A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
Marat540 [252]

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

7 0
3 years ago
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