Ask question

Ask question

All categories

3 years ago

8

Hydroelectric power is energy created by _______.

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

You might be interested in

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