Which transformation of energy occurs in a hydroelectric power plant?

Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on

qaws [65]

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

Δx=760 m
V₀=87 m/s
t=13.6 s
a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

$760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}$

$760=(1183.2)(cos\theta)$

$cos\theta=\frac{760}{1183.2}$

$\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}$

3 0

3 years ago

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

Mila [183]

Answer:

11.4 m/s

Explanation:

The expression for the Centripetal acceleration is :

$a=\frac{v^2}{R}$

Where, a is the accleration

v is the velocity around circumference of circle

R is radius of circle

In the given question,

a = g = Acceleration due to gravity as the car is at top = $9.81\ m/s^2$

v = ?

R = 13.2 m

So,

$9.81=\frac{v^2}{13.2}$

$v^2=9.81\times {13.2}$

8 0

3 years ago

What's is meant by relative motion?

saveliy_v [14]

Relative motion means a motion relative to a reference point. We can also say, relative motion means motion referred or observed from a reference point.

For example, Alex is in a train and Ace is at the station, when the train starts moving, for Ace it is moving at a speed of 10 m/s, but for Alex it is moving at 0 m/s, or we can say that it is at rest for Alex, but in motion for Ace. This is relative motion.

4 0

3 years ago

Read 2 more answers

Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.18 m away from a waterfall 0.294 m in heigh

Karolina [17]

Answer:

v = 7.65 m/s

t = 0.5882 s

Explanation:

We are told that the salmon started downstream, 3.18 m away from a waterfall.

Thus, range = 3.18 m

Since the horizontal velocity component is constant, then;

Range = vcosθ × t

Thus,

vcosθ × t = 3.18 - - - (eq 1)

We are told the salmon reached a height of 0.294 m

Thus, using distance equation;

s = v_y•t + ½gt²

g will be negative since motion is against gravity.

s = v_y•t - ½gt²

Thus;

0.294 = v_y•t - ½gt²

v_y = vsinθ

Thus;

0.294 = vtsinθ - ½gt² - - - (eq 2)

From eq(1), making v the subject, we have;

v = 3.18/tcosθ

Plugging into eq 2,we have;

0.294 = (3.18/tcosθ)tsinθ - ½gt²

0.295 = 3.18tanθ - ½gt²

We are given g = 9.81 m/s² and θ = 45°

0.295 = (3.18 × tan 45) - ½(9.81) × t²

0.295 = 3.18 - 4.905t²

3.18 - 0.295 = 4.905t²

4.905t² = 2.885

t = √2.885/4.905

t = 0.5882 s

Thus;

v = 3.18/(0.5882 × cos45)

v = 7.65 m/s

8 0

3 years ago

A current-carrying wire 1.50 m long is positioned perpendicular to a uniform magnetic field. If the current is 10 A and there is

LenaWriter [7]

Answer:

0.2

Explanation:

F= BIL sin©

3= B×10×1.5 sin90

B=3/15

B= 0.2

8 0

2 years ago