Ask question

Ask question

All categories

3 years ago

10

Niagara Falls is a set of very large waterfalls located on the border between New York and Ontario, Canada. Over 200,000 cubic f

eet of fast-moving waterfalls approximately 180 feet every second. Water at the top of the Falls possesses.
a. kinetic energy and gravitational potential energy.
b. only kinetic energy.
c. only gravitational potential energy.
d. neither gravitational potential energy nor kinetic energy.

2 answers:

PSYCHO15rus [73]3 years ago

8 0

Answer:

a) kinetic energy and gravitational potential energy

Explanation:

The water at the top of the fall possesses kinetic energy because the water is in motion, moving.

The water at the top also possess gravitational potential energy because it is falling under the influence of gravity. It is in a gravitational field.

krek1111 [17]3 years ago

5 0

Answer:

a. kinetic energy and gravitational potential energy.

Explanation:

It has kinetic energy because of its "fast-moving" water that is water in motion.

It has gravitational potential energy because of its height which is hundreds of feet

above the riverbed that is getting ready to fall down to, as soon as

it goes over the edge.

You might be interested in

I NEED ANSWERS!

Neporo4naja [7]

Answer:

A water fall?

Explanation:

It's an example I took in my science class

5 0

3 years ago

Read 2 more answers

A sock stuck to the inside of the clothes dryer spins around the drum once every 2.0 s at a distance of 0.50 m from the center o

Rashid [163]

a) 1.57 m/s

The sock spins once every 2.0 seconds, so its period is

T = 2.0 s

Therefore, the angular velocity of the sock is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{2.0}=3.14 rad/s$

The linear speed of the sock is given by

$v=\omega r$

where

$\omega$ is the angular velocity

r = 0.50 m is the radius of the circular path of the sock

Substituting, we find:

$v=(3.14)(0.50)=1.57 m/s$

B) Faster

In this case, the drum is twice as wide, so the new radius of the circular path of the sock is twice the previous one:

$r' = 2r = 1.00 m$

At the same time, the drum spins at the same frequency as before, therefore the angular frequency as not changed:

$\omega' = \omega = 3.14 rad/s$

Therefore, the new linear speed would be:

$v'=\omega' r' = \omega (2r)$

And substituting,

$v'=(3.14)(1.00)=3.14 rad/s = 2v$

So, we see that the linear speed has doubled.

8 0

3 years ago

The current in a wire is 4.00A. How much time is needed for 1 mole of electrons (6.02×1023 electrons) to pass a point in the wir

scZoUnD [109]

Answer:

t= 24080 s

Explanation:

Given that

Current in the wire ,I = 4 A

The charge ,q = 6.02 x 10²³ e C

We know that

$I=\dfrac{q}{t}$

I=Current

q=Charge

t=time

$t=\dfrac{q}{I}$

Now by putting the values in the above equation we get'

$t=\dfrac{6.02\times 10^{23}\times 1.6\times 10^{-19}}{4}\ s$

t= 24080 s

8 0

3 years ago

A block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The sprin

Mariana [72]

Answer:

The velocity of block = 0.188 $\frac{m}{s}$

Explanation:

Mass m = 5.6 kg

k = 1040 $\frac{N}{m}$

$\mu$ = 0.26

$x_{1} =$ 0.035 m , $v_{1}$ = 0

$x_{2} =$ 0.02 m

From work energy theorem

$K_{1} + U_{1} + W_{other} = K_{2} + U_{2}$ --------- (1)

Kinetic energy

$K = \frac{1}{2} k x^{2}$ ------- (1)

Potential energy

$U = \frac{1}{2} k x^{2}$ ------- (2)

Work done

W = F.s ------ (3)

From Newton's second law

$R_{N}$ = mg

$R_{N}$ = 5.6 × 9.81 = 54.9 N

Friction force = 0.4 × 54.9 = 21.9 N

Now the work done by the friction

$W_{f}$ = - 21.9 × 0.015

$W_{f}$ = - 0.329 J

Now kinetic energy

At point 1

$K_{1} = \frac{1}{2} m v^{2} _{1}$

$K_{1} = 0$

$U_{1} = \frac{1}{2} k x^{2}$

$U_{1} = \frac{1}{2} (1040) 0.035^{2}$

$U_{1} =$ 0.637 J

At point 2

$K_{2} = \frac{1}{2} (5.6) v^{2} _{2}$

$K_{2} = 2.8 v_{2} ^{2}$

Potential energy

$U_{2} = \frac{1}{2} k x_2^{2}$

$U_{2} = \frac{1}{2} (1040) 0.02^{2}$

$U_{2} = 0.208$ J

From equation (1) we get

0 + 0.637 - 0.329 = 2.8 $v_{2} ^{2}$ + 0.208

2.8 $v_{2} ^{2}$ = 0.1

$v_{2} =$ 0.188 $\frac{m}{s}$

This is the velocity of block.

6 0

4 years ago

A 40 foot beam that weighs 125 pounds is supported at the two ends by walls. It also supports a 600 pound AC unit 5 feet from th

Dmitrij [34]

Answer:

A) 588 pounds

Explanation:

According to the given conditions, we assume the beam to be simply supported at the ends carrying a uniformly distributed load of 125 pounds per feet and a point load of 600 pounds acting at 5 feet from the right support.

Referring the schematic:

<u>Moment about any point will be zero in equilibrium condition. </u>

∴Take moment about point L

$F_r\times 40=125\times 20+35\times 600$

$F_r=587.5 lb$

8 0

4 years ago