Ask question

Ask question

All categories

kolbaska11 [484]

2 years ago

11

If a sled has at the top of 10 m hill had 1000 J of potential energy what would happen to the PE if the sled were to moved to a

5 m hill?

1 answer:

nataly862011 [7]2 years ago

7 0

Answer:

500J

Explanation:

there are many students who can not get answer step by step and in required time. so

there are a wats group where you can get your answer by trusted experts.

join this wats up group and be smart

You might be interested in

The engine oil at 150 degree Celsius is cooled to 80 degree Celsius in a parallel flow heat exchanger by water entering at 25 de

Setler [38]

Answer:

Explanation:

Attach is the solution

6 0

3 years ago

During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 33.5 rad/s. Find the angular displ

zavuch27 [327]

Answer: angular displacement in rad = 3038.45 rad

angular displacement in rev = 483.589 rev

Explanation: mathematically

Angular velocity = angular displacement / time taken.

Angular velocity = 33.5 rad/s, time taken = 90.7s

33.5 = angular displacement /90.7

Angular displacement = 33.5 * 90.7 = 3038.45 rad

But 1 rev =2π

Hence 3038.45 rad to rev is

3038.45/2π = 483.599 rev

7 0

3 years ago

Read 2 more answers

A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round

Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

$u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s$

$v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s$

Now we find the centripetal acceleration which is given by

$a_c=\frac{v^2}{r}$

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

$a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2$

we also have a tangential acceleration, which is given by

$a_t = \frac{v-u}{t}$

where

t = 17.0 s

Substituting values,

$a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2$

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

$a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2$

6 0

2 years ago

Read 2 more answers

What would you have loved to press the pause button on so you could go deeper

kiruha [24]

Answer:

~Banana Fish~

3 0

3 years ago

What is the magnitude of the gravitational force acting on the sun due to the earth?The earth does not exert any gravitational f

Sholpan [36]

I haven't worked on Part-A, and I don't happen to know the magnitude of the gravitational force that the Sun exerts on the Earth.

But whatever it is, it's exactly, precisely, identical, the same, and equal to the magnitude of the gravitational force that the Earth exerts on the Sun.

I think that's the THIRD choice here, but I'm not sure of that either.

4 0

2 years ago