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densk [106]
3 years ago
12

A box is sliding along a frictionless surface and gets to a ramp. Disregarding friction, how fast should the box be going on the

ground in order to slide up the ramp to a height of 2.5 meters, where it stops? (Use g = 9.8 m/s2.)
Physics
1 answer:
horsena [70]3 years ago
6 0

Answer:

7.0 m/s

Explanation: I just did it

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PLEASE HELP QUICKLY!!!!!
Grace [21]

Answer:

D air

Explanation:

it is not found on the periodic table

brainliest plsssssssssssssssss

3 0
3 years ago
True or false gravity is a force
mezya [45]
The answer is true because the invention ofthe beto
4 0
3 years ago
Newton first law of motion ?​
maw [93]

Answer:

The law of inertia

Explanation:

A body at rest will remain at rest, and a body in motion will remain in motion unless it is acted upon by an external force

8 0
3 years ago
Read 2 more answers
An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average mag
Afina-wow [57]

Answer:

The induced emf in the loop is 7.35\times 10^{-4}\ V

Explanation:

Given that,

Length of the wire, L = 1.22 m

It changes its shape is changed from square to circular. Then the side of square be its circumference, 4a = L

4a = 1.22

a = 0.305 m

Area of square, A=a^2=(0.305)^2=0.0930\ m^2

Circumference of the loop,

C=2\pi r=L\\\\r=\dfrac{L}{2\pi}\\\\r=\dfrac{1.22}{2\pi}=0.194\ m

Area of circle,

A'=\pi r^2\\A'=\pi (0.194)^2\\\\A'=0.118\ m^2

The induced emf is given by :

\epsilon=\dfrac{\d\phi}{dt}\\\\\epsilon=\dfrac{\d(BA)}{dt}\\\\\epsilon=B\dfrac{A'-A}{t}\\\\\epsilon=0.125 \times \dfrac{0.118-0.0930}{4.25}\\\\\epsilon=7.35\times 10^{-4}\ V

So, the induced emf in the loop is 7.35\times 10^{-4}\ V

8 0
3 years ago
A mass of .1539 kg moves down a 5 meter ramp in 2 seconds. What
enot [183]

Answer:

Velocity=2.5m/s

KE=4809.375J

Explanation:

Velocity=5m/2s=2.5m/s

KE=½×1539kg×(2.5m/s)²

KE=4809.375J

4 0
3 years ago
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