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Rom4ik [11]
3 years ago
13

1. A 65 kg student, starting from rest, slides down an 16.2 m high water slide. On the way down, friction does 5700 J of work on

him. How fast is he going at the bottom of the slide
Physics
1 answer:
GuDViN [60]3 years ago
4 0

Answer:

11.94

Explanation:

Remark

Find the Potential Energy at the top.

Givens

m = 65 kg

h = 16.2 m

g = 9.81

PE = 65 * 9.81 * 16.2

PE = 10329.93

The tricky part is what do you do about Friction?

Formula

PE = Friction + KE

Solution

PE = 10329.93 Joules

Friction = 5700 Joules

Find the KE

10329.93 = 5700 + KE

KE = 10329.93 - 5700

KE = 4629.93

Find V from the KE formula

KE = 4629.93

m = 65

KE = 1/2 m v^2

KE = 1/2 65 v^2

4629.93 = 1/2 65 v^2

v^2 = 142.46

v = √142.46

v = 11.94

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Velocity is a vector meaning it includes an objects speed and direction.
3 0
3 years ago
The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5
KiRa [710]

Answer:

0.68 s

Explanation:

We are given that

Initial velocity of box=u=13m/s

Final velocity of box=v=11.5 m/s

Distance=d=8.5 m

We have to find the time taken by box to slow by this amount.

We know that

v^2-u^2=2as

Substitute the values

(11.5)^2-(13)^2=2a(8.5)

132.25-169=17a

-36.75=17a

a=\frac{-36.75}{17}=-2.2m/s^2

We know that

Acceleration=a=\frac{v-u}{t}

Substitute the values

-2.2=\frac{11.5-13}{t}

-2.2=\frac{-1.5}{t}

t=\frac{1.5}{2.2}=0.68 s

Hence, the time taken by box to slow by this amount=0.68 s

8 0
3 years ago
In an elastic collision, _______ energy is conserved.
Nonamiya [84]

Answer:

Kinetic energy and momentum are conserved.

Explanation:

4 0
2 years ago
A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e
mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

 ∑ M = 0

mass of the log will be concentrated at the center  

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)

200 \times 0.1 L = 53.5 \times (x - 0.6 L)

0.374 L =x - 0.6 L

       x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

7 0
3 years ago
What are some of the characteristics scientists are looking for when they search for other Earth-like planets?
Yanka [14]
The ability to sustain life
(ie water, shelter, food, basic needs)
Hope this helped!
:-)
4 0
4 years ago
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