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Alik [6]
2 years ago
8

45. 3. (III) A 7.26 kg bowling ball hangs from the end of a 2.5 m rope. The ball is pulled back until the rope makes an angle of

45 with the vertical. a. What is the increase in the ball’s potential energy when it is lifted? b. If the ball is released from rest and swings downward like a pendulum, how much kinetic energy will the ball have at the bottom of its swing? c. How fast will the ball be moving at the bottom of its swing?
Physics
1 answer:
docker41 [41]2 years ago
3 0

Explanation:

For the first question look at the picture. It should be pretty clear. Remember that cos45° = sqrt2 / 2.

2) We calculate it thanks to the conservation of mechanical energy:

Em1 = Em2

U1 + K1 = U2 + K2

K1 = 0J

U2 = 0j

U1 = K2

3) mgh = 1/2mv^2

v = sqrt(2gh)

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Waves can transfer energy through
lina2011 [118]

Answer:

electromagnetic waves

Explanation:

"wave" is a common term for a number different ways in which energy is transferred

3 0
3 years ago
A girl throws a ball of mass 0.8 kg against a wall. The ball strikes the wall horizontally with a speed of 11 m/s, and it bounce
Karolina [17]

Answer:

F = 352 N

Explanation:

we know that:

F*t = ΔP

so:

F*t = MV_f-MV_i

where F is the force excerted by the wall, t is the time, M the mass of the ball, V_f the final velocity of the ball and V_i the initial velocity.

Replacing values, we get:

F(0.05s) = (0.8 kg)(11m/s)-(0.8 kg)(-11m/s)

solving for F:

F = 352 N

 

3 0
3 years ago
How did the continental Shelf form? Please help, thank you!! :)
shutvik [7]
The movements of the tectonic plates
6 0
3 years ago
The amount of time required for 2 successive wave crests to pass a fixed point is called wave ________.
mihalych1998 [28]
I believe this is known as wave period.
hope this helps!
7 0
2 years ago
A particle starts from rest and has an acceleration function 5 − 10t m/s2 . (a) What is the velocity function? (b) What is the p
frez [133]

Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

a(t)=(5-10t)\ m/s^2

(a) Since, a=\dfrac{dv}{dt}

v = velocity

dv=a.dt

v=\int(a.dt)

v=\int(5-10t)(dt)

v=5t-\dfrac{10t^2}{2}=5t-5t^2

(b) v=\dfrac{dx}{dt}

x = position

x=\int v.dt

x=\int (5t-5t^2)dt

x=\dfrac{5}{2}t^2-\dfrac{5}{3}t^3

(c) Velocity function is given by :

v=5t-5t^2

5t-5t^2=0

t = 1 seconds

So, at t = 1 second the velocity of the particle is zero.

7 0
2 years ago
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