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GrogVix [38]
2 years ago
5

A pendulum of 50 cm long consists of small ball of 2kg starts swinging down from height of 45cm at rest. the ball swings down an

d strikes a bigger ball. what is the maximum kinetic energy of the 2kg bob
Physics
1 answer:
Ket [755]2 years ago
5 0

Assuming that all energy of the small ball is transferred to the bigger ball upon impact, then we can say that:

Potential Energy of the small ball = Kinetic Energy of the bigger ball

Potential Energy = mass * gravity * height

Since the small ball start at 45 cm, then the height covered during the swinging movement is only:

height = 50 cm – 45 cm = 5 cm = 0.05 m

Calculating for Potential Energy, PE:

PE = 2 kg * 9.8 m / s^2 * 0.05 m = 0.98 J

Therefore, maximum kinetic energy of the bigger ball is:

<span>Max KE = PE = 0.98 J</span>

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Suppose you are talking by interplanetary telephone to your friend, who lives on the Moon. He tells you that he has just won a n
Savatey [412]

Answer:

The friend on moon will be richer.

Explanation:

We must calculate the mass of gold won by each person, to tell who is richer. For that purpose we will use the following formula:

W = mg

m = W/g

where,

m = mass of gold

W = weight of gold

g = acceleration due to gravity on that planet

<u>FOR FRIEND ON MOON</u>:

W = 1 N

g = 1.625 m/s²

Therefore,

m = (1 N)/(1.625 m/s²)

m(moon) = 0.6 kg

<u>FOR ME ON EARTH</u>:

W = 1 N

g = 9.8 m/s²

Therefore,

m = (1 N)/(9.8 m/s²)

m(earth) = 0.1 kg

Since, the mass of gold on moon is greater than the mass of moon on earth.

<u>Therefore, the friend on moon will be richer.</u>

7 0
2 years ago
In a power plant, pipes transporting superheated vapor are very common. Superheated vapor flows at a rate of 0.3 kg/s inside a p
grigory [225]

Answer:h=160.84 W/m^2-K

Explanation:

Given

mass flow rate=0.3 kg/s

diameter of pipe=5 cm

length of pipe=10 m

Inside temperature=22

Pipe surface =100

Temperature drop=30

specific heat of vapor(c)=2190 J/kg.k

heat supplied Q=mc\Delta T=0.3\times 2190\times (30)

Heat due to convection =hA(100-30)

A=\pi d\cdot L

A=\pi 0.05\times 10=1.571 m^2

Q_{convection}=h\times 1.571\times (100-22)=122.538 h

Q=Q_{convection}

19,710=122.538 h

h=160.84 W/m^2-K

5 0
2 years ago
There are two forces on the 2 kg box in the overhead view of the following figure, but only one is shown. For F1=20N, a= 12 m/s2
maw [93]

Answer:

second force = 32.784

Magnitude =\sqrt{32.784

θ = -90°

Explanation:

a)

Fnet = ma

F1 + F2 = ma

20N + F2 = 2(12 × cos30° + 12 ×sin30°)

F2 = 2 × 12 ( sin 30° + cos 30°)

    = 24 × ( 1 + √3 )÷ 2

  =12 (1 +√3 )

  = 32.784

b) \sqrt{12(1 +\sqrt{3}}

= \sqrt{12 ( 1+ 1.732)}

= \sqrt{12 (2.732)}

= \sqrt{32.784}  

c)

θ = 30° + 180°

θ = 210°

210° - 300°

θ = -90°

8 0
2 years ago
Calculate a pendulum's frequency of oscillation (in Hz) if the pendulum completes one cycle in 0.5 s.
Marina86 [1]
Time taken to complete one oscillation for a pendulum is Time Period, T = 0.5 s 
Frequency of the pendulum oscillation = 1 / Time Period => f = 1 / T = 1 / 0.5  
Frequency f = 2 Hz
3 0
2 years ago
A 1.00-kilogram ball is dropped from the top of a building. just before striking the ground, the ball's speed is 12.0 meters per
Anarel [89]
During the fall, the potential energy stored in the ball is converted into kinetic energy.
Thus,
PE = KE before hitting the ground
= 1/2 • mv^2
= 1/2 • 1 • 12^2
= 72J
6 0
2 years ago
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