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GrogVix [38]
3 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]3 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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How many football fields (including the 10 yards in each end zone) would it take to make a mile?
Inessa05 [86]

Answer:

Explanation:

1760 yd/mi / 120 yd/field = 14⅔ fields/mi

4 0
3 years ago
50 kg of water at 75o C is cooled to 25o C. How much heat was given off?
attashe74 [19]

Answer:

b the answer is b

Explanation:

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8 0
2 years ago
What kind of friction exists between solid objects moving in water?
eduard
Fluid Friction exists when it is acted upon an object when in fluid.


6 0
3 years ago
A​ right-circular cylindrical tank of height 8 ft and radius 4 ft is laying horizontally and is full of fuel weighing 52 ​lb/ft3
tresset_1 [31]

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ ft^{3} to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = \pi r^{2} =  \pi 4^{2} =16\pi ft^{2}

Now,

Work done required = \int_{0}^{8} 52\times 16\pi (21 - y)dy

                                  = 832\pi \int_{0}^{8} (21 - y)dy

                                  = 832\pi([ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\)

                                  = 113152\pi = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

7 0
4 years ago
In the lab, you submerge 100 g of 40°C nails in 200 g of 20°C water. (The specific heat of iron is 0.12 cal/g # °C.) Equate the
Deffense [45]

Answer:

Final temperature of the mixture becomes

T = 21.13^oC

Explanation:

Here we know that heat given by the nail is equal to the heat absorbed by water

So here we know that heat to change the temperature is given as

Q = ms\Delta T

so by equating the heat we have

m_{iron}s_{iron}\Delta T = m_{water}s_{water}\Delta T

now we have

100 (0.12) (40 - T) = 200 (1) (T - 20)

4.8 - 0.12 T = 2T - 40

44.8 = 2.12 T

T = 21.13^oC

5 0
3 years ago
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