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RUDIKE [14]
3 years ago
11

A green ball has a mass of 0.525 kg and a blue ball has a mass of 0.482 kg. A croquet player strikes the green ball and it gains

an initial velocity of 2.26 m/s. It then strikes the blue ball, which is initially at rest. After the collision, the green ball has a velocity of 1.14 m/s in the same direction. If the balls roll on a frictionless surface and the collision is head-on, what is the final velocity of the blue ball? (Round your answer to the nearest hundredths place.)
Physics
1 answer:
Gelneren [198K]3 years ago
4 0

Answer:

v' = 1.21 m/s

Explanation:

Mass of a green ball, m = 0.525 kg

Mass of a blue ball, m' = 0.482 kg

Initial velocity of green ball, u = 2.26 m/s

Initial velocity of blue ball, u' = 0 (at rest)

After the collision,

The final velocity of the green ball, v = 1.14 m/s

We need to find the final velocity of the blue ball after the collision if the collision is head on. Let v' is the final velcity of the blue ball. Using the conservation of momentum to find it :

mu+m'u'=mv+m'v'\\\\0.525 (2.26)+0=0.525 (1.14)+0.482v'\\\\0.588=0.482v'\\\\v'=\dfrac{0.588}{0.482}\\\\v'=1.21\ m/s

So, the final velocity of the blue ball is 1.21 m/s.

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When air is blown across the top of an open
Tanzania [10]
When air is blown across the top of an open <span>water bottle, air molecules in the bottle vibrate at a particular frequency and sound is produced in a process called "refraction". 
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8 0
4 years ago
A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 r
sammy [17]

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana m_{b} = 8.97 kg

at distance r = 4/5  { radius of platform }

mass of monkey m_{m} = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + m_{b} (\frac{4}{5} R)² + m_{m}R² ]w

m/2 × ω₀ = [ m/2 + m_{b} (\frac{4}{5} )² + m_{m} ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

8 0
3 years ago
What is the average power output pf a weight lifter who can lift 250 kg 2.0 m in 2.0 s
liq [111]

Answer:

Power = 2.45Kw or 2450 Watts.

Explanation:

<u>Given the following data;</u>

Mass, m = 250kg

Height, h = 2m

Time, t = 2secs

We know that acceleration due to gravity, g is equal to 9.8m/s²

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

Power = \frac {Energy}{time}

But Energy = mgh

Substituting into the equation, we have

Power = \frac {mgh}{time}

Power = \frac {250*9.8*2}{2}

Power = \frac {4900}{2}

Power = 2450 Watts

To convert to kilowatt (Kw), we would divide by 1000

Power = 2450/1000

Power = 2.45Kw.

Therefore, the average power output of the weightlifter is 2.45 Kilowatts.

8 0
3 years ago
Why are household wall sockets alternating current (AC) instead of direct current (DC)? Select all that apply.
Bond [772]

Answer:

It is easier to scale the voltage of AC from high to low and low to high than with DC

Explanation:

typically power is used far away from the place where it's generated so to ensure that transmission losses( copper losses) are minimized voltage has to be stepped up during transmission..but due to the fact that most house hold equipment requires low voltage levels it has to be stepped down once it reaches a household/ domestic load...it's easier to do this for Ac than for DC.

8 0
2 years ago
A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.
LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

W = \frac{1}{2}(50)(5.61^2) = 786.8025 J

Now, we can define work:

\large\boxed{W = Fdcos\theta}}

Now, plug in the values:

786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta

Solve for theta:

cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}

4 0
3 years ago
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