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OLEGan [10]
3 years ago
12

A 10.0kg water balloon is dropped from a height of 12.0m. Calculate the speed of the balloon just before it hits the ground

Physics
1 answer:
kolbaska11 [484]3 years ago
5 0

Answer:

15.5 m/s.

Explanation:

Potential energy of the balloon has been converted to kinetic energy.

potential energy = kinetic energy.

mgh = ½mv².

10* 10* 12= ½ *10 *v²

1200 = 5v²

v²=1200÷5

v=√240

v= 15.49~15.5 m/s.

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An organ pipe of length 3.0 m has one end closed. The longest and next-longest possible wavelengths for standing waves inside th
iris [78.8K]

Answer:

The longest wavelength for closed at one end and open at the other is

y / 4      where y is the wavelength - that is node - antinode

The next possible wavelength is 3 y / 4 -    node - antinode - node -antinode

y / 4 = 3 m     y = 12 meters    the longest wavelength

3 y / 4 = 3 m      y = 4 meters   1 / 3 times as long

6 0
3 years ago
You drop a ball off the roof. It takes 2.5s for the ball to hit the bottom. How fast is it going when it hits the ground?
Anni [7]

Answer:

1.5 thats how long it is

Explanation:

good luck

6 0
2 years ago
Two children are riding on the edge of a merry-go-round that has a mass of 100.kg and radius of 1.60m and is rotating at 20.0rpm
Gre4nikov [31]

Here since both children and merry go round is our system and there is no torque acting on this system

So we will use angular momentum conservation in this

I_1\omega_1 = I_2\omega_2

now here we have

I_1 = \frac{MR^2}{2} + m_1R^2 + m_2R^2

I_1 = \frac{100(1.60)^2}{2} + (22 + 28)(1.60)^2

I_1 = 256

Now when children come to the position of half radius

then we will have

I_2 = \frac{MR^2}{2} + m_1(\frac{R}{2})^2 + m_2(\frac{R}{2})^2

I_2 = \frac{100(1.6)^2}{2} + (28 + 22)(0.8)^2

I_2 = 160

now from above equation we have

256 (20.0 rpm) = 160(\omega_2)

\omega_2 = 32 rpm

8 0
3 years ago
A lightbulb has an efficiency of 13.1%.How much light energy (not heat energy) is generated by the bulb every second if the bulb
maw [93]

Answer:

1.048 W

Explanation: Given that a lightbulb has an efficiency of 13.1%.How much light energy (not heat energy) is generated by the bulb every second if the bulb has a power pf 8W?

The input power = 8W

The output power = ?

The efficiency = 13.1%

Using the formula below:

Efficiency = (power output / power input) × 100

13.1 = (Power output / 8 ) × 100

0.131 = power output / 8

Cross multiply to make the power output the subject of formula

Power output = 0.131 × 8

Power output = 1.048 W

Therefore, the light energy (not heat energy) generated by the bulb every second is 1.048 W

4 0
3 years ago
If the activation energy for a given compound is found to be 103 kJ/mol, with a frequency factor of 4.0 × 1013 s-1, what is the
Crank

Answer: The Rate constant is 1.209

Explanation:

in the attachment

7 0
3 years ago
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