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Morgarella [4.7K]
3 years ago
15

A ball of mass of 0.5kg is dropped from a height of 2m.

Physics
1 answer:
erastovalidia [21]3 years ago
3 0

Answer:

6.3m/s

Explanation:

Given parameters:

Mass of the ball  = 0.5kg

Height  = 2m

Unknown:

Velocity when the ball hits the ground  = ?

Solution:

Since the potential  energy is transformed into kinetic energy;

            P.E  = K.E

              mgh  = \frac{1}{2} m v²

     cancelling m;

               gh  =  \frac{1}{2} v²

                v² = 2gh

                v  = √2gh

Insert the parameters and solve;

               v  = √2gh  = √ 2 x 9.8 x 2  = 6.3m/s

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You're in your room blasting music with door shut, your mom opens your door. Now music is heard through out your home. this is e
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Answer:

A

Explanation:

Diffraction, as the waves spread out (specifically spread to the whole house) by passing the door.

5 0
2 years ago
A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow
shusha [124]

Answer:

150000000

\dfrac{F_e}{F_g}=0.00000203873598369

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

1\ C=6.25\times 10^{18}\ electrons

Number electrons is

n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000

The number of electrons added or removed was 150000000

Force is given by

F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N

The ratio is

\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369

The ratio is \dfrac{F_e}{F_g}=0.00000203873598369

Balancing the forces we get

Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C

The electric field required is 49050000 N/C

4 0
3 years ago
Which kind of good might not be bought when prices rise?
IgorC [24]

luxury goods might not be bought

5 0
3 years ago
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Suppose a light source is emitting red light at a wavelength of 700 nm and another light source is emitting ultraviolet light at
klasskru [66]

Answer:

b) twice the energy of each photon of the red light.

Explanation:

\lambda = Wavelength

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

c = Speed of light = 3\times 10^8\ m/s

Energy of a photon is given by

E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}

Let \lambda_1 = 700 nm

\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}

For red light

E_1=\dfrac{hc}{\lambda_1}

For UV light

E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}

Dividing the equations

\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1

Hence, the answer is  b) twice the energy of each photon of the red light.

7 0
3 years ago
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I need help with this please
Firdavs [7]
C is the answer to the question
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3 years ago
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