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Korolek [52]
3 years ago
10

Determine the acceleration that result when a 12-N net force is applied to a 3-kg object and then to a 6-kg object

Physics
1 answer:
user100 [1]3 years ago
3 0

For 3 kg mass: <em>F = \lpha ma,a = \frac{F}{m} = \frac{12N}{3kg} = 4\lpha m / s^{2}</em>

For 6 kg mass: <em>F = \lpha ma,a = \frac{F}{m} = \frac{12N}{6kg} = 2\lpha m / s^{2}</em>


Hope this helped, have a great day!~

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If a scale on Earth reads 650 N, what is your mass?
OlgaM077 [116]
If the scale reads 650N, then the mass of whoever it is standing on the scale is

         (weight) / (gravity)  =  (650N) / (9.8 m/s²)  =  66.3 kilograms  .

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4 0
3 years ago
A bird sits on top of a 639 m tall tower. If it's gravitational potential energy up there is 2033 J, what is its mass?
iris [78.8K]

The mass of the bird is 0.32 kg.

<u>Explanation:</u>

Gravitational potential energy, the energy exhibited by an object at rest due to the influence of gravitational force. So the increase in distance of object from the surface of earth leads to increase in the gravitational potential energy. Thus,

       \text {Gravitational potential energy}=m \times \text { Acceleration } \times \text { Distance of bird from bottom }

So, as the gravitational potential energy is given as 2033 J and the position of bird placed on the tall tower is 639 m away from the bottom, then the mass (m) of the bird can be found as below.

       m o f \text { bird }=\frac{\text {Gravitational potential energy}}{a \times \text {Distance}}=\frac{2033}{9.8 \times 639}=\frac{2033}{6262.2}

So, finally we get the bird's mass as,

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What is the velocity of a wave with a frequency of 760Hz and a wavelength of 0.45m?
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A rotating paddle wheel is inserted in a closed pot of water. The stirring action of the paddle wheel heats the water. During th
fgiga [73]

Answer:

the final energy of the system is 35.5 kJ.

Explanation:

Given;

initial energy of the system, E₁ = 10 kJ

heat transferred to the system, q₁  30 kJ

Heat lost to the surrounding, q₂ = 5kJ

heat gained by the system, Q = q₁ - q₂ = 30 kJ - 5kJ = 25 kJ

work done on the system, W = 500 J = 0.5 kJ

Apply first law of thermodynamic,

ΔU = Q + W

where;

ΔU  is change in internal energy

Q is the heat gained by the system

W is work done on the system

ΔU = 25kJ + 0.5 kJ

ΔU = 25.5 kJ

The final energy of the system is calculated as;

E₂ = E₁ + ΔU

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E₂ =  35.5 kJ.

Therefore, the final energy of the system is 35.5 kJ.

3 0
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