Complete question:
A 0.50 kilogram frog is at rest on the bank surrounding a pond of water. As the frog leaps from the bank, the magnitude of the acceleration of the frog is 4.0 meters per second^2. Calculate The magnitude of the net force exerted on the frog as it leaps.
Answer:
2.0N
Explanation:
Given that,
Mass, m of the frog = 0.5 kg
The acceleration of the frog = 4.0 m/s².
We have been asked To find,
The magnitude of the net force exerted on the frog as it leaps.
So
We calculate this using the formula below :
F = ma
When we insert the values into the formula, we have:
F = 0.5 kg × 4 m/s²
F = 2.0 N
Therefore, the magnitude of net force is 2.0 N.
The Three fission reactions will produce nine neutrons
<h3>Meaning of Fission</h3>
Fission can be defined as a nuclear process that involves the breaking of a whole nuclear matter into smaller bits.
In a fission reaction, the matter is broken down into simpler bits.
In conclusion, The Three fission reactions will produce nine neutrons
Learn more about fission:brainly.com/question/3992688
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Answer:
Please find the answer in the explanation.
Explanation:
To describe a situation in which you exert a force on something and it does not move, you need to Identify the action force and the reaction force, the relative size of the forces, and the objects they act on.
The magnitude of the static friction between the object and the ground is larger than the magnitude of the force applied.
The object will not move if the magnitude of the static friction is greater than the magnitude of the force applied.
The action force is the force applied while the reaction force is equal to the frictional force between the object and the ground.
Answer:
Can you do this questions?
Answer:
0.05 m
Explanation:
From the question given above, the following data were obtained:
Mass of first object (M1) = 9900 kg
Gravitational force (F) = 12 N
Mass of second object (M2) = 52000 kg
Distance apart (r) =?
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Thus, we can obtain the distance between the two objects as shown below:
F = GM1M2/r²
12 = 6.67×10¯¹¹ × 9900 × 52000 /r²
Cross multiply
12 × r² = 6.67×10¯¹¹ × 9900 × 52000
Divide both side by 12
r² = (6.67×10¯¹¹ × 9900 × 52000)/12
Take the square root of both side
r = √[(6.67×10¯¹¹ × 9900 × 52000)/12]
r = 0.05 m
Therefore, the distance between the two objects is 0.05 m
