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 efficiency of the engine at the given useful energy and input energy is 28%.
The given parameters:
- <em>Input energy of the engine, = 125 J</em>
- <em>Useful energy of the engine, 35 J</em>
The efficiency of the engine is defined as the ratio of the output energy (useful energy) to the input energy and it is calculated as follows;
Thus, the efficiency of the engine at the given useful energy and input energy is 28%.
Learn more about efficiency of engine here: brainly.com/question/6751595
Answer:
sure i'll help. but with what?
i'll help
Answer:
44.3 m/s
Explanation:
Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.
What is the magnitude of its velocity just before it strikes the ground ?
The parameters given are:
Height H = 100m
Since the ball is thrown from a top of a building, initial velocity U = 0
Let g = 9.8m/s^2
Using third equation of motion
V^2 = U^2 + 2gH
Substitute all the parameters into the formula
V^2 = 2 × 9.8 × 100
V^2 = 200 × 9.8
V^2 = 1960
V = 44.27 m/s
Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately
Answer:
t = (ti)ln(Ai/At)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Explanation:
Let Ai represent the initial amount and At represent the final amount of beryllium-11 remaining after time t
At = Ai/2^n ..... 1
Where n is the number of half-life that have passed.
n = t/half-life
Half life = 14
n = t/14
At = Ai/2^(t/14)
From equation 1.
2^n = Ai/At
Taking the natural logarithm of both sides;
nln(2) = ln(Ai/At)
n = ln(Ai/At)/ln(2)
Since n = t/14
t/14 = ln(Ai/At)/ln(2)
t = 14ln(Ai/At)/ln(2)
Ai = 800
At = 50
t = 14ln(800/50)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Let half life = ti
t = (ti)ln(Ai/At)/ln(2)
