Yurem is pulling a wagon across the playground with a force of 10 N. He asks Elianna to help. She agrees and pushes the back of
2 answers:
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NB: The diagram of the pulley system is not shown but the information provided is sufficient to answer the question
Answer:
Power = 2702.56 W
Explanation:
Let the power consumed be P
Energy expended = E = mgh
height, h = 5 m
E = 80 * 9.8 * 5
E = 3920 J
To calculate the time, t
From F = ma
F = 900 N
900 = 80 a
a = 900/80
a = 11.25 m/s²
From the equation of motion,
The drill head starts from rest, u = 0 m/s
Power, P = E/t
P = 3920/0.0.943
P = 4157.79 W
But Efficiency, E = 0.65
P = 0.65 * 4157.79
Power = 2702.56 W
Answer:
L = μ₀ n r / 2I
Explanation:
This exercise we must relate several equations, let's start writing the voltage in a coil
= - L dI / dt
Let's use Faraday's law
E = - d Ф_B / dt
in the case of the coil this voltage is the same, so we can equal the two relationships
- d Ф_B / dt = - L dI / dt
The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil
n d Ф_B = L dI
we can remove the differentials
n Ф_B = L I
magnetic flux is defined by
Ф_B = B . A
in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product
n B A = L I
the loop area is
A = π R²
we substitute
n B π R² = L I (1)
To find the magnetic field in the coil let's use Ampere's law
∫ B. ds = μ₀ I
where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil
s = 2π R
we solve
B 2ππ R = μ₀ I
B = μ₀ I / 2πR
we substitute in
n ( μ₀ I / 2πR) π R² = L I
n μ₀ R / 2 = L I
L = μ₀ n r / 2I
Answer:
Option (a)
Explanation:
Given that,
Mass of a car, m = 1200 kg
Force exerted by the engine, F = 600 N
Noe force,F = ma
a is the acceleration of the engine
So, the acceleration of the car is 0.5 m/s².