A) Two workers are trying to move a heavy crate. One pushes onthe crate with a force A, which has amagnitude of 445 newtons and
1 answer:
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The period of a simple pendulum is given by:
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
(1)
We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.
We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
And its period is the reciprocal of its frequency:
So now we can use eq.(1) to find the gravitational acceleration of the planet:
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.
We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
And its period is the reciprocal of its frequency:
So now we can use eq.(1) to find the gravitational acceleration of the planet:
Given
Car 1
m1 = 1300 kg
v1 = 20 m/s
m2 = 900 kg
v2 = -15 m/s
(Negative sign shows that direction of car 2 is opposite to car 1)
Procedure
As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,
Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.
Answer:
magnitude of force on charge 2Q =
Direction of force on charge = 61 ⁰
Explanation:
The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q i.e x-component of the net force and the y-component of the net force
║F║ = = after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW
magnitude of force acting on 2Q =
The direction of the force on charge 2Q is calculated as
tan ∅ = = 1.8284
therefore ∅ = 1.8284
= 61⁰