A bird accelerating from rest at a constant rate,experiences a displacement of 28 m in 11s.What is its acceleration
1 answer:
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To solve for force, you need to get the product of mass and acceleration.
F = ma
Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms
As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration.
Acceleration is the change in velocity over a period of time.
a = change in velocity/time
To get the change in velocity, you get the difference between the initial velocity and final velocity:
The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s.
So we can put it into our formula now:
WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s.
x 
= 
So your new time is 0.02s. Now we put this time into the formula:
As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop.
So now we have our acceleration. Now using this, we can get our force.
Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is
but as you can see our mass given is in grams. So again, before you put them into the equation we need to change it into kg first.
Our new mass is 0.150kg.
To make things clearer, let us write down all our new values:
m = 0.150kg
a = -2,500
Now that all our units match, we can put that into our formula:
The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign.
F = ma
Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms
As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration.
Acceleration is the change in velocity over a period of time.
a = change in velocity/time
To get the change in velocity, you get the difference between the initial velocity and final velocity:
The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s.
So we can put it into our formula now:
WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s.
So your new time is 0.02s. Now we put this time into the formula:
As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop.
So now we have our acceleration. Now using this, we can get our force.
Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is
Our new mass is 0.150kg.
To make things clearer, let us write down all our new values:
m = 0.150kg
a = -2,500
Now that all our units match, we can put that into our formula:
The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign.
Answer:
a) F_net = 6.48 10⁻¹⁸ ( ), b) x = 0.15 m
Explanation:
a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.
The electric force is given by Coulomb's law
F =
Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.
F_net= ∑ F = F₁ + F₂
let's locate a reference system in the load that is on the left side, the distances are
left side - electron r₁ = x
right side -electron r₂ = d-x
let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q
we substitute
F_net = k q Q ( )
F _net = kqQ ( )
let's substitute the values
F_net = 9 10⁹ 1.6 10⁻¹⁹ 4.50 10⁻⁹ ( )
F_net = 6.48 10⁻¹⁸ ( )
now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values of the distance (x) and the net force are given
x (m) F (N)
0.05 27.0 10-16
0.10 8.10 10-16
0.15 5.76 10-16
0.20 8.10 10-16
0.25 27.0 10-16
b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m
