1) 12 cm
2) 3 N
Explanation:
1)
The relationship between force and elongation in a spring is given by Hooke's law:
where
F is the force applied
k is the spring constant
x is the elongation
For the spring in this problem, at the beginning we have:
So the spring constant is
Later, the force is tripled, so the new force is
Therefore, the new elongation is
2)
In this second problem, we know that the elongation of the spring now is
From part a), we know that the spring constant is
Therefore, we can use the following equation to find the force:
And substituting k and x, we find:
So, the force to produce an elongation of 6 cm must be 3 N.
<span>1) The differential equation that models the RC circuit is :
(d/dt)V_capacitor </span>+ (V_capacitor/RC) = (V_source/<span>RC)</span>
<span>Where the time constant of the circuit is defined by the product of R*C
Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s
2) C<span>harge of the capacitor 1.57 time constants
1.57*(2.742) = 4.3048 s
The solution of the differential equation is
</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T) + </span>V_capac(∞)
Since the capacitor is initially uncharged V_capac(0) = 0
And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery V_capac(∞) = 9V
This means,
V_capac (t) = (-9V)e ^(-t /T) + 9V
The charge in a capacitor is defined as Q = C*V
Where C is the capacitance and V is the Voltage across
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V = -1.87V +9V
V_capac (4.3048 s) = 7.1275 V
Q (4.3048 s) = 89.9mF*(7.1275V) = 0.6407 C
3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V
Q (∞) = 89.9mF*(9V) = 0.8091 C
ANSWER IS A)
THE SPEED IS 25 M/S FOR BOTH PEOPLE
The magnitude of the displacement of the car from the starting point to halfway around the track is 256 m.
Answer:
Explanation:
Since the race track is a circular track, the distance for one lap will be equal to the circumference of the circular track. And the circumference will be equal to the circumference of the circle.
Since the radius of the track is given as 200 m, then the circumference of the circular track will be
Circumference = 2πr = 2 × 3.14 × 200
So the circumference of the circular track = 1256 m.
So the starting point or position of the track is considered as zero and if the car has traveled half way means, the car has covered half of the circumference of the track.
As the circumference = 1256 m, then half of the circumference of the circle = 1256/2 = 256 m.
So the displacement is the measure of difference between the final position and initial position. As here the initial position is zero and the final position is the halfway around the track which is equal to 256 m.
Then Displacement = Final-Initial = 256-0= 256 m.
So the magnitude of the displacement of the car from the starting point to halfway around the track is 256 m.
Answer:843.75kg
Explanation:
kinetic energy(ke)=432000j
Velocity(v)=32m/s
Mass(m)=?
Ke=(mxv^2)/2
432000=(mx32^2)/2
432000=(mx32x32)/2
Cross multiplying we get
432000x2=(mx1024)
864000=1024m
Divide both sides by 1024
864000/1024=1024m/1024
843.75=m
m=843.75kg
