Answer:
1)a. It is constant the whole time the ball is in free-fall.
2)b. = 14 m/s
3) e. = 19.6 m/s
Explanation:
1) given that the only force acting on the ball is gravity, gravity acts along the vertical axis. Since no other force acts on the ball then the horizontal velocity will remain constant all through the flight since there is no horizontal force acting on the ball.
2) speed = distance/time
horizontal distance = 56m
Time = 4 seconds
Speed = 56m/4s = 14m/s
3) acceleration due to gravity g = 9.8m/s^2
Initial vertical velocity = u
Final vertical velocity = v = -u
Using the law of motion;
v = u + at
a = acceleration = -g = -9.8m/s^2
t = time of flight = 4
Substituting the values;
-u = u - 4(9.8)
-2u = -4(9.8)
u = -4(9.8)/-2
u = 2(9.8) = 19.6 m/s
Initial vertical velocity = u = 19.6 m/s
Answer:
1.) 274.5v
2.) 206.8v
Explanation:
1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.
The potential difference and charge across EACH capacitor will be
V = Voe
Where Vo = initial voltage
e = natural logarithm = 2.718
For the first capacitor 2.50 µF,
V = Vo × 2.718
746 = Vo × 2.718
Vo = 746/2.718
Vo = 274.5v
To calculate the charge, use the below formula.
Q = CV
Q = 2.5 × 10^-6 × 274.5
Q = 6.86 × 10^-4 C
For the second capacitor 6.80 µF
V = Voe
562 = Vo × 2.718
Vo = 562/2.718
Vo = 206.77v
The charge on it will be
Q = CV
Q = 6.8 × 10^-6 × 206.77
Q = 1.41 × 10^-3 C
B.) Using the formula V = Voe again
165 = Vo × 2.718
Vo = 165 /2.718
Vo = 60.71v
Q = C × 60.71
Q = C
Scientific Notation: 4.580 x 10^-4
Scientific e Notation: 4.580e-4
Answer:
(a) r = 1.062·R
= 
(b) r = 
(c) Zero
Explanation:
Here we have escape velocity v
given by
and the maximum height given by
Therefore, when the initial speed is 0.241v we have
v = 
so that;
v² = 
v² = 
is then
Which gives
or
r = 1.062·R
(b) Here we have
Therefore we put 
in the maximum height equation to get
From which we get
r = 1.32·R
(c) The we have the least initial mechanical energy, ME given by
ME = KE - PE
Where the KE = PE required to leave the earth we have
ME = KE - KE = 0
The least initial mechanical energy to leave the earth is zero.
Answer:
the object will travel 0.66 meters before to stop.
Explanation:
Using the energy conservation theorem:
The work done by the friction force is given by:
![W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]](https://tex.z-dn.net/?f=W_f%3DF_f%2Ad%5C%5CW_f%3D%5C%C2%B5%2Am%2Ag%2Ad%5C%5CW_f%3D0.35%2A4%2A9.81%2Ad%5C%5CW_f%3D13.7d%5BJ%5D)
so:
