Y - yo = Vo*t - g * (t^2) / 2
Vo = - 9.0 m/s
t = 0.50 s
=> y - yo = -9.0 m/s * 0.5 s - 9.8 m/s^2 * (0.5s)^2 / 2 = - 4.5m - 1.225m = - 5.725 m.
Answer: option c) - 5.7
Answer:
The canon B hits the ground fast.
Explanation:
Given that,
Speed of cannon A = 85 m/s
Speed of cannon B= 100 m/s
Speed of cannon C = 75 m/s
We need to calculate the cannonballs will hit the ground with the greatest speed
Using conservation of energy
The final kinetic energy of canon depends on initial kinetic energy and potential energy.
The final velocity depends upon initial velocity and initial height.
So, the initial velocity of canon B is high.
Hence, The canon B hits the ground fast.
If I were to go from the United States to China in one second, that's a large distance in an incredibly short time. I'd say that's pretty fast.
If I were to go from my room to the door of my room in a year, then that would be unbearably slow.
Answer:
5.78amps
Explanation:
Given data
Time t= 57 seconds
Charge Q= 330C
Current I= ??
The expression for the electric current is given as
Q= It
Substituting we have
330= I*57
I= 330/57
I=5.78 amps
Hence the current is 5.78amps
From the information given, cannon ball weighs 40 kg and has a potential energy of 14000 J.
We need to find its height.
We will use the formula P.E = mgh
Therefore h = P.E / mg
where P.E is the potential energy,
m is mass in kg,
g is acceleration due to gravity (9.8 m/s²)
h is the height of the object's displacement in meters.
h = P.E. / mg
h = 14000 / 40 × 9.8
h = 14000 / 392
h = 35.7
Therefore the canon ball was 35.7 meters high.
