The force of the tripped catch exerted on the 2.5 Kg ball moving at 8.5 m/s to the Left is 160 N
<h3>Data obtained from the question </h3>
- Initial velocity (u) = 8.5 m/s
- Final velocity (v) = 7.5 m/s
- Time (t) = 5 ms = 0.25 s
- Mass (m) = 2.5 Kg
- Force (F) = ?
<h3>How to determine the force</h3>
The force exerted on the ball can be obtained as follow:
F = m(v + u) / t
F = [2.5(7.5 + 8.5)]/ 0.25
F = 40 / 0.25
F = 160 N
Thus, the force exerted on the ball is 160 N
Learn more about momentum:
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7) PE= Fwh = (72 m) (966 N) = 69552 Joules 7
8) zero PE=mass*g*height
I hope it’s correct
All these resistors are in series so we can take the sum of them by:
Rtotal = R1 + R2 + R3......
So...
Rtotal = 2 + 3 + 4 + 6
Rtotal = 15
So now the total resistance in the circuit is 15 ohms and the potential difference applied to the circuit is 45 volts
Now we can use:
V = IR
Isolate for I
V/R = I
45/15 = I
I = 3 amps (A)
I think its friction and gravity change the motion hope this helps :)
Answer: 180N/m(to 2 significant figures)
Explanation:
According to hooked law which states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically, F = ke where;
F is the applied force in newtons
k is the elastic/spring constant in N/m
e is the extension in meters
Given applied force = 35N
extension = 20cm = 0.2m
Since F = ke,
k = F/e = 35/0.2
k = 175N/m
The spring constant is 175N/m
= 180N/m (to 2significant figures)