<span>14.79 m/s
At the top of the loop, there's 2 opposing forces. The centripetal force that's attempting to push the roller coaster away and the gravitational attraction. These 2 forces are in opposite directions and their sum is 0.80 mg where m = mass and g = gravitational attraction. So let's calculate the amount of centripetal force we need.
0.80 = F - 1.00
1.80 = F
So we need to have a centripetal force that's 1.8 times the local gravitational attraction which is 9.8 m/s^2. So
1.8 * 9.8 m/s^2 = 17.64 m/s^2
The formula for centripetal force is
F = mv^2/r
where
F = force
m = mass
v = velocity
r = radius
We can eliminate mass from the equation since the same mass is being affected by both the centripetal force and gravity. So:
F = v^2/r
17.64 m/s^2 = v^2/12.4 m
218.736 m^2/s^2 = v^2
14.78972616 m/s = v
So the velocity at the top of the loop (rounded to 2 decimal places) is 14.79 m/s.</span>
The minimum speed needed a non-propelled object to escape from the gravitational influence of a massive body, that is to achieve an infinite distance from it. I think
<h3><u>Answer;</u></h3>
Average velocity
<h3><u>Explanation;</u></h3>
- <em><u>Average Velocity of a moving body or an object is the ratio of total displacement covered to the time interval during which the displacement occurred.</u></em>
- That is; Change in displacement divided by time.
- The formula, therefore will be;
Average velocity = Displacement/ the time interval
- Instantaneous Velocity on the other hand is the velocity of an object or a body at some instance.
Answer:
Explanation:
Rx = -28.2 units
Ry = 19.6 units
magnitude of R = √ [( - 28.2 )² + ( 19.6 ) ]
= √ ( 795.24 + 384.16 )
= 34.34 units
If θ be the angle measured counterclockwise from the +x-direction
Tanθ = 19.6 / - 28.2 = -0.695
θ = 180 - 34.8
= 145.2° .
Answer:
Q=107μC
Explanation:
Given an RC series circuit
Input voltage V=8V
Resistor =90Ω
And the capacitor=20 μF
Charge Q? After t=0.002
In an RC circuit,
The capacitor voltage is given as
Vc=V(1-e^-t/τ )
Where τ is time constant
τ =RC
τ =90×20×10^-6
τ=0.0018s
Therefore,
Vc=V(1-e^-t/τ)
Given that V=8V and t=0.002
Then,
Vc=8(1-e^-0.002/0.0018)
Vc=8(1-e^-1.111)
Vc=8(1-0.329)
Vc=8×0.671
Vc=5.37Volts
Then the voltage across the capacitor is 5.37V
From the relation
Q=CV, we can calculate the charge
Q=20×10^-6×5.37
Q=1.07×10^-4 C
Q=107μC
