Answer:
56.86153 N
Explanation:
t =Time taken
F = Force
Power

Work done

The magnitude of the force that is exerted on the handle is 56.86153 N
Explanation:
average speed = total distance travelled / total time travelled
time to travel the first 6km: 6 / 50 = 3/25 (h)
time to travel the next 6km: 6 / 90 = 1/15 (h)
[I think there's problem in the question 'cause 900km/h sounds impossible for normal person to travel in normal condition]
The total time: 3/25 + 1/15 = 14/75 (h)
Average speed over the 12 km drive will be:

Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC
Answer:N=322.53 rpm
Explanation:
Given
Linear velocity (v)=1.25 m/s
Position from center is 3.7 cm
we know



and 


N=322.53 rpm
Answer:
d) 1/32 microgram
Explanation:
First half life is the time at which the concentration of the reactant reduced to half.
Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.
Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.
Forth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.
Fifth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/32.
The initial mass of the sample = 1 microgram
After 5 half-lives, the mass should reduce to 1/32 of the original.
So the concentration left = 1/32 of 1 microgram = 1/32 microgram