Given :
Initial speed , u = 0 m/s .
Final speed , v = 91 km/h = 25.28 m/s .
To Find :
a) Average acceleration .
b ) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.3 s .
Solution :
a )
We know ,by equation of motion :
b)
Also , by equation of motion :
Hence , this is the required solution .
Salutations!
If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, _______________ is being done.
<span>If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, work is being done. Energy being transferred and the object begins to move is called work.
Thus, your answer is option B.
Hope I helped (:
Have a great day!</span>
Answer:
The datapoint 9.0 ppm is outlier at the 90% confidence level.
Explanation:
The old data has following values
mean=10.5 mm
standard deviation 0.2 mm
Now the mean of new values is calculated as following
So the value as 9.0 ppm can be considered easily as outlier in this regard.
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
