A night-light uses 3 W.If the voltage is 120 V, what is the current? A.0.025A B.40 A C.360A D.0.25A
1 answer:
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The formula for the period of wave is: wave period is equals to 1 over the frequency.
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.
wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:
wave period= 1/ 200/s
get the reciprocal form of 200/s which is s/200
then you can start the actual computation:
wave period= 1 x s divided by 200
this would give us an answer of 0.005 s.
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.
wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:
wave period= 1/ 200/s
get the reciprocal form of 200/s which is s/200
then you can start the actual computation:
wave period= 1 x s divided by 200
this would give us an answer of 0.005 s.
a) 893 N
b) 8.5 m/s
c) 3816 W
d) 69780 J
e) 8030 W
Explanation:
a)
The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:
where
m is Bolt's mass
a is the acceleration
In the first 0.890 s of motion, we have
m = 94.0 kg (Bolt's mass)
(acceleration)
So, the net force is
And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).
b)
Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:
where
v is the final speed
u is the initial speed
a is the acceleration
t is the time
In the first phase of Bolt's race we have:
u = 0 m/s (he starts from rest)
(acceleration)
t = 0.890 s (duration of the first phase)
Solving for v,
c)
First of all, we can calculate the work done by Bolt to accelerate to a speed of
v = 8.5 m/s
According to the work-energy theorem, the work done is equal to the change in kinetic energy, so
where
m = 94.0 kg is Bolt's mass
v = 8.5 m/s is Bolt's final speed after the first phase
is the initial kinetic energy
So the work done is
The power expended is given by
where
t = 0.890 s is the time elapsed
Substituting,
d)
First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.
In the first 0.890 s, the force exerted was
We know that the average force for the whole race is
Which can be rewritten as
And solving for , we find the average force exerted by Bolt on the ground during the second phase:
The net force exerted by Bolt during the second phase can be written as
(1)
where D is the air drag.
The net force can also be rewritten as
where
is the acceleration in the second phase, with
u = 8.5 m/s is the initial speed
v = 12.4 m/s is the final speed
t = 8.69 t is the time elapsed
Substituting,
So we can now find the average drag force from (1):
So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:
where
d is Bolt's displacement in the second part, which can be found by using suvat equation:
And so,
e)
The power that Bolt must expend just to voercome the drag force is given by
where
is the increase in internal energy due to the air drag
t is the time elapsed
Here we have:
t = 8.69 s is the time elapsed
Substituting,
And we see that it is about twice larger than the power calculated in part c.