Power delivered = (energy delivered) / (time to deliver the energy)
Power delivered = (4,000 J) / (0.5 sec)
Power delivered = 8,000 watts
I'm a little surprised to learn that Electro draws his power from the mains. This is VERY good news for Spiderman ! It means that Spiderman can always avoid tangling with Electro ... all he has to do is stay farther away from Electro than the length of Electro's extension cord.
But OK. Let's assume that Electro draws it all from the mains. Then inevitably, there must be some loss in Electro's conversion process, between the outlet and his fingertips (or wherever he shoots his bolts from).
The efficiency of Electro's internal process is
<em>(power he shoots out) / (power he draws from the mains) </em>.
So, if he delivers energy toward his target at the rate of 8,000 watts, he must draw power from the mains at the rate of
<em>(8,000 watts) / (his internal efficiency) . </em>
The equator has no continental borders.
Answer:
The frequency is 
Explanation:
From the question we are told that
The frequency of the tuning fork is 
The beat period is 
Generally the beat frequency is mathematically represented as
The beat frequency is also represented mathematically as
Where 
is the frequency of the piano
So
Answer:
Explanation:
a )
Reaction force of the ground
R = mg
= 160 N
Maximum friction force possible
= μ x R
= μ x 160
= .4 x 160
= 64 N .
b )
160 N will act at middle point . 740N will act at distance of 3 / 5 m from the wall ,
Taking moment about top point of ladder
160 x 1.5 + 740 x 3/5 + f x 4 = 900 x 3
240 + 444 + 4f = 2700
f = 504 N
c )
Let x be the required distance.
Taking moment about top point of ladder
160 x 1.5 + 740 x 3 x / 5 + .4 x 900 x 4 = 900 x 3 ( .4 x 900 is the maximum friction possible )
240 + 444 x + 1440 = 2700
x = 2.3 m
so man can go upto 2.3 at which maximum friction acts .
Answer: D. 5cm
Explanation:
Given the following :
Focal length (f) = - 6.0 cm
Height of object = 15.0cm
Distance of object from mirror (u) = 12.0cm
Height of image produced by the mirror =?
Firstly, we calculate the distance of the image from the mirror.
Using the mirror formula
1/f = 1/u + 1/v
1/v = 1/f - 1/u
1/v = 1/-6 - 1/12
1/v = - 1/6 - 1/12
1/v = (- 2 - 1) / 12
1/v = - 3 / 12
v = 12 / - 3
v = - 4
Using the relation :
(Image height / object height) = (- image distance / object distance)
Image height / 15 = - (-4) / 12
Image height / 15 = 4 / 12
Image height = (15 × 4) / 12
Image height = 60 / 12
Image height = 5cm
