Answer:
I = 3.785 W/m²
Explanation:
given,
distance from the light, r = 1.45 m
Power of bulb = 100 W
The area is that of a sphere with a radius equal to the distance from the light bulb.
The formula for the area of a sphere of radius 1.45 meters is:
A = 4 π r²
A = 4 x π x 1.45²
A = 26.42 m²
Intensity of the light is equal to
I = 3.785 W/m²
Intensity of the light at 1.45 m is equal to I = 3.785 W/m²
Work = (force) x (distance.
The force required to lift the load is its weight.
Weight = (mass) x (gravity)
so Work = (mass) x (gravity) x (distance)
Now Power = (work) / (time)
so Power = (mass) x (gravity) x (distance) / (time)
= (700kg) x (9.8 m/s²) x (2 m) / (0.4 sec)
= ( 700 x 9.8 x 2) / (0.4) (kg-m²/sec²) / (sec)
= ( 34,300 ) (joule) / (sec)
= 34,300 watts .
This is one of those exercises where the math and the physics
are air-tight and bullet-proof but the answer is absurd.
34,300 watts is about 46 horsepower. I don't care how many
Wheaties Power Lifter Paul had for breakfast today, he is NOT
snatching a barbell that weighs 1,543 pounds (0.77 ton !)
to the height of the top of his head in less than 1/2 second !
Answer
The Little Bee will have a greater force of 12N
<u>Explanation</u>
As we know that Force is equal to the product of mass and acceleration (F=ma)
The Force with which the Big Bee will hit the windshield = (5kg x 2m/s^2)
Force of Big Bee= 10 Newton (N)
The Force with which the Little Bee will hit the windshield = (3kg x 4m/s^2)
Force of Little Bee = 12N
Hence, the little bee will hit the windshield with a greater force.
Answer:
the answer is 11 N left
Explanation:
there is more force being applied in the direction left, so the ball will move left. to find the net force that the ball will move in that direction subtract the force being applied in the opposite direction. so, 16N-5N=11N. your answer in 11 N left.
Answer:
a,b,c and d.
Explanation:
Of the following statements the correct are:
a) It is possible to induce a current in a closed loop of wire located in a uniform magnetic field by either increasing or decreasing the area enclosed by the loop.
b) It is possible to induce a current in a closed loop of wire without the aid of a power supply or battery.
d) It is possible to induce a current in a closed loop of wire by change the orientation of a magnetic field enclosed by the wire.
e) It is possible to induce a current in a closed loop of wire by changing the strength of a magnetic field enclosed by the wire.
