Answer:
The value of current generated would increase.
Explanation:
Electromagnetic induction is the process by which an electromotive force is induced due to a variation of magnetic field.
The induced current is directly proportional to rate at which the coil cuts the magnetic field. Using more powerful battery in the experiment would increase the rate at the the coil cuts the magnetic field, therefore increasing the rate of variation in the magnetic field. This effect would cause a greater deflection on the galvanometer's scale, showing an increase in the current generated.
This experiment proves that an alternating current can be produced from magnetic field.
Given parameters;
Time taken to complete a lap = 8.667s
Radius of flower = 13.9cm
convert to SI unit of m, 100cm = 1m
13.9cm gives
= 0.139m
Unknown = speed
To solve this problem, we need to first find the circumference of the flower.
Circumference of the circular flower = 2 π r
where r is the radius of the flower;
Circumference = 2 x 3.142 x 0.139 = 0.87m
Now to find the how fast the bug is travelling,
Speed = 
Since the bug covered 1 lap, the distance is 0.87m
Now input the parameters and solve for speed;
Speed =
= 0.1m/s
The bug is travelling at a speed of 0.1m/s
The water pressure at this depth and the total pressure due to water and atmosphere are 10.3 x 10⁵ Pa and 11.31× 10⁵ Pa.
<h3>What is pressure?</h3>
The pressure is the amount of force applied per unit area.
Atmospheric pressure, Patm =1.01×10⁵ Pa
Density of water, ρ=1030 kg/m³
Depth, h=100 m
Pressure =ρgh
P = 1030×10×100
P = 10.3 x 10⁵ Pa.
Total pressure, P=Po +ρgh
P=1.01×10⁵ + 1030×10×100
P=11.31× 10⁵ Pa
Hence, total pressure is 11.31× 10⁵ Pa.
Learn more about pressure.
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Answer:
The correct option is C
Explanation:
According to third equation of motion, v
2
=u
2
+2ax
Here, u=0 m/s
a=−g and x=−h
Negative sign indicates downward direction. Displacement and acceleration both are downwards.
So,v=±
2(−g)(−h)
We take minus sign because it is downwards.
v=−
2gh
After bouncing. velocity becomes 80% of v, i.e.,
v
′
=+0.8
2gh
(positive sign because the direction of ball has reversed after bouncing and is upwards.
Applying third equation of motion again, for u=v
′
, v=0 and a=−g
v
2
=u
2
+2×a×x
Thus,
0=0.64(2gh)+2(−g)x
or
x=0.64h