Answer:
<h2>153J</h2>
Explanation:
Step one:
given data
mass m= 8.5kg
radius r= 10cm in meters = 0.01m
velocity v= 6m/s
The work done will be the same as the energy possessed by the sphere at such speed
Required
the work one in Joules
we know that
KE=1/2mv^2
KE=1/2*8.5*6^2
KE= 1/2*8.5*36
KE=306/2
KE=153J
Answer:
Ribosomes are found 'free' in the cytoplasm or bound to the endoplasmic reticulum (ER) to form rough ER.
Answer:
Explanation:
Radius of ring r = .05 m
Electric field due to a uniformly charged ring
E = k Q x / ( x² + r² )³/² ; Q is charge on the ring , x is distance of point from the center of the ring and r is radius of the ring
electric field due to first ring at middle point or at x = 7.5 cm
E = 9 x 10⁹ x 33 x 10⁻⁹ x 7.5 x 10⁻² / ( 7.5² + 5² )³/² x 10⁻³
= 9 x 10⁹ x 33 x 10⁻⁹ x 7.5 / ( 7.5² + 5² )³/² x 10⁻¹
= 9 x 10⁹ x 33 x 10⁻⁹ x 7.5 / 73.23
= 30.41 N/C
The same field will be created by the other ring at the middle point because charge on the ring is same in magnitude . Due to negative charge on the second ring , field due to both the rings will align in the same direction.
Total field = 2 x 30.41
= 60.82 N/C
Answer:
The distance traveled by the ball is 8.5 m
Explanation:
Initial height of the ball, h₁ = 1.5 m above the ground
final height of the ball, h₂ = 5m
Upward distance = distance traveled by the ball from a height of 1.5m to 5m = 5m - 1.5m = 3.5 m
Downward distance = distance traveled by the ball from 5m height to the ground =5m - 0 = 5m
Total distance traveled = upward distance + downward distance
Total distance traveled = 3.5 m + 5m = 8.5 m
Therefore, the distance traveled by the ball is 8.5 m
