The answers To your question is c
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
Answer:
Explanation:
Given
Resistor A has length 
and Resistor B has Length 
and Resistance is given by
Considering 
and A to be constant thus
because 
(a)When they are connected in series
As the current in series is same and power is 
therefore 
as R is greater for second resistor
(b)if they are connected in Parallel
In Parallel connection Voltage is same
resistance of 2 is greater than 1 thus Power delivered by 1 is greater than 2
Answer:
Explanation:
one end of tank will be circular in shape . Area of circle A
= π r² , r is radius of the circle
= 3.14 x 3²
A = 28.26 ft³
To calculate force on the circular area , we first find pressure at the center of the circle which is at depth equal to r
pressure at the center = h d g ' here h = depth = r , d = density of milk
pressure = 3 x 64.6 x 32 poundal / ft²
= 6201.6 poundal / ft²
total force on circular face = pressure at the center x area of circle
= 6201.6 x 28.26
= 175257.21 poundal .
