(a) 392 N/m
Hook's law states that:
(1)
where
F is the force exerted on the spring
k is the spring constant
is the stretching/compression of the spring
In this problem:
- The force exerted on the spring is equal to the weight of the block attached to the spring:

- The stretching of the spring is

Solving eq.(1) for k, we find the spring constant:

(b) 17.5 cm
If a block of m = 3.0 kg is attached to the spring, the new force applied is

And so, the stretch of the spring is

And since the initial lenght of the spring is

The final length will be

- Height (h) = 10 m
- Density (ρ) = 1000 Kg/m^3
- Acceleration due to gravity (g) = 10 m/s^2
- We know, pressure in a fluid = hρg
- Therefore, the pressure exerted by a column of fresh water
- = hρg
- = (10 × 1000 × 10) Pa
- = 100000 Pa
<u>Answer</u><u>:</u>
<u>1000</u><u>0</u><u>0</u><u> </u><u>Pa</u>
Hope you could understand.
If you have any query, feel free to ask.
Answer: a) 139.4 μV; b) 129.6 μV
Explanation: In order to solve this problem we have to use the Ohm law given by:
V=R*I whre R= ρ *L/A where ρ;L and A are the resistivity, length and cross section of teh wire.
Then we have:
for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω
and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω
Finalle we calculate the potential difference (V) for both wires:
Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V
V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V
I wanna say B Red-billed oxpeckers eat ticks off of impalas
Explanation:
It is given that,
Diameter of the circular loop, d = 1.5 cm
Radius of the circular loop, r = 0.0075 m
Magnetic field, 
(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :



I = 32.22 A
(b) The magnetic field on a current carrying wire is given by :



r = 0.00238 m

Hence, this is the required solution.