Answer:
The bottom of the sea is 25 m below sea level.
Explanation:
Given data
Mass = 6.1 × ![10^{8} \ kg](https://tex.z-dn.net/?f=10%5E%7B8%7D%20%5C%20kg)
![\rho_{sea} = 1020\ \frac{kg}{m^{3} }](https://tex.z-dn.net/?f=%5Crho_%7Bsea%7D%20%3D%201020%5C%20%20%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%20%7D)
We know that Buoyant force on the tank is equal to gravity force of the tank.
![F_B = F_g](https://tex.z-dn.net/?f=F_B%20%3D%20F_g)
![(\rho_{Fluid}) (g) (V_{disp}) = m g](https://tex.z-dn.net/?f=%28%5Crho_%7BFluid%7D%29%20%28g%29%20%28V_%7Bdisp%7D%29%20%3D%20m%20g)
![(\rho_{Fluid}) (V_{disp}) = m](https://tex.z-dn.net/?f=%28%5Crho_%7BFluid%7D%29%20%20%28V_%7Bdisp%7D%29%20%3D%20m)
1020 ×
= 6.1 × ![10^{8}](https://tex.z-dn.net/?f=10%5E%7B8%7D)
= 598039.21 ![m^{3}](https://tex.z-dn.net/?f=m%5E%7B3%7D)
We know that
= W × L × H
598039.21 = 300 × 80 × H
H = 25 m
Therefore the bottom of the sea is 25 m below sea level.
Answer:
<em>The magnetic field through the coil at first increases steadily up to its maximum value, and then decreases gradually to its minimum value.</em>
<em></em>
Explanation:
At first, the magnet fall towards the coils; inducing a gradually increasing magnetic field through the coil as it falls into the coil. At the instance when half the magnet coincides with the coil, the magnetic field magnitude on the coil is at its maximum value. When the magnet falls pass the coil towards the floor, the magnetic field then starts to decrease gradually from a strong magnitude to a weak magnitude.
This action creates a changing magnetic flux around the coil. The result is that an induced current is induced in the coil, and the induced current in the coil will flow in such a way as to oppose the action of the falling magnet. This is based on lenz law that states that the induced current acts in such a way as to oppose the motion or the action that produces it.
Force can alter its direction,slow or stop it you could say it can change its velocity
Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together. This is done because the friction force is going to have to be compensated for. We will need that much more force than we otherwise would to achieve the desired acceleration:
![F_{NoFric}=ma=0.39kg \times0.18 \frac{m}{s^2} =0.0702N](https://tex.z-dn.net/?f=F_%7BNoFric%7D%3Dma%3D0.39kg%20%5Ctimes0.18%20%5Cfrac%7Bm%7D%7Bs%5E2%7D%20%20%3D0.0702N)
The friction force will be given by the normal force times the coefficient of friction. Here the normal force is just its weight, mg
![F_{Fric}=0.39kg \times 9.8 \frac{m}{s^2} \times 0.21=0.803N](https://tex.z-dn.net/?f=F_%7BFric%7D%3D0.39kg%20%5Ctimes%209.8%20%5Cfrac%7Bm%7D%7Bs%5E2%7D%20%5Ctimes%200.21%3D0.803N%20)
Now the total force required is:
0.0702N+0.803N=0.873N
Answer:
Explanation:
3.4 m/s due North, -1.1 m/s due East