Answer:
you absolute buffoon Use Ohms' Law: V = RI
V = (1x10^3)(5x10^-3) = 5 volts
Yes, this is in the range of normal household voltages.
Explanation:
(D)
Explanation:
The more massive an object is, the greater is the curvature that they produce on the space-time around it.
Answer:
The velocity of water at the bottom, 
Given:
Height of water in the tank, h = 12.8 m
Gauge pressure of water, 
Solution:
Now,
Atmospheric pressue, 
At the top, the absolute pressure, 
Now, the pressure at the bottom will be equal to the atmopheric pressure, 
The velocity at the top,
, l;et the bottom velocity, be
.
Now, by Bernoulli's eqn:

where

Density of sea water, 



Answer: A chemical equation describes a chemical reaction. Reactants are starting materials and are written on the left-hand side of the equation. Products are the end-result of the reaction and are written on the right-hand side of the equation.
Explanation:
Answer:
a) 2 m/s
b) i) 
ii) 
Explanation:
The function for distance is 
We know that:
Velocity = 
Acceleration = 
To find speed at time t = 0, we derivate the distance function:

Substitute t = 0 in velocity function:

Velocity at t = 0 will be 2 m/s.
To find the function for Kinetic Energy of the box at any time, t.

We know that 

