Answer:
3 seconds
Explanation:
Applying,
Applying,
v = u±gt................ Equation 1
Where v = final velocity, u = initial velocity, t = time, g = acceleration due to gravity.
From the question,
Given: v = 0 m/s ( at the maximum height), u = 30 m/s
Constant: g = -10 m/s
Substitute these values into equation 1
0 = 30-10t
10t = 30
t = 30/10
t = 3 seconds
Answer:
1.01 × 10⁵ Pa
Explanation:
At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.
We need to find the total pressure on the air in the lungs of a person to a depth of 1 meter.
Pressure at a depth is given by :

Where
is the density of air, 
So,

Total pressure, P = Atmospheric pressure + 12 Pa
= 1.013 × 10⁵ Pa + 12 Pa
= 1.01 × 10⁵ Pa
Hence, the total pressure is 1.01 × 10⁵ Pa.
Via half-life equation we have:

Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:

There is 6.25 grams left of Ra-229 after 12 minutes.
Answer:
Explanation:
A rectifier is an electrical device that converts alternating current (AC) to direct current (DC), a process known as rectification. Rectifiers have many uses including as components of power supplies and as amplitude modulation detectors (envelope detectors) of radio signals. Rectifiers are most commonly made using solid state diodes but other type of components can be used when very high voltages or currents are involved. When only a single diode is used to rectify AC (by blocking the negative or positive portion of the waveform), the difference between the term diode and the term rectifier is simply one of usage. The term rectifier describes a diode that is being used to convert AC to DC. Most rectifier circuits contain a number of diodes in a specific arrangement to more efficiently convert AC power to DC power than is possible with only a single diode.
Answer:
Part a)
%
Part b)
%
Explanation:
As we know that total power used in the room is given as

here we have






Part a)
Since power supply is at 110 Volt so the current obtained from this supply is given as


now resistance of transmission line



now power loss in line is given as



Now percentage loss is given as


%
Part b)
now same power must have been supplied from the supply station at 110 kV, so we have


now power loss in line is given as



Now percentage loss is given as


%