Answer:
C. 28.09 amu
Explanation:
The natural occurring element exist in 3 isotopic forms: namely X-28 (27.977 amu, 92.23% abundance), X-29 (28.976 amu, 4.67% abundance) and X-30 (29.974 amu, 3.10% abundance).
The atomic weight of elements depends on the isotopic abundance. If you know the fractional abundance and the mass of the isotopes the atomic weight can be computed.
The atomic weight is computed as follows:
atomic weight = mass of X-28 × fractional abundance + mass of X-29 × fractional abundance + mass of X-30 × fractional abundance
atomic weight = 27.977 × 0.9223 + 28.976 × 0.0467 + 29.974 × 0.0310
atomic weight = 25.8031871 + 1.3531792 + 0.929194
atomic weight = 28.0855603 amu
To 2 decimal place atomic weight = 28.09 amu
Given Information:
Power = P = 100 Watts
Voltage = V = 220 Volts
Required Information:
a) Current = I = ?
b) Resistance = R = ?
Answer:
a) Current = I = 0.4545 A
b) Resistance = R = 484 Ω
Explanation:
According to the Ohm’s law, the power dissipated in the light bulb is given by

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.
Re-arranging the above equation for current I yields,

Therefore, 0.4545 A current is flowing through the light bulb.
According to the Ohm’s law, the voltage across the light bulb is given by

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.
Re-arranging the above equation for resistance R yields,

Therefore, the resistance of the bulb is 484 Ω
Answer:
0.66
Explanation:
By using the formula
u = v^2 / r g
Where u is coefficent of friction
u = 23.5 × 23.5 / (85 × 9.8)
u = 0.66
To solve this problem we will use the concepts related to power, defined as the amount of energy applied over a period of time.
The energy in this case is the accumulated in the form of potential energy, over a period of time. Thus we will have that the mathematical expression of the power can be expressed as

Here,
E = Energy
t = time
As the energy is equal to the potential Energy we have tat

The weight (mg) of the man is 700N, the height (h) is 8m and the time is 10s, then:


Therefore the correct answer is A.
Tides are influenced by the force of gravity exerted by the earth, moon and the sun. The sun has a larger mass than the moon and as such has a greater gravitational pull on the earth. the moon however has greater influence over the tides because they are caused by the difference in gravity fields. This means that the moon is the dominant influence due to the fact that the fractional difference in its force across the earth is greater than that seen from the sun.