Answer:
A
Explanation:
A. The pencil is on the table in broad daylight
Answer:
Density (φ) = 0,8827 Kg/L
Specific weight (Ws) = 8,65 N/L
Specific gravity (Gs) = 0,8827 (without unit)
Explanation:
The density formula: φ = 
I know the mass "m", I need to find out the volume of the cylinder (V)
V = π* r²*h
The radius "r" is equal to half the diameter (150mm) = 75mm
Now I can find out the density (φ)
φ = 
= 0,8827 Kg/L
The specific weight (Ws) is the relationship between the weight of substance (oil) and its volume. We apply the following formula:
Ws = φ*g
(g = gravity = 9,8 m/s²)
Finally, specific gravity (Gs) is the ratio between the density of a substance (oil) "φ(o)" and the density of water "φ(w)" :
Gs = φ(o) / φ(w)
(φ(w) = 1 Kg/L
Hope this can help you !!
Answer:
Explanation:
<u>Horizontal Launch</u>
When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it eventually hits the ground.
The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:
vx=v
The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:
The horizontal component of the velocity is always the same:
The vertical component at t=5.5 s is:
Answer:
R₁ = 50.77 Ω
Explanation:
Since, we know that:
Electric Power = P = VI
but from Ohm's Law:
V = IR
(or) I = V/R
Therefore,
P = V²/R
(OR) R = V²/P
where,
V = Battery Voltage
R = Resistance of combination
FOR SERIES COMBINATION:
R = Rs = (57 V)²/48 W
Rs = 67.69 Ω
but, we know that:
Rs = R₁ + R₂
R₁ + R₂ = 67.69 Ω
R₁ = 67.69 Ω - R₂ __________ eqn (1)
FOR PARALLEL COMBINATION:
R = Rp = (57 V)²/256 W
Rp = 12.69 Ω
but, we know that:
Rp = (R₁R₂)/(R₁ + R₂) = 12.69 Ω
using eqn (1) and value of R₁ + R₂, we get
Rp = 12.69 = R₂(67.69 - R₂)/67.69
859.08 = 67.69 R₂ - R₂²
R₂² - 67.69 R₂ + 859.08 = 0
Solving this quadratic equation we get the answers:
Either, R₂ = 50.76 Ω
Either, R₂ = 16.92 Ω
Since, it is stated in the question that R₁ > R₂. Therefore, we choose the second value. So,
<u>R₂ = 16.92 Ω</u>
using this value in eqn (1), we get:
R₁ = 67.69 Ω - 16.92 Ω
<u>R₁ = 50.77 Ω</u>
Answer:
a) 1450watts
b) 564watts
c) 1.11
Explanation:
Power consumed = IV
I is the current rating
V is the operating voltage
If a blow-dryer and a vacuum cleaner each operate with a voltage of 120 V and the current rating of the blow-dryer is 12 A, while that of the vacuum cleaner is 4.7 A then their individual power rating is calculated thus;
a) For blow-dryer
Operating voltage = 120V
Its current rating = 12A
Power consumed = IV
= 120×12
= 1440watts
b) For vacuum cleaner:
Operating voltage is the same as that of blow dryer = 120V
Its current rating = 4.7A
Power consumed = IV
= 120×4.7
= 564watts
c) Energy used = Power consumed × time taken
Energy used = Power × time
Energy used by blow dryer = 1440×20×60
= 1,728,000Joules
Energy used up by vacuum cleaner = 564×46×60
= 564×2760
= 1,556,640Joules
Ratio of the energy used by the blow-dryer in 20 minutes to the energy used by the vacuum cleaner in 46 minutes will be 1,728,000/1,556,640 = 1.11
