Answer:
Δ h = 52.78 m
Explanation:
given,
Atmospheric pressure at the top of building = 97.6 kPa
Atmospheric pressure at the bottom of building = 98.2 kPa
Density of air = 1.16 kg/m³
acceleration due to gravity, g = 9.8 m/s²
height of the building = ?
We know,
Δ P = ρ g Δ h
(98.2-97.6) x 10³ = 1.16 x 9.8 x Δ h
11.368 Δ h = 600
Δ h = 52.78 m
Hence, the height of the building is equal to 52.78 m.
Answer:
1.551×10^-8 Ωm
Explanation:
Resistivity of a material is expressed as shown;.
Resistivity = RA/l
R is the resistance of the material
A is the cross sectional area
l is the length of the wire.
Given;
R = 0.0310 Ω
A = πd²/4
A = π(2.05×10^-3)²/4
A = 0.000013204255/4
A = 0.00000330106375
A = 3.30×10^-6m
l = 6.60m
Substituting this values into the formula for calculating resistivity.
rho = 0.0310× 3.30×10^-6/6.60
rho = 1.023×10^-7/6.60
rho = 1.551×10^-8 Ωm
Hence the resistivity of the material is 1.551×10^-8 Ωm
The density of the object is the ratio of its mass and volume. From the given dimensions above, we determine the volume through the equation,
V = L x W x H
Substituting,
V = (3 cm)(2 cm)(1 cm) = 6 cm³
From the idea presented above,
d = m/V
Substituting the known values,
d = (30 g)/ (6 cm³) = 5 g/cm³
ANSWER: 5 g/cm³
Answer:
2.32 s
Explanation:
Using the equation of motion,
s = ut+g't²/2............................ Equation 1
Where s = distance, u = initial velocity, g' = acceleration due to gravity of the moon, t = time.
Note: Since Onur drops the basket ball from a height, u = 0 m/s
Then,
s = g't²/2
make t the subject of the equation,
t = √(2s/g')...................... Equation 2
Given: s = 10 m, g' = 3.7 m/s²
Substitute this value into equation 2
t = √(2×10/3.7)
t = √(20/3.7)
t = √(5.405)
t = 2.32 s.
By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s
Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:
mass m = 0.170 kg
initial speed u = 6 m/s.
Distance covered s = 61 m
To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V
To do this, let us first calculate the kinetic energy at which the ball move.
K.E = 1/2m
K.E = 1/2 x 0.17 x 
K.E = 3.06 J
The work done on the ball is equal to the kinetic energy. That is,
W = K.E
But work done = Force x distance
F x S = K.E
F x 61 = 3.06
F = 3.06/61
F = 0.05 N
From here, we can calculate the acceleration of the ball from Newton second law
F = ma
0.05 = 0.17a
a = 0.05/0.17
a = 0.3 m/
To calculate the final velocity, let us use third equation of motion.
= + 2as
= + 2 x 0.3 x 61
= 36 + 36
= 72
V = 
V = 8.485 m/s
Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.
Learn more about dynamics here: brainly.com/question/402617
