Which example describes a nonrenewable resource?
2 answers:
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Answer
3
Explanation:
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Answer:
1450.4 KN
Explanation:
Pressure = ρhg
where: ρ is the density of the liquid, h is the height and g the force of gravity.
Total pressure exerted by the liquids at the base = Pressure of oil + Pressure of water + Pressure of mercury
So that,
i. Pressure of oil = ρhg
(ρ = 0.8 g/cm³ = 800 kg/m³)
= 800 x 5 x 9.8
= 39200
Pressure of oil = 39200 N
ii. Pressure of water = ρhg
(ρ = 1 g/cm³ = 1000 kg/m³)
= 1000 x 8 x 9.8
= 78400
Pressure of water = 78400 N
ii. Pressure of mercury = ρhg
(ρ = 13.6 g/cm³ = 13600 kg/m³)
= 13600 x 10 x 9.8
= 1332800
Pressure of mercury = 1332800 N
So that,
Total pressure exerted by the liquids at the base = 39200 + 78400 + 1332800
= 1450400
= 1450.4 KN
Total pressure exerted by the liquids at the base is 1450.4 KN.
Answer:
B = 1058.4 N
Explanation:
Given that,
The volume of a metal block, V = 0.09 m³
The density of fluid, d = 1200 kg/m³
We need to find the buoyant force when it's Completely immersed in brine. The formula for the buoyant force is given by :
g is acceleration due to gravity
So, the required buoyant force is 1058.4 N.
Answer:
1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s
2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions
3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s
4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²
Explanation:
The given parameters are;
The initial velocity of the fan, n = 0.220 rev/s
The magnitude of the angular acceleration = 0.920 rev/s²
The direction of the angular acceleration and the angular velocity = Clockwise
The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m
1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s
The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²
The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;
ω = ω₀ + α·t
From which we have;
ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s
The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s
2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;
θ = ω₀·t + 1/2·α·t²
θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians
2·π radians = 1 revolution
∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution
The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions
3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;
The tangential speed, = ω × r = ω × D/2
At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;
= ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413
The tangential speed, = 1.0338604413 m/s
The tangential speed ≈ 1.034 m/s
4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;
a = α × r = α × D/2
a = 5.78053048261 × 0.800/2 = 2.31221219304
The tangential acceleration, a ≈ 2.312 m/s²