When Isabella pushes a box of mass 5.25 kilograms with a force of 15.7 newtons, it accelerates by a rate of 2.50 meters/second2.
2 answers:
Answer:
The coefficient of friction is 0.05
Explanation:
Given that,
When Isabella pushes a box of mass 5.25 kilograms with a force of 15.7 newtons.
It accelerates by a rate of 2.50 m/s²
Let coefficient of friction be μ
Force (F) = 15.7 N
Acceleration (a) = 2.5 m/s²
Mass (m) = 5.25 kg
Gravity (g) = 9.8 m/s²
Balanced equation of box to move
F - μmg = ma
15.7 - μ(5.25)(9.8) = (5.25)(2.50)
Hence, The coefficient of friction is 0.05
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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
Answer:
The voltage across a semiconductor bar is 0.068 V.
Explanation:
Given that,
Current = 0.17 A
Electron concentration
Electron mobility
Length = 0.1 mm
Area = 500 μm²
We need to calculate the resistivity
Using formula of resistivity
Put the value into the formula
We need to calculate the resistance
Using formula of resistance
We need to calculate the voltage
Using formula of voltage
Put the value into the formula
Hence, The voltage across a semiconductor bar is 0.068 V.