In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller i
s 6.00 m, and, while being driven into rotation around a fixed axis, its angular position is expressed as
1 answer:
Question:
In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller is 6.00 m, and, while being driven into rotation around a fixed axis, its angular position is expressed as θ = 2.70t² − 0.900t3³
where θ is in radians and t is in seconds.
(a) Find the maximum angular speed of the roller.
(b) What is the maximum tangential speed of a point on the rim of the roller?
(c) At what time t should the driving force be removed from the roller so that the roller does not reverse its direction of rotation?
(d) Through how many rotations has the roller turned between t = 0 and the time found in part (c)?
Answer:
a. 2.7rad/s
b. 8.1 m/s
c. 2 s
d. 3.6 rad
Explanation:
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Refer to the diagram shown.
There are twelve 5-minute divisions.
Each 5-minute division is equal to 360°/12 = 30°.
By convention, angles are measured counterclockwise from the positive x-axis.
The angular position of the minute hand at 2:55 is
θ = 90° + 30° = 120°
Because 360° = 2π radians, therefore
θ = (120/360)*2π = (2π)/3 radians = 2.0944 radians
Answer: (2π)/3 radians ofr 2.0944 radians.
There are twelve 5-minute divisions.
Each 5-minute division is equal to 360°/12 = 30°.
By convention, angles are measured counterclockwise from the positive x-axis.
The angular position of the minute hand at 2:55 is
θ = 90° + 30° = 120°
Because 360° = 2π radians, therefore
θ = (120/360)*2π = (2π)/3 radians = 2.0944 radians
Answer: (2π)/3 radians ofr 2.0944 radians.
Answer:
1) The total charge of the top plate is 0.008 C
b) The total charge of the bottom plate is -0.008 C
2) The electric field at the point exactly midway between the plates is 0
3) The electric field between plates is approximately 1.1294 × 10¹² N/C
4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N
Explanation:
The given parameters of the parallel plate capacitor are;
The dimensions of the plates = 4 × 2 cm
The distance between the plates = 10 cm
The surface charge density of the top plate, σ₁ = 10 C/m²
The surface charge density of the bottom plate, σ₂ = -10 C/m²
The surface area, A = 0.04 m × 0.02 m = 0.0008 m²
1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C
b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C
2) The electrical field at the point exactly midway between the plates is given as follows;
Therefore, we have;
The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m
The electric field at the point exactly midway between the plates, = 0
3) The electric field, 'E', between plates is given as follows;
E ≈ 1.1294 × 10¹² N/C
The electric field between plates, E ≈ 1.1294 × 10¹² N/C
4) The force on an electron in the middle of the two plates
The charge on an electron, e = -1.6 × 10⁻¹⁹ C
The force on an electron in the middle of the two plates, = E × e
∴ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N
The force on an electron in the middle of the two plates, ≈ 1.807 × 10⁻⁷ N
Answer:
The maximum speed of car will be 20.5m/sec
Explanation:
We have given mass of car = 1100 kg
Radius of curve = 82.3 m
Static friction
We have to find the maximum speed of car
We know that at maximum speed centripetal force will be equal to frictional force
So the maximum speed of car will be 20.5m/sec