A rectangular box has side 10cm x 5cm x 2cm. If it lies on a horizontal surface and has weight 1.00N, calculate the maximum and
1 answer:
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For rotational equilibrium of the door we can say that torque due to weight of the door must be counter balanced by the torque of external force
here weight will act at mid point of door so its distance is half of the total distance where force is applied
here we know that
now we will have
so our applied force is 72.5 N
Answer:
i) The period of the particle is 0.3 seconds
ii) The angular velocity is approximately 20.94 rad/s
iii) The linear velocity is approximately 1.047 m/s
iv) The centripetal acceleration is approximately 6.98 m/s²
Explanation:
The given parameters are;
The number of revolution of the particle, n = 800 revolution
The time it takes the particle to make 800 revolutions = 4 minutes
The dimension of the circle = 5 cm = 0.05 m
Given that the dimension of the circle is the radius of the circle, we have;
i) The period of the particle, T = The time to complete one revolution
T = 1/(The number of revolutions per second)
∴ T = 1/(800 rev/(4 min × 60 s/min)) = 3/10 s
The period, T = 3/10 seconds = 0.3 seconds
ii) The angular velocity, ω = Angle covered/(Time)
800 revolutions in 4 minutes = Angle of (800 × 2·π) in 4 minutes
∴ ω = (800 × 2·π)/(4 × 60) = 20·π/3
The angular velocity, ω = 20·π/3 rad/s ≈ 20.94 rad/s
iii) The linear velocity, v = r × ω
∴ The linear velocity, v = 0.05 m × 20·π/3 rad/s = π/3 m/s ≈ 1.047 m/s
iv) The centripetal acceleration, = v²/r
∴ The centripetal acceleration, = (π/3)²/(0.05) = 20·π/9
The centripetal acceleration, = 20·π/9 m/s² ≈ 6.98 m/s²
The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is .
The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.
The given data:
time, t = 1.1 s
initial speed, u = 1000 km/h =
final speed, v = 0 m/s
So we will be using the equation of motion, that is,
v = u + at
Hence , the deceleration of the rocket is .
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Answer:
Explanation:
Given that , we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:
#To find the particular solution:
Hence the charge at any time, t is