Natural causes are the reason for the recent warming trend of the average global temperature.

A bucket filled with water has a mass of 54 kg and is hanging from a rope that is wound around a 0.050 m radius stationary cylin

Lubov Fominskaja [6]

Answer:

The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.

Explanation:

Given that,

Mass of bucket = 54 kg

Radius = 0.050 m

We need to calculate the magnitude of the torque the bucket produces around the center of the cylinder

Using formula of torque

$\tau=F\times r$

$\tau=mg\times r$

Where, m = mass

g = acceleration due to gravity

r = radius

Put the value into the formula

$\tau=54\times9.8\times0.050$

$\tau=26.46\ N-m$

Hence, The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.

3 0

3 years ago

You are in your car driving on a highway at 23 m/s when you glance in the passenger-side mirror (a convex mirror with radius of

Irina-Kira [14]

Answer:

v = 26. 88 m/s +23 m/s

Explanation:

u = 23 m/s, r = 150 cm, u₁ = 2.0 m/s, s =2.0 m

$\frac{1}{s} +\frac{1}{s'} = \frac{2}{R}$

$\frac{1}{2.0 m} +\frac{1}{s'} = \frac{2}{1.50 m}$

Solve s'

$\frac{1}{s'} = \frac{2}{1.50 m} - \frac{1}{2.0 m}$

$\frac{1}{s'} = 1.833 m$

$s' = - 0.545 m$

To determine the speed of the trick to the highway

$\frac{ds}{dt}= \frac{s^2* \frac{ds}{dt}}{s' ^2} =\frac{2.0 ^2m * 2.0 m/s}{0.545^2m}$

$\frac{ds}{dt} = 26.88 m/s$

Now to determine the velocity highway is going to be

v = ds/dt + u

v = 26. 88 m/s +23 m/s

8 0

4 years ago

5. How does privatization contribute to productivity?

tangare [24]

Answer:

increased competition

Explanation:

3 0

2 years ago

Which statement best describes the concept of Einstein’s equation E = mc2?

faltersainse [42]

What are the statements please

5 0

4 years ago

A merry-go-round accelerating uniformly from rest achieves itsoperating speed of 2.5rpm in five revolution.

g100num [7]

Answer:

(A) The magnitude of its angular acceleration 1.09 x 10⁻³ rad/s²

(B) Time of motion 240.2 s

Explanation:

Given;

final angular speed, ωf = 2.5 RPM

angular distance, θ = 5 rev = 5 x 2π = 10π rad

initial angular speed, ωi = 0

final angular speed in rad/s;

$\omega_f = \frac{2.5 \ rev}{min} \times \ \frac{2\pi}{1 \ rev} \times \ \frac{1 \min}{60 s} = 0.2618 \ rad/s \\$

(A) the magnitude of its angular acceleration(rad/s^2);

$\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\(0.2618)^2 = 0 + (2\times 10\pi)\alpha\\\\0.0685 = 20\pi \alpha\\\\\alpha = \frac{0.0685}{20\pi} \\\\\alpha = 1.09\times 10^{-3} \ rad/s^2$

(B) Time of motion;

$\omega_f = \omega_i + \alpha t\\\\0.2618 = 0 + 1.09\times 10^{-3} t\\\\t = \frac{0.2618}{1.09\times 10^{-3}} \\\\t = 240.2 \ s$

7 0

3 years ago