The standard unit of brightness is called the candela.<br> True<br> False

A cylindrical block of mass M=50kg and height h=0.2m is hanging on a rope and is in equilibrium. Any difference in atmospheric p

agasfer [191]

Answer:

$\Delta P = 1961.4\,Pa$

Explanation:

The difference of pressure is given by gauge pressure:

$\Delta P = \rho_{w}\cdot g \cdot \Delta h$

$\Delta P = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.2\,m)$

$\Delta P = 1961.4\,Pa$

8 0

4 years ago

Read 2 more answers

A student throws a set of keys vertically upward to his fraternity brother, who is in a window 3.60 m above. The brother's outst

Contact [7]

Answer:

$v_{i}=10.10 m/s$

Explanation:

The equation of the position is:

$y=y_{i}+v_{i}t-0.5gt^{2}$

Where:

v(i) is the initial velocity

The initial position y(i) will be zero and the final position y = 3.60 m.

So, we just need to solve this equation for v(i).

$v_{i}=\frac{y+0.5gt^{2}}{t}$

$v_{i}=\frac{3.6+0.5*9.81*1.6^{2}}{1.6}$

$v_{i}=10.10 m/s$

Therefore, the initial velocity is 10.10 m/s upwards.

I hope it helps you!

5 0

3 years ago

Which statement is true about an object moving in a circular motion due to centripetal force, F, when the mass is doubled? A. It

Allushta [10]

Force = [Mass x (velocity)^2]/Radius
Therefore,
The mass and the force are directly proportional.If you double the mass,the force will be doubled too.

A. Its centripetal force would <span>be doubled.</span>

3 0

3 years ago

With a total force of 20,000 newtons, an airplane has a mass of 1,200 kg. What is its acceleration?

swat32

Answer:

<h3>The answer is 16.67 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$m = \frac{20000}{1200} = \frac{200}{12} \\ = 16.66666...$

We have the final answer as

Hope this helps you

8 0

3 years ago

At time t=0 a grinding wheel has an angular velocity of 30.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until

Rama09 [41]

Answer:

(A) 570 rad

(B) 10 s

(C) 12.5 rad/s²

Explanation:

The equations of motion for circular motions are used.

Initial angular velocity, $\omega_0 = 30.0 \text{ rad/s}$

Angular acceleration, $\alpha =35.0 \text{ rad/s}^2$

(A)

At <em>t</em> = 2.00 s, the angular displacement, <em>θ</em>, is given by

$\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}$

After this time, it decelerates through an angular displacement of 440 rad.

Total angular displacement = 130 + 440 rad = 570 rad

(B)

At the time the circuit breaker tips, the angular velocity is given by

$\omega = \omega_0+\alpha t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}$

This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:

$\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t$