The illustration shows a light ray striking an object.

A rod of length L is hinged at one end. The moment of inertia as the rod rotates around that hinge is ML2 /3. Suppose a 2.00-m r

Stels [109]

To solve this problem we will apply the concepts related to the moment of inertia and Torque, the latter both its translational and rotational expression.

According to the information given the moment of inertia of the body would be

$I = \frac{1}{3} mL^2$

Replacing we have

$I = \frac{1}{3} (3kg)(2m)^2$

$I = 4 kg * m^2$

Now the translational torque would be the product between the force applied (Its own Weight) and the distance (Its center of mass at the middle)

$\tau = F*r$

$\tau = mg (\frac{L}{2})$

$\tau = (3)(9.8)(\frac{2}{2})$

$\tau = 29.4N\cdot m$

Now the rotational torque is defined as the product between the moment of inertia and the angular acceleration, then,

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

Replacing,

$\alpha = \frac{29.4}{4}$

$\alpha =7.35 rad / s^2$

Therefore the angular acceleration is $7.35rad/s^2$

4 0

3 years ago

(a) Calculate the total force of the atmosphere acting on the top of a table that measures 1.5 m 2.2 m.

frozen [14]

A)
Atmospheric pressure is 101325 Pa.
Pressure = Force / Area
Area of table = 1.5 x 2.2
= 3.3 m²
Force = 101325 x 3.3
= 334 kN

B)
According to Newton's third law, every action has an equal and opposite reaction. Thus, the upward force is equal to the force acting downward on the table and is being balanced. The force is 334 kN in the upward direction.

8 0

4 years ago

Larry weighs 300 N at the surface of Earth. What is the weight of Earth in the gravitational field of Larry

Setler [38]

The gravitational force of attraction between two objects is directly proportional to the product of the two masses and inversely proportional to the square of the distance between them.

$F = \frac{GMm}{r^2}$

Where,

G = Gravitational Universal Constant

M = Mass of the Planet

m = Mass of the object

r = Distance

Therefore the amount of force exerted by the first object on the second object is equal to the amount of the force exerted on the second object by the first.

The gravitational force exerted by Larry on the Earth is same as the force exerted on Larry by the Earth.

That is 300N.

4 0

3 years ago

A backyard swimming pool with a circular base of diameter 6.00m is filled to depth 1.50m. (b) Two persons with combined mass 150

Mashutka [201]

The pressure increase at the bottom of the pool after they enter the pool and float is 106.103 Pa.

<h3>What is absolute pressure?</h3>

Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool,

mass of the two person = 150 Kg

weight of water level displaced exists equal to the weight of person.

$\rho \mathrm{Vg}=2 \mathrm{mg} \\$