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mestny [16]
3 years ago
8

A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The moment of inertia of t

he rod about this axis is given by (1/3)ML2. The rod is released from rest at an angle of 30° above the horizontal. What is the angular acceleration of the rod at the instant it is released?
Physics
1 answer:
emmasim [6.3K]3 years ago
5 0

Answer:

Angular acceleration = 6.37rad/sec²

Approximately, Angular acceleration =

6.4 rad/sec²

Explanation:

Length of the rod = 2.0m long

Inclination of the rod (horizontal) = 30°

Mass of the rod is not given so we would refer to it as = M

Rotational Inertia of the Rod(I) = 1/3ML²

Angular Acceleration = ?

There is an equation that shows us the relationship between Torque and Angular acceleration.

The equation is :

Torque(T) = Inertia × Angular Acceleration

Angular acceleration = Torque ÷ Inertia

Where:

Torque = L/2(MgCosθ)

Where M = Mass

L = Length = 2.0m

θ = Inclination of the rod (horizontal) = 30°

g = Acceleration due to gravity = 9.81m/s²

Inertia = 1/3ML²

Angular Acceleration =  (Mass × g × Cos (30°) × (L÷2)) ÷ 1/3ML²

Angular Acceleration =

(3 × g × cos 30°) ÷ 2× L

Angular Acceleration = (3 × 9.81m/s² × cos 30°) ÷ 2× L

Angular Acceleration = 3 × 9.81m/s² × cos 30°) ÷ 2× 2.0m

Angular Acceleration = 6.37rad/sec²

Approximately Angular Acceleration =

6.4rad/sec²

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sergij07 [2.7K]

Answer:

The workdone is  W = 1.712 *10^{-20 } \  J  

Explanation:

From the question we are told that

    The potential difference is  V  =  107 mV =  107 *10^{-3} \  V

Generally the charge on  Na^{+} is  Q_{Na^{+}} = 1.60 *10^{-19 } \  C

 Generally the workdone is mathematically represented as

         W =  Q_{Na^{+}}V

=>     W = 1.60 *10^{-19 } *  107 *10^{-3}    

=>     W = 1.712 *10^{-20 } \  J    

8 0
3 years ago
How can you increase the momentum of an object
Olin [163]

Answer:

You can change the momentum of an object by giving the object more force or less force.

Explanation:

Think about a ball. It is going slow, you push it and you give it more momentum.

5 0
3 years ago
a college student produces about 100 kcal of heat per hour on the average what is the rate of energy production and joules
Bond [772]

Given:

Amount of heat produced = 100 kcal per hour

Let's find the rate of energy production in joules.

We know that:

1 calorie = 4.184 Joules

1 kcal = 4.184 Joules

To find the rate of energy production in Joules, we have:

\begin{gathered} Rate=100\ast4.184 \\  \\ \text{Rate}=418.4\text{ KJ/hour} \end{gathered}

Therefore, the rate of energy production in joules is 418.4 kJ/h which is equivalent to 418400 Joules

ANSWER:

418.4 kJ/h

6 0
11 months ago
You are trying to overhear a juicy conversation, but from your distance of 23.0 m , it sounds like only an average whisper of 40
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4 0
3 years ago
A water tower is a familiar sight in many towns. The purpose of such a tower is to provide storage capacity and to provide suffi
Mademuasel [1]

The image of the water tower and the houses is in the attachment.

Answer: (a) P = 245kPa;

(b) P = 173.5 kPa

Explanation: <u>Gauge</u> <u>pressure</u> is the pressure relative to the atmospheric pressure and it is only dependent of the height of the liquid in the container.

The pressure is calculated as: P = hρg

where

ρ is the density of the liquid, in this case, water, which is ρ = 1000kg/m³;

When it is full the reservoir contains 5.25×10⁵ kg. So, knowing the density, you know the volume:

ρ = \frac{m}{V}

V = ρ/m

V = \frac{5.25.10^{5}}{10^{3}}

V = 525 m³

To know the height of the spherical reservoir, its diameter is needed and to determine it, find the radius:

V = \frac{4}{3}.\pi.r^{3}

r = \sqrt[3]{ \frac{3}{4\pi } .V}

r = \sqrt[3]{\frac{525.3}{4\pi } }

r = 5.005 m

diameter = 2*r = 10.01m

(a) Height for House A:

h = 15 + 10.01

h = 25.01

P = hρg

P = 25.01.10³.9.8

P = 245.10³ Pa or 245kPa

(b) h = 25 - 7.3

h = 17.71

P = hρg

P = 17.71.1000.9.8

P = 173.5.10³ Pa or 173.5 kPa

4 0
3 years ago
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