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2 years ago

A uniform rod of length l and mass m is pivoted around a line perpindicular to the rod at location l/3

Physics

1 answer:

beks73 [17]2 years ago

3 0

Answer:

I for a rod about its center is I = M L^2 / 12

A = L (1/2 - 1/3) = L / 6 distance from middle to location L/3

A^2 = L^2 / 36

I = M L^2 (1/12 + 1/36) = M L^2 (4 / 36) = M L^2 / 9

I about given location by parallel axis theorem

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Scientists are able to use body waves to determine what makes up the different layers of the Earth's interior. What characterist

k0ka [10]

Answer:

P-waves travel through liquids and solid while S-waves only travel through solids.

Explanation:

Scientists are able to use the fact that P-waves travel through both solids and liquids and waves travel through only solids to determine what makes the different layers of the Earth.

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2 years ago

True or False: The rems of a particular type of radiation is influenced by the Q value of the radiation and the rads of the radi

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3 years ago

An ideal gas in a sealed container has an initial volume of 2.50 L. At constant pressure, it is cooled to 21.00 ∘C, where its fi

goblinko [34]

<u>Answer:</u> The initial temperature of the system comes out to be 147 °C

<u>Explanation:</u>

To calculate the initial temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=2.50L\\T_1=?K\\V_2=1.75L\\T_2=25^oC=(25+273)K=294K$

Putting values in above equation, we get:

$\frac{2.50L}{T_1}=\frac{1.75L}{294K}\\\\T_1=420K$

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

$T(K)=T(^oC)+273$

$420=T(^oC)+273\\T(^oC)=147^oC$

Hence, the initial temperature of the system comes out to be 147 °C

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4 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

Mariana [72]

a) $F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

- Vertical direction: the equation of the forces is

$R-Fsin \theta - mg = 0$ (1)

where

R is the normal reaction of the floor (upward)

$F sin \theta$ is the component of the force F in the vertical direction (downward)

mg is the weight of the crate (downward)

- Horizontal direction: the equation of the forces is

$F cos \theta - \mu_k R = 0$ (2)

where

$F cos \theta$ is the horizontal component of the force F (forward)

$\mu_k R$ is the force of friction (backward)

From (1) we get

$R=Fsin \theta +mg$

And substituting into (2)

$F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

b) $\mu_s=cot(\theta)$

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

$R-Fsin \theta - mg = 0$ (1)

$F cos \theta - \mu_s R = 0$ (2)

In this second case, we want to find the critical value of $\mu_s$ such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, $F cos \theta$ .

Therefore, using the same procedure as before,

$R=Fsin \theta +mg$

$F cos \theta - \mu_s (Fsin \theta +mg) = 0$

And solving for $\mu_s$ ,

$F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}$

Now we analyze the expression that we found. We notice that if the force applied F is very large, $F sin \theta >> mg$ , therefore we can rewrite the expression as

$\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)$

So, this is the critical value of the coefficient of static friction.

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3 years ago

5. Work is being done when a force

Artyom0805 [142]

Answer:

work is done when a force is applied. the force must not be zero and should make the object move through a certain distance.

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2 years ago

Read 2 more answers