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3 years ago

8459299

Physics

1 answer:

sergeinik [125]3 years ago

3 0

Answer:

Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. Mechanical energy can be either kinetic energy (energy of motion) or potential energy

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The coefficient of linear expansion of ordinary glass is three times that of Pyrex glass. An ordinary glass rod and a Pyrex glas

DochEvi [55]

Answer:

b. 3 times

Explanation:

Lets take

Coefficient for ordinary glass = α₁

Coefficient for pyrex glass = α₂

Given that α₁ = 3 α₂

Initial length of both glasses are equal = L

Change in the temperature is also same .= ΔT

We know that change in the length given as

ΔL = L α ΔT

Therefore

$\dfrac{\Delta L_1}{\Delta L_2}=\dfrac{L\alpha_1\Delta T}{L\alpha_2\Delta T}$

$\dfrac{\Delta L_1}{\Delta L_2}=\dfrac{3\alpha_2}{\alpha_2}$

ΔL₁ = 3ΔL₂

Therefore change in the length of original glass is three time of pyrex glass.

b. 3 times

6 0

3 years ago

When Jim and Rob ride bicycles, Jim can only accelerate at three-quarters the acceleration of Rob. Both startfrom rest at the bo

Natali5045456 [20]

Answer:

46.4 s

Explanation:

5 minutes = 60 * 5 = 300 seconds

Let g = 9.8 m/s2. And $\theta$ be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim . Both of their motions are subjected to parallel component of the gravitational acceleration $gsin\theta$

Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]

Jim equation of motion is $s = a_Jt_J^2/2 = (3a/4)t_J^2/2 = 3at_J^2/8$

As both of them cover the same distance

$45000a = 3at_J^2/8$

$t_J^2 = 45000*8/3 = 120000$

$t_J = \sqrt{120000} = 346.4 s$

So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time

7 0

4 years ago

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

7 0

3 years ago

If a person climbed Mt. Everest has a mass of 105 kg and a weight of 625 N what would be the acceleration due to gravity?

DanielleElmas [232]

The acceleration due to gravity would be 5.95 m/s²

A force is known to be a push or pull and it is the change in momentum per time. It can be expressed by using the relation.

Force = mass × acceleration.

From the parameters given:

Mass = 105 kg
Force = 625 N

By replacing the given values into the above equation, we can determine the acceleration.

∴

625 N = 105 kg × acceleration.

$\mathbf{acceleration = \dfrac{625 \ N}{105 \ kg}}$

acceleration = 5.95 N/kg

Since 1 N/kg = 1 m/²

acceleration = 5.95 m/s²

Learn more about acceleration(a) here:

brainly.com/question/14344386

3 0

3 years ago

If the pressure on an ideal gas is increased, what will happen to the volume of the gas?

iogann1982 [59]

<span>B). it will decrease.

But, you should keep the temperature constant, 'cause according to Boyle's law, pressure of the ideal gases is indirectly proportional to it's volume but at constant temperature. So, don't confuse in that.

Hope this helps!

</span>

7 0

3 years ago