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3 years ago

The kinetic energy of the pendulum bob in figure 15-1 increases the most between locations

Physics

2 answers:

Ket [755]3 years ago

7 0

Answer:A and C

Explanation:

Fudgin [204]3 years ago

3 0

Answer:

A and C

Explanation:

Kinetic energy of a body is given by the expression, $KE =\frac{1}{2}mv^2$ , where v is the velocity and m is the mass of body.

In case of a simple pendulum the maximum velocity is at it's stationary position.

So maximum velocity i at position C, which means maximum kinetic energy is at position A.

The velocity of pendulum reduces as the angle between vertical stationary and current position increases.

So, velocity at E and A are the minimum, so at A and E the kinetic energy value is minimum.

Now examining the options given, we will understand the kinetic energy increase is maximum in case of A and C.

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Which best describes the beginning of the Big Bang Theory?

maxonik [38]

A. is your best bet lol

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3 years ago

Read 2 more answers

Calculate A, E, μ, cv and S for 1 mole of Kr at 298 K and 1 atm (assuming ideal behavior)

vaieri [72.5K]

Answer:

internal energy E 3716.35 j

cv = 12.47 J/K

S = 12.47 J/K

A = 0.29 J

$\mu =3716.35 J/mole$

Explanation:

given data:

Kr atomic number = 36

degree of freedom = 3

1) internal energy E = $\frac{f}{2} n R T$

= $\frac{3}{2} *1*8.314*298 = 3716.35 j$

2) $cv = \frac{E}{T}$

$= \frac{3716.35}{298} = 12.47 J/K$

3) $S = cv =\frac{E}{T} = 12.47 J/K$

4) A, Halmholtz free energy = E -TS = 37146.35 - 12.47*298 = 0.29 J

5) $chemcial potential \mu = \frac{energy}{mole} = 3716.35 J/mole$

6 0

3 years ago

If youre walking from point a to b, the magnitude of your displacement will always be equal or less than or greater than your di

xenn [34]

The magnitude of your displacement can be equal to the distance you covered, or it can be less than the distance you covered. But it can never be greater than the distance you covered.

This is because displacement is a straight line, whereas distance can be a straight line, a squiggly line, a zig-zag line, a line with loops in it, a line with a bunch of back-and-forths in it, or any other kind of line.

The straight line is always the shortest path between two points.

3 0

3 years ago

An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below

grin007 [14]

<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let $P$ be the pressure at a point.

Let $p$ be the density fluid at a point.

Let $v$ be the velocity of fluid at a point.

Bernoulli's equation states that $P+\frac{1}{2}pv^{2}+pgh=constant$ for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let $p_{up}$ be the pressure of a point just above the wing.

Let $p_{do}$ be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

$\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}$

So, $p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa$

Force is given by the product of pressure difference and area.

Given that area is $27ms^{2}$ .

So,lifting force is $16398.48\times 27=442758.96N$

6 0

4 years ago

A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 70.0 from the horizontal w

professor190 [17]

Answer: 211.059 m

Explanation:

We have the following data:

$\theta=70\°$ The angle at which the ball leaves the bat

$V_{o}=55 m/s$ The initial velocity of the ball

$g=-9.8 m/s^{2}$ The acceleration due gravity

We need to find how far (horizontally) the ball travels in the air: $x$

Firstly we need to know this velocity has two components:

<u>Horizontally:</u>

$V_{ox}=V_{o}cos \theta$ (1)

$V_{ox}=55 m/s cos(70\°)=18.811 m/s$ (2)

<u>Vertically:</u>

$V_{oy}=V_{o}sin \theta$ (3)

$V_{oy}=55 m/s sin(70\°)=51.683 m/s$ (4)

On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

$V=V_{o}+gt=0$ (5)

Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (6)

$t=\frac{-55 m/s}{-9.8 m/s^{2}}$ (7)

$t=5.61 s$ (8)

Now that we have the time it takes to the ball to travel half of is path, we can find the total time $T$ it takes the complete parabolic path, which is twice $t$ :

$T=2t=2(5.61 s)=11.22 s$ (9)

With this result in mind, we can finally calculate how far the ball travels in the air:

$x=V_{ox}T$ (10)

Substituting (2) and (9) in (10):

$x=(18.811 m/s)(11.22 s)$ (11)

Finally:

$x=211.059 m$

8 0

3 years ago