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3 years ago

Potential and kinetic energy

Physics

1 answer:

AveGali [126]3 years ago

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What is the Question????

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Which one of these are considered as mechanical energy?

MA_775_DIABLO [31]

Answer:

Kinetic energy and potential energy.

The term 'mechanical energy' refers to the sum of the kinetic energy and the gravitational potential energy of an object,

8 0

2 years ago

PLZZZ HELP 100 POINTS WILL MARK BRAINELST

Kitty [74]

Explanation:

Crust...molten

a. Oceanic, iron

b. Continental, silicates

c. less

3. Mantle, Denser

a. Lithosphere

b. Asthenosphere

4. Core

a. elements, rocks

b. liquid, magnetic

(I guess the liquid should come after the is)

Couldn't answer all but wanted to help

3 0

3 years ago

Read 2 more answers

Now consider a wave which is paired with seven other waves into seven pairs. The two waves in each pairing are identical, except

kupik [55]

The pair BCEG will interfere constructively, while the pair ADF will interfere destructively.

Constructive and destructive interference:

For interference, the waves must be coherent.

Two coherent waves interfere constructively when the path difference is equal to an integral multiple of the wavelength.

That is the path difference must be mλ

where m = 0,1,2,3.... is an integer and λ is the wavelength

So pair BCEG interfere constructively

Two coherent waves interfere destructively when the path difference is equal to a half-integral multiple of the wavelength.

That is the path difference must be (m+1/2)λ

where m = 0,1,2,3.... is an integer and λ is the wavelength

Therefore, the interference is in the pair ADF which is interfere destructively.

Learn more about interference: brainly.com/question/1194044

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[NOTE: THE COMPLETE QUESTION IS:

Now consider a wave which is paired with seven other waves into seven pairs. The two waves in each pairing are identical, except that one of them is shifted relative to the other in the pair by the distance shown: A. -(1/2) ?B. 2?C. -5?D. (3/2)?E. 0F. (17/2)?G. (6/2)?Identify which of the seven pairs will interfere constructively and which will interfere destructively. Each letter represents a pair of waves. Enter the letters of the pairs that correspond to constructive interference in alphabetical order and the letters of the pairs that correspond to pairs that interfere destructively in alphabetical order separated by a comma. For example if pairs A, B and D interfere constructively and pairs C and F interfere destructively enter ABD,CF.]

4 0

1 year ago

Read 2 more answers

A ship's propeller of diameter 3 m makes 10.6 revolutions in 30s. What is the angular velocity of the propeller?

ycow [4]

Answer:

The angular velocity of the propeller is 2.22 rad/s.

Explanation:

The angular velocity (ω) of the propeller is:

$\omega = \frac{\Delta \theta}{\Delta t}$

Where:

θ: is the angular displacement = 10.6 revolutions

t: is the time = 30 s

$\omega = \frac{\Delta \theta}{\Delta t} = \frac{10.6 rev*\frac{2\pi rad}{1 rev}}{30 s} = 2.22 rad/s$

Therefore, the angular velocity of the propeller is 2.22 rad/s.

I hope it helps you!

5 0

2 years ago

The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T

zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

f) The two trains avoid collision if the acceleration of the freight train is at least $0.35 m/s^2$

Explanation:

We can describe the position of the passenger train at time t with the equation

$x_p(t)=u_p t + \frac{1}{2}at^2$

where

$u_p = 25.0 m/s$ is the initial velocity of the passenger train

$a=-0.100 m/s^2$ is the deceleration of the train

On the other hand, the position of the freight train is given by

$x_f(t)=x_0 + v_f t$

where

$x_0=200 m$ is the initial position of the freight train

$v_f = 15.0 m/s$ is the constant velocity of the train

The collision occurs if the two trains meet, so

$x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0$

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

$x_f(t)=x_0 +v_ft$

And substituting t = 22.5 s, we find:

$x_f(22.5)=200+(15)(22.5)=537.5 m$

We can verify that the passenger train is at the same position at the time of the collision:

$x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m$

So, the two trains collide at x = 537.5 m.

In the graph in attachment, the position-time graph of each train is represented. We have:

The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

In order to avoid the collision, the freight train must have a initial position of

$x_0'$

such that the two trains never meet.

We said that the two trains meet if:

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t$

Re-arranging,

$\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0$

Substituting the values for the acceleration and the velocity,

$0.05t^2-10t+x_0'=0$

The solution of this equation is given by the formula

$t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}$

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

$10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m$

Therefore, the freight train must have a head start of 500 m.

In this case, we want to find the acceleration $a'$ of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t$

Re-arranging

$\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0$

Substituting,

$-0.5at^2-10t+200=0$

The solution to this equation is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}$

Again, the two trains never meet if the discriminant is negative, so

$10^2-4\cdot (-0.5a') \cdot (200)$

So, the deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

$x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2$

where $a_f$ is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2$

Re-arranging,

$\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0$

And the solution is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}$

Again, the two trains avoid collision if the discriminant is negative, so

$10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2$

Learn more about accelerated motion:

brainly.com/question/9527152

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3 years ago