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SCORPION-xisa [38]

3 years ago

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24 A person stands at the side of a straight railway track. A train moves towards the person and emits sound from its whistle. T

he person hears a sound of frequency 1690 Hz as the train approaches him. The person then hears sound of frequency 1500 Hz as the train moves away from him. The speed of sound in air is 340 m s–1. What is the speed of the train? A 20 m s–1 B 38 m s–1 C 41 m s–1 D 43 m s–

1 answer:

Andrei [34K]3 years ago

6 0

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For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is s=29.0cm long)

Lesechka [4]

The question is missing a diagram of the ray reflection. I attached a diagram which comes from a similar question in the answer section. The full question should be as follows:

Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point d = 10.0cmfrom their point of intersection, as shown in the figure. For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is s=29.0cm long) after reflecting from the first mirror?

Answer:

34.6°

Explanation:

To strike the midpoint of the second mirror, the ray light will have to travel half of the distance vertically

i.e. 29/2 = 14.5

We can solve this through trigonometry.

Let the angle between the ray and the vertical plane mirror is known as α

tan α = 10/14.5

α = $tan^{-1} (10/14.5)$ = 34.6°

The angle of incidence is the angle between the ray and the normal line of the mirror.

Let angle of incidence of first mirror be β

β = α = 34.6

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3 years ago

Which statement best describes the main idea of the poem

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Explanation:

option D ) is correct the speaker explain a difficult decision he had to make

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3 years ago

50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t

r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

$PE=mgh$

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

$KE=\frac{1}{2}mv^2$

where v is the speed of the object

When the ball is sitting on the top of the building, we have

$h=40 m$ , therefore the potential energy is not zero
$v=0$ , since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

$h=20 m$

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

$E=PE+KE=const.$

At the top of the building,

$E=PE_{top}$

While halfway through the fall,

$PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}$

And the mechanical energy is

$E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}$

which means

$KE_{half}=\frac{E}{2}$

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

The height of the ball relative to the ground is now zero: $h=0$ . This means that the potential energy of the ball is zero: $PE=0$

The kinetic energy, instead, is not zero: in fact, the ball has gained speed during the fall, so $v\neq 0$ , and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

4)

The potential energy of the ball as it sits on top of the building is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

$PE=(2)(9.8)(40)=784 J$

5)

The potential energy of the ball as it is halfway through the fall is given by

$PE=mgh$

where:

m = 2 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

$PE=(2)(9.8)(20)=392 J$

6)

The kinetic energy of the ball halfway through the fall is given by

$KE=\frac{1}{2}mv^2$

where

m = 2 kg is the mass of the ball

v = 19.8 m/s is the speed of the ball when it is halfway through the fall

Substituting the values into the equation, we find the kinetic energy of the ball when it is halfway through the fall:

$KE=\frac{1}{2}(2)(19.8)^2=392 J$

We notice that halfway through the fall, half of the initial potential energy has converted into kinetic energy.

7)

The kinetic energy of the ball just before hitting the ground is given by

$KE=\frac{1}{2}mv^2$

where:

m = 2 kg is the mass of the ball

v = 28 m/s is the speed of the ball just before hitting the ground

Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

$KE=\frac{1}{2}(2)(28)^2=784 J$

We notice that when the ball is about to hit the ground, all the potential energy has converted into kinetic energy.

Learn more about kinetic and potential energy:

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4 0

3 years ago

What is the area called where the particles are spread apart?

Ostrovityanka [42]

Answer: Because of the longitudinal motion of the air particles, there are regions in the air where the air particles are compressed together and other regions where the air particles are spread apart. These regions are known as compressions and rarefactions respectively

Explanation:

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3 years ago

A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo

horrorfan [7]

Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

$s=r_{2}-r_{1}$

Where, $r_{1}$ = initial position

$r_{2}$ = final position

Put the value into the formula

$s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)$

$s= -6.80i+6.20j-0.1k$

(a). We need to calculate the work done on the object

Using formula of work done

$W=F\cdot s$

Put the value into the formula

$W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)$

$W=-13.6+55.8-0.53$

$W=41.67\ J$

(b). We need to calculate the average power due to the force during that interval

Using formula of power

$P=\dfrac{W}{t}$

Where, P = power

W = work

t = time

Put the value into the formula

$P=\dfrac{41.67}{5.40}$

$P=7.71\ Watt$

(c). We need to calculate the angle between vectors

Using formula of angle

$\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})$

Put the value into the formula

$\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})$

$\theta=79.7^{\circ}$

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c). The angle between vectors is 79.7°

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4 years ago