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2 years ago

Give two ways in which the water vapour changes as it passes down the glass tube in the condenser

1 answer:

6 0

<span>It changes from a liquid into a solid which means it is changing states. It also is changing temperature due to the condenser increasing or decreasing the temperature in order for the vapor to be transformed in liquid.</span>

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The speed of sound in air is 320 ms-1 and in water it is 1600 ms-1. It takes 2.5 s for sound to reach a certain distance from th

Nonamiya [84]

Answer:

Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.

Explanation:

Given:

Speed of sound in air = 320 m/s

Speed of sound in water = 1600 m/s

Time taken to reach certain distance in air = 2.5 sec

We have to find the distance traveled by sound in air.

Distance = Product of speed and time.

⇒ $Distance = Speed\times time\ taken$

⇒ $Distance = 320\times 2.5$

⇒ $Distance = 800$ meters.

Now we have to find how much time the sound will take to travel in water.

⇒ Time = Ratio of distance and speed

⇒ $Time =\frac{distance}{speed}$

⇒ $Time =\frac{800}{1600}$ <em> ...distance = 800 m and speed = 1600 m/s</em>

⇒ $Time =\frac{1}{2}$

⇒ $Time =0.5$ seconds.

Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.

7 0

2 years ago

A rectangular conducting loop of wire is approximately half-way into a magnetic field B (out of the page) and is free to move. S

Black_prince [1.1K]

Answer:

. The loop is pushed to the right, away from the magnetic ﬁeld

Explanation

This decrease in magnetic strength causes an opposing force that pushes the loop away from the field

8 0

2 years ago

A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.

fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

$F=\dfrac{mv^2}{r}-mg$

$F=m(\dfrac{v^2}{r}-g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)$

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

$F=\dfrac{mv^2}{r}+mg$

$F=m(\dfrac{v^2}{r}+g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)$

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

$T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}$

$T=\dfrac{m}{r}(g(r-h)+v^2)$

Since, $v^2=u^2-2gh$

$T=\dfrac{m}{r}(u^2-3gh+gr)$

Hence, this is the required solution.

6 0

2 years ago

Dawn and Aram have stretched a slinky between them and begin experimenting with waves. As the frequency of the waves is doubled

m_a_m_a [10]

Answer:

halved

Explanation:

The velocity of the a wave is obtained by multiplying the frequency and wavelength.

$v=f\lambda\\\Rightarrow f=\frac{v}{\lambda}\\\Rightarrow \lambda=\frac{v}{f}$

Where

v = Velocity

f = Frequency

$\lambda$ = Wavelength

The velocity here is constant. So, if the frequency is doubled the wavelength is halved.

6 0

2 years ago

A car is traveling at 20.0 m/s on tires with a diameter of 70.0 cm. The car slows down to a rest after traveling 300.0 m. If the

cupoosta [38]

Answer: deceleration of $1.904\ rad/s^2$

Explanation:

Given

Car is traveling at a speed of u=20 m/s

The diameter of the car is d=70 cm

It slows down to rest in 300 m

If the car rolls without slipping, then it must be experiencing pure rolling i.e. $a=\alpha \cdot r$

Using the equation of motion

$v^2-u^2=2as\\$

Insert $v=0,u=20,s=300$

$0-(20)^2=2\times a\times 300\\\\a=\dfrac{-400}{600}\\\\a=-\dfrac{2}{3}\ m/s^2$

Write acceleration as $a=\alpha \cdot r$

$-\dfrac{2}{3}=\alpha \times 0.35\\\\\alpha =-\dfrac{2}{1.05}\\\\\alpha =-1.904\ rad/s^2$

So, the car must be experiencing the deceleration of $1.904\ rad/s^2$ .

4 0

2 years ago