All categories

3 years ago

If you place a balloon in the freezer, it will __________.

Physics

2 answers:

Free_Kalibri [48]3 years ago

6 0

Answer: deflate because volume is directly proportional to temperature

Explanation:

Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

$V\propto T$ (At constant pressure and number of moles)

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

$T_1$ = initial temperature of gas

$T_2$ = final temperature of gas

As volume is directly proportional to temperature, as the balloon is kept in freezer, the temperature would decrease and thus the volume will decrease and it will deflate.

tester [92]3 years ago

5 0

The answer is "Deflate because volume is directly proportional to temperature"

You might be interested in

Drag the item from the item bank to its corresponding match.

tigry1 [53]

1- first law
2- third law
3- first law
4- second law
5- third law
6- second law

3 0

3 years ago

Read 2 more answers

Three light bulbs are connected to a battery in a series circuit.How will the bulbs behave if the circuit is closed

Drupady [299]

The answer is

<em>All the three light bulbs will glow</em>

hope this helps

6 0

4 years ago

Read 2 more answers

The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p

leva [86]

1) $1.86\cdot 10^6 rad/s^2$

2) 2418 rad/s

3) $27000 m/s^2$

4) 36.3 m/s

Explanation:

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

$\theta=\omega_i t +\frac{1}{2}\alpha t^2$

where:

$\theta$ is the angular displacement

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

$\theta=90^{\circ} = \frac{\pi}{2}rad$ is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

Solving for $\alpha$ , we find:

$\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2$

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

$\omega_f = \omega_i + \alpha t$

where

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

Therefore, the final angular speed is:

$\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s$

The tangential acceleration is related to the angular acceleration by the following formula:

$a_t = \alpha r$

where

$a_t$ is the tangential acceleration

$\alpha$ is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

$a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2$

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

$v=u+at$

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

$a=27900 m/s^2$ (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

$v=0+(27900)(0.0013)=36.3 m/s$

5 0

3 years ago

How large a band of frequencies does each television broadcasting channel get ?

k0ka [10]

Answer:

Since the waves must carry a great deal of visual as well as audio information, each channel requires a larger range of frequencies than simple radio transmission. TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz. (The entire FM radio band lies between channels 88 MHz and 174 MHz.)

8 0

3 years ago

Read 2 more answers

In general, the ________ of a simple machine is the ratio of the distance over which the force is applied to the distance over w

aksik [14]

Mechanical advantage, i hope i helped you!

3 0

3 years ago

Read 2 more answers