All categories

2 years ago

9. When a balloon rises into the air, it

Physics

1 answer:

Ksenya-84 [330]2 years ago

4 0

Answer:

Answer 3: When a balloon goes up higher in the air, its size will increase. Since there's less air in the upper atmosphere, there's less stuff pushing back on the balloon, and hence the pressure is lower, which allows the balloon to expand

Answer: C

Explanation:

As the balloon rises, the gas inside the balloon expands because the atmospheric pressure surrounding the balloon drops. The atmosphere is 100 to 200 times less dense at the float altitudes than on the ground. and as the air is heated inside the balloon it causes it to rise upwards (because it is lighter than the cooler air on the outside). When the pilot needs to bring the balloon down again, he simply reduces the temperature of the air inside the balloon causing it to slowly descend.

You might be interested in

This diagram shows how a certain type of precipitation is formed. Water drops are caught in up-drafts and down-drafts, over and

Svet_ta [14]

It would be B. Hail.

5 0

3 years ago

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

elena55 [62]

Answer:

Thunderbird is 995.157 meters behind the Mercedes

Explanation:

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird decided to take a pit stop and slows down for 250 m. She spent 5 seconds in the pit stop.

Here final velocity $v=0 \ m/s$

initial velocity $u= 71 m/s$ distance

Distance covered in the slowing down phase = $250 m$

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

$a= \frac {(v^2-u^2)}{2s}$

$a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2$

$v=u+at$

$t= \frac {(v-u)}{a}$

$= \frac {(0-71)}{(-10.082)}=7.042 s$

$t_1=7.042 s$

The car is in the pit stop for 5s $t_2=5 s$

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

$a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2$

$v=u+at$

$t= \frac{(v-u)}{a}$

$t= \frac{(71-0)}{6.81}= 10.425 s$

$t_3=10.425 s$

total time= $t_1 +t_2+t_3=7.042+5+10.425=22.467 s$

Distance covered by the Mercedes Benz during this time is given by $s=vt=71 \times 22.467= 1595.157 m$

Distance covered by the Thunderbird during this time= $250+350=600 m$

Difference between distance covered by the Mercedes and Thunderbird

= $1595.157-600=995.157 m$

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

6 0

2 years ago

A steel ball and a piece of clay have equal mass. They are dropped from the same height on a horizontal steel platform. The ball

emmasim [6.3K]

Answer: The ball (option A)

Explanation: change in momentum is defined by the formulae m(v - u) where m = mass of object, v = final velocity and u = initial velocity.

For the ball, it hits the ground and bounces back with the same speed, that's final velocity equals initials (v = - u)

Change in momentum = m( -u- u) = m(-2u) = m(-2u) = -2mu

For the clay, it final velocity is zero since it sticks to the floor, hence (v =0)

m(v - u) = m(0 - u) = - mu.

-2mu (change in momentum from the ball) is greater than - mu ( change in momentum of clay)

6 0

3 years ago

According to structural functionalists, what is government like in a mass society?

kvv77 [185]

The answer here is The government is expansive

8 0

3 years ago

Read 2 more answers

A river flows down towards a lake along an incline. initially the river is 90 m above the lake flowing at a rate of 3 m/s. at th

lubasha [3.4K]

Total energy =kinetic energy +potential energy

Change in energy =change in (kinetic energy +potential energy)

potential energy, $P.E=mgh$

where m is the mass, g is acceleration due to gravity and h is the height.

potential energy per unit mass =gh

change in potential energy per unit mass = $\Delta P.E.=g\Delta h$

where, h is the height.

kinetic energy= $K.E. =\frac{1}{2}mv^2$

change in kinetic energy per unit mass, $\Delta K.E. =\frac{1}{2}\Delta v^2$

In the given question:

Height varies from 90 m to zero as river flows from 90 m height to lake at 0 m

Velocity varies from 3m/s at top to o m/s at bottom.

Therefore,

$\Delta E =\Delta K.E.+\Delta P.E.\\ \Rightarrow \Delta E/m=\frac{1}{2}\Delta v^2+g\Delta h=\frac{1}{2}(0^2-3^2) m^2s^{-2}+9.8 (0-9)m^2s^{-2}=(-\frac{9}{2}-88.2)m^2s^{-2}=-92.7 m^2s^{-2}$

Here, it was mentioned in the question internal energy of the water is constant and there is no change in the pressure at the inlet and outlet.

4 0

2 years ago