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3 years ago

Explain the difference between cool air and warm air

Physics

1 answer:

Naddik [55]3 years ago

3 0

Answer: ok

Explanation:

The molecules in hot air are moving faster than the molecules in cold air. Because of this, the molecules in hot air tend to be further apart on average, giving hot air a lower density. That means, for the same volume of air, hot air has fewer molecules and so it weighs less.

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a boy throw a ball upward from the top of a cliff 73m high.Calculate the time in which ball will fall on the ground and the velo

LekaFEV [45]

The ball rises for v/g seconds; which equals 14.7/9.8=1.5 seconds . After this time, it’s height will be:

h(t)=g/2(1.5)²+14.7(1.5)

=-4.9 x 2.25 + 22.05

=11.025m

The ball then falls for 49+11.025=60.025m, which takes:

g/2t²=60.025

t²=12.25

t=3.5 secs

Total time: 1.5+3.5=5 seconds

3 0

4 years ago

Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

adoni [48]

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q = magnitude of electronic charge = $1.609 * 10^{-19} c$

mass of an electronic charge = $9.10 * 10^{-31} kg$

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = $\frac{mv^{2} }{2}$ , potential energy = qV

hence, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

7 0

3 years ago

A student's life was saved in an automobile accident because an airbag expanded in front of his head. If the car had not been eq

dexar [7]

Answer:

Compared to windshield the airbag exerts much lesser force

Explanation:

Impulse is defined as change in momentum of the object when it is acted upon by a force during interval of time

Impulse = Impulsive force *time

I = F*Δt

If the object should be bought to rest from certain velocity there should be change in momentum. If the duration in which the momentum is increased then there would be less force applied and hence less damage.

Airbags are used to reduce the force experience by the people when they are met with accident by extending the time required to stop the momentum.

During the collision, the passenger is carried towards the windshield and if they are stopped by collision with wind shield the force will be larger and more damage.But if they are hit with airbag then the force will be less due to increased time.

The change is momentum will be the same with or without momentum but its the time that decides the impact of force.By making it longer the force become less.

Thus compared to the windshield the airbag exerts much lesser force.

 

8 0

3 years ago

A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori

olasank [31]

Explanation:

The given data is as follows.

k = 130 N/m, $\Delta x$ = 17 cm = 0.17 m (as 1 m = 100 cm)

mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

Energy = $\frac{1}{2}kx^{2}$

Now, spring energy gets converted into kinetic energy when the box is launched.

So, $\frac{1}{2}kx^{2}$ = $\frac{1}{2}mv^{2}$

$\frac{1}{2} \times 130 \times (0.17)^{2}$ = $\frac{1}{2} \times 2.8 \times v^{2}$

$v^{2} = \frac{3.757}{2.8}$

= 1.34

v = 1.15 m/sec

Now,

Frictional force = $\mu \times mg$

= $0.15 \times 2.8 \times 9.8$

= 4.116 N

Also, Kinetic energy = work done by friction

$\frac{1}{2}mv^{2} = F_{f} \times d$

$\frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d$

1.8515 = $4.116 \times d$

d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

8 0

3 years ago

The density of water is 1.00 g/cm3. What is its density in kg/m3?

r-ruslan [8.4K]

1 g = 1 ÷ 1000 kg
= 0.001 kg

1 cm³ = 1 ÷ 100 ÷ 100 ÷ 100 m³
= 0.000001 m³

1 g/cm³ = 1 g / 1 cm³
= 0.001 kg / 0.000001 m³
= 1000 kg/m³

The density is 1000 kg/m³.

3 0

3 years ago