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3 years ago

13

Does heat rise or fall

2 answers:

muminat3 years ago

8 0

It is called convection, hot air rises and cool air sinks

Aleksandr-060686 [28]3 years ago

3 0

Hot air is less dense than cold air, causing it to displace and rise to the top.

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Why are large astronomical bodies round

-Dominant- [34]

<span>Matter of all types have gravity, which causes it to attract to each other. The most efficient way for all this matter to congregate is the sphere. As they consolidate, they form the shape. As they compress, temperatures in the center of the mass start to go up and if it hits the proper point, it can ignite and become a star.</span>

7 0

4 years ago

Read 2 more answers

A mass m at the end of a spring vibrates with a frequency

Wittaler [7]

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

$F = -kx$

And Newton's Second Law also states that

$F = ma = m\frac{d^2x}{dt^2}$

Combining two equations yields

$a = -\frac{k}{m}x$

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

$\omega = \sqrt{\frac{k}{m}}$

And given that ω = 2πf

the relation between frequency and mass becomes

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ .

Let's apply this to the variables in the question.

$0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg$

6 0

3 years ago

A glass tube 1 mm in diameter is dipped into glycerin. The density of the glycerin is 1260 kg/m3, surface tension is 6.3x10-2 N/

Sunny_sXe [5.5K]

Answer:

The capillary rise of the glycerin is most nearly $y = 0.0204 \ m$

Explanation:

From the question we are told that

The diameter of the glass tube is $d = 1 \ mm = 0.001 \ m$

The density of glycerin is $\rho = 1260 \ kg /m^3$

The surface tension of the glycerin is $\sigma = 6.3 *10^{-2} \ N /m$

The capillary rise of the glycerin is mathematically represented as

$y = \frac{4 * \sigma * cos (\theta )}{ \rho * g * d}$

substituting value

$y = \frac{4 * 6.3 *10^{-2} * cos (0 )}{ 1260 * 9.8 * 0.001}$

$y = 0.0204 \ m$

Therefore the height of the glass tube the glycerin was able to cover is

$y = 0.0204 \ m$

4 0

3 years ago

. Two identical vehicles traveling at the same speed are made to collide with barriers in an insurance company collision test. T

Serggg [28]

Answer:

F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

Explanation:

For this exercise let's use the relationship between momentum and momentum.

I = F t = Δp

in this case the final velocity is zero

F t = 0 -m v₀

F = m v₀ / t

in order to answer the question we must assume that the two vehicles have the same mass and speed

concrete barrier

F₁ = -p₀ / 0.1

F₁ = - 10 p₀

barrier collapses

F₂ = -p₀ / 1

let's look for the relationship of the forces

F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

5 0

3 years ago

You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation

Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, $\omega = 25.13 rad/s$

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = $\frac{1}{T}$

f = $\frac{1}{0.25}$

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

$\omega = 2\pi \times f$

$\omega = 2\pi \times 4$

$\omega = 25.13 rad/s$

5 0

4 years ago