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4 years ago

What is it called when velocity changes over time

Physics

2 answers:

Vedmedyk [2.9K]4 years ago

6 0

It's called Acceleration .

natka813 [3]4 years ago

3 0

Well,

When an object's velocity changes, we call it acceleration.

Acceleration: The time rate of change in an object's velocity

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Water is most dense at 4 degrees Celsius. Since at this temperature 1 ml of water has a mass of 1 g. What is its density?

sveticcg [70]

The density is 1. 1/1=1

7 0

3 years ago

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A particular spiral galaxy can be approximated by a thin disk-like volume 62 Thousand Light Years in radius and 7 Hundred Light

SVETLANKA909090 [29]

Answer:

Approximate linear dimension is 2 light years.

Explanation:

Radius of the spiral galaxy r = 62000 LY

Thickness of the galaxy h = 700 LY

Volume of the galaxy = πr²h

= (3.14)(62000)²(700)

= (3.14)(62)²(7)(10)⁸

= 84568×10⁸

= $8.45\times 10^{12}$ (LY)³

Since galaxy contains number of stars = 1078 billion stars ≈ $1.078\times 10^{12}$

Now volume covered by each star of the galaxy = $\frac{\text{Total volume of the galaxy}}{\text{Number of stars}}$

= $\frac{8.45\times 10^{12} }{1.078\times 10^{12}}$

= 7.839 Light Years

Now the linear dimension across the volume

= $(\text{Average volume per star})^{\frac{1}{3}}$

= $(7.839)^{\frac{1}{3}}$

= 1.99 LY

≈ 2 Light Years

Therefore, approximate linear dimension is 2 light years.

8 0

3 years ago

A 2.45 g table tennis ball has the KE of 31.5 J. At what velocity is the tennis ball traveling?

Ket [755]

KE =
$\frac{1}{2} mv {}^{2}$
Convert g in kg
Find v:
2KE = mv^2
v=√2KE/m
v=√2*31.5/0.00245
v=160 m/s

3 0

3 years ago

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If a sound is 30 dB and its absolute pressure was 66 x 10-9 Pa, what must have been the reference pressure?

Ksenya-84 [330]

Given:

I = 30dB

P = 66 × $10^{-9}$ Pa

Solution:

Formula used:

I = $20\log_{10}(\frac{P}{P_{o}})$ (1)

where,

I = intensity of sound

P = absolute pressure

$P_{o}$ = reference pressure

Using Eqn (1), we get:

$30 = 20\log _{10}\frac{66\times 10^{-9}}{P^{o}}$

$P_{o}$ = $\frac{66\times 10^{-9}}{10^{1.5}}$

$P_{o}$ = 2.08 × $10^{-9}$ Pa

4 0

3 years ago

In the fall of 2002, a group of scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237

Llana [10]

The concepts required to solve this problem are those related to density, as a function of mass and volume. In turn, we will use the geometric concept defined for the volume.

The relationship between volume, density and mass is given under the function

$\rho = \frac{m}{V}$

Here,

m = Mass

V = Velocity

Rearranging for the Volume,

$V = \frac{m}{\rho}$

With our information the volume is

$V = \frac{60kg}{19.5g/cm^3 (\frac{10^3kg/m^3}{1g/cm^3})}$

$V = 3.0769*10^{-3}m^3$

Now the volume of sphere is expressed as

$V = \frac{4}{3} \pi r^3$

Here r is the radius of Sphere, then rearranging to find the radius we have

$r = \sqrt[3]{\frac{3V}{4\pi}}$

$r = \sqrt[3]{\frac{3(3.0769*10^{-3})}{4\pi}}$

$r = 0.0902m$

Therefore the radius of a sphere made of this material that has a critical mass is 9.02cm

8 0

3 years ago