Determine the average speed of a runner who runs 15 Km in one hour and 30 minutes

How does this diagram demonstrate the law of superposition?

Mandarinka [93]

Answer:

well, as u can tell the top layer will always be the youngest layer aka the newest layer. The farther u go down the older the layers get. So the deeper u dig the farther back in time we see.

Explanation:

8 0

3 years ago

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What does a main sequence star become after it uses up its hydrogen in its core

Alexandra [31]

After the hydrogen fuel at the core has been consumed, the star evolves away from the main sequence on the HR diagram. The behavior of a star now depends on its mass, with stars below 0.23 M☉ becoming white dwarfs directly, whereas stars with up to ten solar masses pass through a red giant stage.

6 0

3 years ago

Part 1: A rope has one end tied to a vertical support. You hold the other end so that the rope is horizontal. If you move the en

Anit [1.1K]

1) c. 2 m/s

Explanation:

The relationship between frequency, wavelength and speed of a wave is

$v=\lambda f$

where

v is the speed

$\lambda$ is the wavelength

f is the frequency

For the wave in this problem,

f = 4 Hz

$\lambda=0.5 m$

So, the speed is

$v=(0.5 m)(4 Hz)=2 m/s$

2) a. 2.8 m/s

The speed of the wave on a string is given

$v=\sqrt{\frac{T}{\mu}}$

where

T is the tension in the string

$\mu$ is the linear mass density

In this problem, we have:

$T=2 \cdot 4 N=8 N$ (final tension in the rope, which is twice the initial tension)

$\mu = 1 kg/m$ --> mass density of the rope

Substituting into the formula, we find

$v=\sqrt{\frac{8 N}{1 kg/m}}=2.8 m/s$

5 0

3 years ago

A uniform solid disk and a uniform ring are place side by side at the top of a rough incline of height h.

nadezda [96]

Explanation:

velocity of disc $=\sqrt((gh)/0.75)$

lets call (h) 1 m to make it simple.

= 3.614 m/s

$\sqrt((4/3) x 1 x 9.8) = 3.614$ m/s pointing towards this:

$4×V_d=\sqrt(4/3hg)$

$V_h=\sqrt(hg)$

velocity of hoop= $\sqrt(gh)$

lets call (h) 1m to make it simple again.

$\sqrt(9.8 x 1) = 3.13$ m/s

$\sqrt(gh) = sqrt(hg)so [tex]4×V_d= \sqrt(4/3hg)V_h=\sqrt(hg)$

The disc is the fastest.

While i'm on this subject i'll show you this:

Solid ball $=0.7v^2= gh$

solid disc $= 0.75v^2 = gh$

hoop $=v^2=gh$

The above is simplified from linear KE + rotational KE, the radius or mass makes no difference to the above formula.

The solid ball will be the faster of the 3, like above i'll show you.

solid ball: velocity $=\sqrt((gh)/0.7)$

let (h) be 1m again to compare.

$\sqrt((9.8 x 1)/0.7) = 3.741$ m/s

solid disk speed $=\sqrt((gh)/0.75)$

uniform hoop speed $=\sqrt(gh)$

solid sphere speed $=\sqrt((gh)/0.7)$

8 0

3 years ago

A billiard ball moving at 6.00 m/s strikes a stationary ball of the same mass. after the collision, the first ball moves at 5.39

kodGreya [7K]

To do this question, we need to know that momentum is conserved, meaning the overall velocity of the two balls has to be the same before and after the collision. 

After collision... 

Ball 1: 4.33m/s *cos 30 = 3.75 m/s (x-component) 
4.33m/s * sin 30 = 2.165 m/s ( y-component) 

Ball 2 (struck ball): 5 m/s - 3.75m/s = 1.25 m/s (x-component) 
-2.165 m/s (y-component) note: it has to be in the opposite direction to conserve momentum 

tan-1(2.165/1.25) = 60 degrees 
Struck ball's velocity = sqrt(1.25^2 + 2.165^2) = 2.5 m/s at 60 degree with respect to the original line of motion. 

Hope you understand!

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3 years ago

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