Very high-energy objects and events spit out very high-energy photons, so the instrument you need in order to detect them is the X-ray telescope. <em>(C) </em>
Inconveniently, X-ray telescopes only work when they're up in orbit, because X-rays get seriously soaked up in Earth's atmosphere, and most of them never make it down to the surface ... (lucky for us !) .
Answer:
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Explanation:
The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:
(1)
Where:
- Impulse, in kilogram-meters per second.
- Mass, in kilograms.
- Initial velocity of the hockey park, in meters per second.
- Final velocity of the hockey park, in meters per second.
If we know that 
, ![\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right]](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D_%7B1%7D%20%3D%20-10%5C%2C%5Chat%7Bi%7D%5C%2C%5Cleft%5B%5Cfrac%7Bm%7D%7Bs%7D%5Cright%5D)
and ![\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right]](https://tex.z-dn.net/?f=%5Cvec%20%7Bv_%7B2%7D%7D%20%3D%2025%5C%2C%5Chat%7Bi%7D%5C%2C%5Cleft%5B%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%5D)
, then the impulse applied by the stick to the park is approximately:
![I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]](https://tex.z-dn.net/?f=I%20%3D%20%280.2%5C%2Ckg%29%5Ccdot%20%5Cleft%2835%5C%2C%5Chat%7Bi%7D%5Cright%29%5C%2C%5Cleft%5B%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%5D)
![I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]](https://tex.z-dn.net/?f=I%20%3D%207%5C%2C%5Chat%7Bi%7D%5C%2C%5Cleft%5B%5Cfrac%7Bkg%5Ccdot%20m%7D%7Bs%7D%20%5Cright%5D)
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Answer: heat
Insulation Traps Heat. Keeping the cold air out
Explanation:
Answer:
B
Explanation:
Reason is that the suitcase is exerted downward and when it moves downward the equation is mgsin tita
Answer:
Given that
V2/V1= 0.25
And we know that in adiabatic process
TV^န-1= constant
So
T1/T2=( V1 /V2)^ န-1
So = ( 1/0.25)^ 0.66= 2.5
Also PV^န= constant
So P1/P2= (V2/V1)^န
= (1/0.25)^1.66 = 9.98
A. RMS speed is
Vrms= √ 3RT/M
But this is also
Vrms 2/Vrms1= (√T2/T1)
Vrms2=√2.5= 1.6vrms1
B.
Lambda=V/4π√2πr²N
So
Lambda 2/lambda 1= V2/V1 = 0.25
So the mean free path can be inferred to be 0.25 times the first mean free path
C. Using
Eth= 3/2KT
So Eth2/Eth1= T2/T1
So
Eth2= 2.5Eth1
D.
Using CV= 3/2R
Cvf= Cvi
So molar specific heat constant does not change
