Answer:
The answer to your question is
Explanation:
Data
mass = 0.5kg
T1 = 35
T2 = ?
Q = - 6.3 x 10⁴ J = - 63000 J
Cp = 4184 J / kg°C
Formula
Q = mCp(T2 - T1)
T2 = T1 + Q/mCp
Substitution
T2 = 35 - 63000/(0.5 x 4184)
T2 = 35 - 63000/2092
T2 = 35 - 30.1
T2 = 4.9 °C
Answer:
Yes,in fact visible 'light' is a form of radiation, which can be defined as an energy that travels in the form of electromagnetic waves. It can also be described as a flow of particle-like 'wave-packets', called photons, that travel constantly at the speed of light (about 300 000 kilometres per second).
Explanation:
Interference and diffraction are the phenomena that support only the wave theory of light. Options 2 and 3 are correct.
<h3 /><h3>What is the interference of waves?</h3>
The result of two or more wave trains flowing in opposite directions on a crossing or coinciding pathways. This phenomenon is known as the interference of waves.
The phenomenon of interference occurs when two wave pulses are traveling along a string toward each other.
The light wave hypothesis states that light behaves like a wave. Since light is an electromagnetic wave, it may be transmitted without a physical medium.
Light has magnetic and electric fields, much like electromagnetic waves do.
Transverse waves, such as those seen in light waves, oscillate in the same direction as the wave's path. A wave of light may experience interference as well as diffraction as a result of these properties.
All of the remaining options are the light phenomenon.
Hence, options 2 and 3 are correct.
To learn more about the interference of waves refer to the link;
brainly.com/question/16098226
#SPJ1
In this item, we are asked to determine the speed of the bobsled given the distance traveled and the time it takes to cover the certain distance. This can mathematically be expressed as,
speed = distance / time
Substituting the given values in this item,
speed = (113 m) / (29 s)
speed = 3.90 m/s
<em>ANSWER: 3.90 m/s</em>
Complete Question
A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.
If the acceleration of the projective is : a = c/s m/s2
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?
Answer:
The value of the constant is 
Explanation:
From the question we are told that
The acceleration is 
The initial position of the projectile is s= 1.5m
The final position of the projectile is 
The velocity is 
Generally 
and acceleration is 
so
=> 
integrating both sides
Now for the limit
a = 200 m/s
b = 0 m/s
c = s= 3 m
d =
= 1.5 m
So we have
![[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bv%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20200%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.%20%20%3D%20c%20%5Bln%20s%5D%5Cleft%20%7C%203%7D%20%5Catop%20%7B1.5%7D%7D%20%5Cright.)
![\frac{200^2}{2} = c ln[\frac{3}{1.5} ]](https://tex.z-dn.net/?f=%5Cfrac%7B200%5E2%7D%7B2%7D%20%20%3D%20%20c%20ln%5B%5Cfrac%7B3%7D%7B1.5%7D%20%5D)
=> 
