Answer:
5.4 J.
Explanation:
Given,
mass of the object, m = 2 Kg
initial speed, u = 5 m/s
mass of another object,m' = 3 kg
initial speed of another orbit,u' = 2 m/s
KE lost after collusion = ?
Final velocity of the system
Using conservation of momentum
m u + m'u' = (m + m') V
2 x 5 + 3 x 2 = ( 2 + 3 )V
16 = 5 V
V = 3.2 m/s
Initial KE = ![\dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmu%5E2%20%2B%20%5Cdfrac%7B1%7D%7B2%7Dm%27u%27%5E2)
= ![\dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%202%5Ctimes%205%5E2%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%203%20%5Ctimes%202%5E2)
= 31 J
Final KE = ![\dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%20%28m%2Bm%27%29V%5E2%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%205%20%5Ctimes%203.2%5E2%20%3D%2025.6%20J)
Loss in KE = 31 J - 25.6 J = 5.4 J.
The formula is:
Work = Force · Displacement
F = m · g
F = 16 kg · 9.8 m/s² = 156.8 N
and we know that:
d = 0.8 m
W = 156.8 N · 0.8 m = 125.44 J
Answer:
W = 125.44 J.
When distance<span> is increased the amount of </span>force<span> needed will depend on the </span>mass<span> of the object. </span>
Answer:
A large piece of charcoal on a grill in the sunlight (if it's burning) will consist of the following types of energy:
- Chemical
- Heat and
- Light
Explanation:
Charcoal is basically carbon which is produced when wood is heated strongly in the absence of oxygen. From a chemistry point of view, charcoal contains combustible carbon whose chemical formula is C. Sometimes,
which is water may be found in it but in very small units.
All matter contains Heat energy. Charcoal is not an exception. As the charcoal burns, the heat energy is produced along with Light energy.
Light comes in many forms such as Infrared rays, Xrays, Visible Spectrum light, etc.
The glow which the coal gives off fall under the visible spectrum of light.
Cheers
Answer
Pressure, P = 1 atm
air density, ρ = 1.3 kg/m³
a) height of the atmosphere when the density is constant
Pressure at sea level = 1 atm = 101300 Pa
we know
P = ρ g h
![h = \dfrac{P}{\rho\ g}](https://tex.z-dn.net/?f=h%20%3D%20%5Cdfrac%7BP%7D%7B%5Crho%5C%20g%7D)
![h = \dfrac{101300}{1.3\times 9.8}](https://tex.z-dn.net/?f=h%20%3D%20%5Cdfrac%7B101300%7D%7B1.3%5Ctimes%209.8%7D)
h = 7951.33 m
height of the atmosphere will be equal to 7951.33 m
b) when air density decreased linearly to zero.
at x = 0 air density = 0
at x= h ρ_l = ρ_sl
assuming density is zero at x - distance
![\rho_x = \dfrac{\rho_{sl}}{h}\times x](https://tex.z-dn.net/?f=%5Crho_x%20%3D%20%5Cdfrac%7B%5Crho_%7Bsl%7D%7D%7Bh%7D%5Ctimes%20x)
now, Pressure at depth x
![dP = \rho_x g dx](https://tex.z-dn.net/?f=dP%20%3D%20%5Crho_x%20g%20dx)
![dP = \dfrac{\rho_{sl}}{h}\times x g dx](https://tex.z-dn.net/?f=dP%20%3D%20%5Cdfrac%7B%5Crho_%7Bsl%7D%7D%7Bh%7D%5Ctimes%20x%20g%20dx)
integrating both side
![P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx](https://tex.z-dn.net/?f=P%20%3D%20g%5Cdfrac%7B%5Crho_%7Bsl%7D%7D%7Bh%7D%5Ctimes%20%5Cint_0%5Eh%20x%20dx)
![P =\dfrac{\rho_{sl}\times g h}{2}](https://tex.z-dn.net/?f=%20P%20%3D%5Cdfrac%7B%5Crho_%7Bsl%7D%5Ctimes%20g%20h%7D%7B2%7D)
now,
![h=\dfrac{2P}{\rho_{sl}\times g}](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7B2P%7D%7B%5Crho_%7Bsl%7D%5Ctimes%20g%7D)
![h=\dfrac{2\times 101300}{1.3\times 9.8}](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7B2%5Ctimes%20101300%7D%7B1.3%5Ctimes%209.8%7D)
h = 15902.67 m
height of the atmosphere is equal to 15902.67 m.