Explanation:
In the given situation two forces are working. These are:
1) Electric force (acting in the downward direction) = qE
2) weight (acting in the downward direction) = mg
Therefore, work done by all the forces = change in kinetic energy
Hence, 
It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.
v = 
= 592999 m/s
Since, the electron is travelling downwards it means that it looses the potential energy.
Answer:
a)W= - 720 J
b)ΔU= 330 J
Explanation:
Given that
P = 0.8 atm
We know that 1 atm = 100 KPa
P = 80 KPa
V₁ = 12 L = 0.012 m³ ( 1000 L = 1 m³)
V₂ = 3 L = 0.003 m³
Q= - 390 J ( heat is leaving from the system )
We know that work done by gas given as
W = P (V₂ -V₁ )
W= 80 x ( 0.003 - 0.012 ) KJ
W= - 0.72 KJ
W= - 720 J ( Negative sign indicates work done on the gas)
From first law of thermodynamics
Q = W + ΔU
ΔU=Change in the internal energy
Now by putting the values
- 390 = - 720 + ΔU
ΔU= 720 - 390 J
ΔU= 330 J
Volume = mass/density
Volume = 15000 g/45 g/cm3 ≈ 333.3 cm<span>3</span>
Answer: d= 0.57* l
Explanation:
We need to check that before ladder slips the length of ladder the painter can climb.
So we need to satisfy the equilibrium conditions.
So for ∑Fx=0, ∑Fy=0 and ∑M=0
We have,
At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal
At the top of ladder, N₂ acting horizontal
And Between somewhere we have the weight of painter acting downward equal to= mg
So, we have N₁=mg
and also mg*d*cosФ= N₂*l*sin∅
So,
d=
* tan∅
Also, we have f₁=N₂
As f₁= чN₁
So f₁= 0.357 * 69.1 * 9.8
f₁= 241.75
Putting in d equation, we have
d= 
* tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip
Geometrically, a screw can be viewed as a narrow inclined plane wrapped around a cylinder. Like the other simple machines a screw can amplify force; a small rotational force (torque) on the shaft can exert a large axial force on a load.
