Answer:
Explanation:
From the data it appears that A is the middle point between two charges.
First of all we shall calculate the field at point A .
Field due to charge -Q ( 6e⁻ ) at A
= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴
= 13.82 x 10⁻⁶ N/C
Its direction will be towards Q⁻
Same field will be produced by Q⁺ charge . The direction will be away
from Q⁺ towards Q⁻ .
We shall add the field to get the resultant field .
= 2 x 13.82 x 10⁻⁶
= 27.64 x 10⁻⁶ N/C
Force on electron put at A
= charge x field
= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶
= 44.22 x 10⁻²⁵ N
Answer:
The weight of Earth's atmosphere exert is 
Explanation:
Given that,
Average pressure 
Radius of earth 
Pressure :
Pressure is equal to the force upon area.
We need to calculate the weight of earth's atmosphere
Using formula of pressure
Where, P = pressure
A = area
Put the value into the formula
Hence, The weight of Earth's atmosphere exert is
Answer:
a) t = 1.8 x 10² s
b) t = 54 s
c) t = 49 s
Explanation:
a) The equation for the position of an object moving in a straight line at constan speed is:
x = x0 + v * t
where
x = position at time t
x0 = initial position
v = velocity
t = time
In this case, the origin of our reference system is at the begining of the sidewalk.
a) To calculate the time the passenger travels on the sidewalk without wlaking, we can use the equation for the position, using as speed the speed of the sidewalk:
x = x0 + v * t
95 m = 0m + 0. 53 m/s * t
t = 95 m/ 0.53 m/s
t = 1.8 x 10² s
b) Now, the speed of the passenger will be her walking speed plus the speed of th sidewalk (0.53 m/s + 1.24 m/s = 1.77 m/s)
t = 95 m/ 1.77 m/s = 54 s
c) In this case, the passenger is located 95 m from the begining of the sidewalk, then, x0 = 95 m and the final position will be x = 0. She walks in an opposite direction to the movement of the sidewalk, towards the origin of the system of reference ( the begining of the sidewalk). Then, her speed will be negative ( v = 0.53 m/s - 2*(1.24 m/s) = -1.95 m/s. Then:
0 m = 95 m -1.95 m/s * t
t = -95 m / -1.95 m/s = 49 s
<h2>For Second Solid Lumped System is Applicabe</h2>
Explanation:
Considering heat transfer between two identical hot solid bodies and their environments -
- If the first solid is dropped in a large container filled with water, while the second one is allowed to cool naturally in the air than for second solid, the lumped system analysis more likely to be applicable
- The reason is that a lumped system analysis is more likely to be applicable in the air than in water as the convection heat transfer coefficient so that the Biot number is less than or equal to 0.1 that is much smaller in air
Biot number = the ratio of conduction resistance within the body to convection resistance at the surface of the body
∴ For a lumped system analysis Biot number should be less than 0.1
Answer:
Explanation:
Given that
J(r) = Br
We know that area of small element
dA = 2 π dr
I = J A
dI = J dA
Now by putting the values
dI = B r . 2 π dr
dI= 2π Br² dr
Now by integrating above equation
Given that
B= 2.35 x 10⁵ A/m³
r₁ = 2 mm
r₂ = 2+ 0.0115 mm
r₂ = 2.0115 mm
By putting the values
