A student rides a bicycle for 15 miles in 3 hours. What is the student's speed? What else would you need to know for the velocit
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Concept: The magnification of spherical mirror can be defined by two ways.
(i) In terms of the height of the object and image.
The magnification of the spherical mirror is defined as the ratio of the height of the image'' to the height of the object '
'. It is denoted by letter 'm'.
Mathematically, it can be written as
(ii) In terms of the object's and image's distances.
The magnification of the spherical mirror is defined as the negative ratio of the image distance'' to the object distance '
'.
Mathematically, it can be written as
Now, from equation (1) and (2) we have,
Given: Spherical Concave Mirror,
We will consider positive sign for object's and image's distance because both are in front of the mirror.
Object distance .
Image distance
Object's height
Image's height
Now, apply equation (3)
Or, hi = - 8 mm
Here; negative sign means, the image will be inverted.
The image's height will be 8 mm.
Answer:
a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,
v/c% = 2.33 10⁻²
Explanation:
a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy
starting point. Where the electrons come out
Em₀ = U = e DV
final point. Where they hit the target
Em_f = K = ½ m v2
energy is conserved
Em₀ = Em_f
e ΔV = ½ m v²
ΔV = mv²/e (1)
If the speed of light is c and this is 100% then 1% is
v = 1% c = c / 100
v = 3 10⁸/100 = 3 10⁶6 m/ s
let's calculate
ΔV =
ΔV = 25.59 V
b) Ask for the potential difference for protons with the same kinetic energy as electrons
K_p = ½ m v_e²
K_p = 9.1 10⁻³¹ (3 10⁶)²
K_p = 40.95 10⁻¹⁹ J
we substitute in equation 1
ΔV = Kp / M
ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹
ΔV = 25.59 V
notice that these protons go much slower than electrons because their mass is greater
c) The speed of the protons is
e ΔV = ½ M v²
v² = 2 e ΔV / M
v² =
v² = 49,035 10⁸
v = 7 10⁴ m / s
Relation
v/c =
v/c= 2.33 10⁻⁴
Answer:
The heat capacity for the second process is 15 J/K.
Explanation:
Given that,
Work = 100 J
Change temperature = 5 k
For adiabatic process,
The heat energy always same.
We need to calculate the number of moles and specific heat
Using formula of heat
Put the value into the formula
We need to calculate the heat
Using formula of heat
Put the value into the formula
We need to calculate the heat capacity for the second process
Using formula of heat
Put the value into the formula
Hence, The heat capacity for the second process is 15 J/K.