Answer:
1. telescope
2.

f- focal length
f- focal length r- the radius of curvature of the mirror

p-the distance of the object from the vertex of the mirror
l-the distance of the figure from the vertex of the mirror
Answer
given,
Speed of car A = 95 Km/h
= 95 x 0.278 = 26.41 m/s
Speed of Car B = 121 Km/h
= 121 x 0.278 = 33.64 m/s
Distance between Car A and B at t=0 = 41 Km
a) Distance travel by car B
d = 26.41 t + 41000
speed of the car A = 33.64 m/s
distance = s x t
26.41 t + 41000 = 33.64 x t
7.23 t = 41000
t = 5670.82 s
time taken by Car B to cross Car A is equal to t = 5670.82 s
distance traveled by car A
D = s x t = 26.41 x 5670.82 = 149766.25 m = 149.76 Km
b) distance travel by the car B in 30 s after overtaking car A
D' = s x t = 33.64 x 30 = 1009.2 m = 1 Km
Answer:
Ф = 2.179 eV
Explanation:
This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.
E = K + Ф
the energy of the photons is given by the Planck relation
E = h f
we substitute
h f = K + Ф
Ф= hf - K
the speed of light is related to wavelength and frequency
c = λ f
f = c /λ
Φ =
let's reduce the energy to the SI system
K = 0.890 eV (1.6 10⁻¹⁹ J / 1eV) = 1.424 10⁻¹⁹ J
calculate
Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹ -1.424 10⁻¹⁹
Ф = 4.911 10⁻¹⁹ - 1.424 10⁻¹⁹
Ф = 3.4571 10⁻¹⁹ J
we reduce to eV
Ф = 3.4871 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
Ф = 2.179 eV
Answer:
the distance traveled by the car is 42.98 m.
Explanation:
Given;
mass of the car, m = 2500 kg
initial velocity of the car, u = 20 m/s
the braking force applied to the car, f = 5620 N
time of motion of the car, t = 2.5 s
The decelaration of the car is calculated as follows;
-F = ma
a = -F/m
a = -5620 / 2500
a = -2.248 m/s²
The distance traveled by the car is calculated as follows;
s = ut + ¹/₂at²
s = (20 x 2.5) + 0.5(-2.248)(2.5²)
s = 50 - 7.025
s = 42.98 m
Therefore, the distance traveled by the car is 42.98 m.