Answer:


Explanation:
Given that
Intensity I


Radius of earth,R = 6370 Km
We know that surface area of earth, A



As we know that pressure due to intensity given as

V =Velocity of light



We know that force F
F = P .A


b)Gravitational force F




So F


Answer:
a. 12.12°
b. 412.04 N
Explanation:
Along vertical axis, the equation can be written as
T_1 sin14 + T_2sinA = mg
T_2sinA = mg - T_1sin12.5 ....................... (a)
Along horizontal axis, the equation can be written as
T_2×cosA = T_1×cos12.5 ......................... (b)
(a)/(b) given us
Tan A = (mg - T_1sin12.5) / T_1 cos12.5
= (176 - 413sin12.5) / 413×cos12.5
A = 12.12 °
(b) T2 cosA = T1 cos12.5
T2 = 413cos12.5/cos12.12
= 412.04 N
Acceleration is the change of velocity, and velocity is the change of distance. The opposite of finding change, or differentiation, is integration.
Acceleration = 1.3 m/s²
Velocity: ∫ 1.3 dx = 1.3x + c m/s
Distance: ∫ 1.3x dx = 1.3x²/2 + c m
Distance run: 1.3*3²/2 = 5.85 m
<em>What</em><em> </em><em>bad</em><em> </em><em>thing</em><em> </em><em>happened</em><em>?</em>
I think the answer is A.........
Answer:
W = 0.842 J
Explanation:
To solve this exercise we can use the relationship between work and kinetic energy
W = ΔK
In this case the kinetic energy at point A is zero since the system is stopped
W = K_f (1)
now let's use conservation of energy
starting point. Highest point A
Em₀ = U = m g h
Final point. Lowest point B
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
mg h = K
to find the height let's use trigonometry
at point A
cos 35 = x / L
x = L cos 35
so at the height is
h = L - L cos 35
h = L (1-cos 35)
we substitute
K = m g L (1 -cos 35)
we substitute in equation 1
W = m g L (1 -cos 35)
let's calculate
W = 0.500 9.8 0.950 (1 - cos 35)
W = 0.842 J