Suvat
we have s, u, v and we want a
the suvat equation with these values in is: v^2 = u^2 - 2as
so a = (-v^2 + u^2)/-2s
plug numbers in
a = (-85^2 + 0^2)/-2*36 = 7225/72 = 100.3... ms^-2
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
Answer:
I = 4.28 [amp]
Explanation:
To solve this type of problems we must have knowledge of the law of ohm, which tells us that the voltage is equal to the product of resistance by current.
Initial data:
v = 1.5 [volt]
R = 0.35 [ohms]
v = I * R
therefore:
I = 1.5 / 0.35
I = 4.28 [amp]
Because a nuclear meltdown can be caused if systems fail.
Answer:
θ = 13.16 °
Explanation:
Lets take mass of child = m
Initial velocity ,u= 1.1 m/s
Final velocity ,v=3.7 m/s
d= 22.5 m
The force due to gravity along the incline plane = m g sinθ
The friction force = (m g)/5
Now from work power energy
We know that
work done by all forces = change in kinetic energy
( m g sinθ - (m g)/5 ) d = 1/2 m v² - 1/2 m u²
(2 g sinθ - ( 2 g)/5 ) d = v² - u²
take g = 10 m/s²
(20 sinθ - ( 20)/5 ) 22.5 = 3.7² - 1.1²
20 sinθ - 4 =12.48/22.5
θ = 13.16 °
