<span>Light waves travels through electromagnetic radiation
when we switch on the lamp, vibrations of electric and magnetic fields will release energy to its surrounding environment. Which the reason why the area around the lamp will feel warmer when we light it up.</span>
Answer:
a) t= 0.515 s
b) vf= 5.047 m/s
Explanation:
Because ball move with uniformly accelerated movement we apply the following formulas:
d= v₀t+ (1/2)*g*t² Formula (1)
vf= v₀+gt Formula (2)
Where:
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
g: acceleration due to gravity in m/s²
Known data
v₀=0
d= 1.30 m
g= 9.8 m/s²
Problem development
a)
We appy the formula 1 to calculate the time (t):
d= v₀t+ (1/2)*g*t²
1.30= 0+ (1/2)* (9.8)*t²
1.30=4.9*t²
t²=1.30/4.9

t= 0.515 s
b) We appy the formula 2 to calculate the final speed (vf):
vf= v₀+gt
vf= 0+(9.8)*(0.515)
vf= 5.047 m/s
Answer:
Δm Δt> h ’/ 2c²
Explanation:
Heisenberg uncertainty principle, stable uncertainty of energy and time, with the expressions
ΔE Δt> h ’/ 2
h’= h / 2π
to relate this to the masses let's use Einstein's relationship
E = m c²
let's replace
Δ (mc²) Δt> h '/ 2
the speed of light is a constant that we can condense exact, so
Δm Δt> h ’/ 2c²
Answer:
x = 1474.9 [m]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.
We must understand that when forces are applied on the body, they tend to slow the body down to stop it.
So as the body continues to move to the left, it is slowing down. Therefore we must calculate this deceleration value using Newton's second law. We must perform a sum of forces on the x-axis equal to the product of mass by acceleration. With leftward movement as negative and rightward forces as positive.
ΣF = m*a
![10 +12*sin(60)= - 6*a\\a = - 3.39[m/s^{2}]](https://tex.z-dn.net/?f=10%20%2B12%2Asin%2860%29%3D%20-%206%2Aa%5C%5Ca%20%3D%20-%203.39%5Bm%2Fs%5E%7B2%7D%5D)
Now using the following equation of kinematics, we can calculate the distance of the block, before stopping completely. The initial speed must be 100 [m/s].

where:
Vf = final velocity = 0 (the block stops)
Vo = initial velocity = 100 [m/s]
a = - 3.39 [m/s²]
x = displacement [m]
![0 = 100^{2}-2*3.39*x\\x=\frac{10000}{2*3.39}\\x=1474.9[m]](https://tex.z-dn.net/?f=0%20%3D%20100%5E%7B2%7D-2%2A3.39%2Ax%5C%5Cx%3D%5Cfrac%7B10000%7D%7B2%2A3.39%7D%5C%5Cx%3D1474.9%5Bm%5D)
Answer:
We are Given:
initial velocity (u) = 0m/s
final velocity (v) = 25 m/s
time (t) = 10 seconds
acceleration (a) = a m/s/s
From the first equation of motion, we know that:
v = u + at
solving for a, we get:
a = (v-u) / t
now, plugging the given values in this equation
a = (25 - 0) / 10
a = 25 / 10
a = 2.5 m/s/s
Therefore, the acceleration of the car is 2.5 m/s/s