Technology can best be defined as what?
2 answers:
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Give u = start velocity
v = end velocity
v = u + at
50 = 400 + a*30
30a = -350
a = -116.67 m/
**Why the accecleration is negative number**
Because displacement, velocity, and acceleration are VECTOR QUANTITIES.
Vector Quantity must have direction.
Answer:
a = 9.81[m/s^2]; v = 18683.5[m/s]
Explanation:
The stament of the problem is:
Suppose that the man pictured on the front side is orbiting the earth (mass = 5.98 x 1024kg) at a distance of 310 miles (1600 meters = 1 mile) above the surface of the earth (radius = 4000 miles).
a. What acceleration does he experience due to the earth's pull?
b. What tangential velocity must he possess in order that he orbit safely (in m/s)?
First we need to convert all the initial data to units of the SI
Rs = 310 [mil] = 498897 [m]
RT= 400 [mil] = 643738 [m]
r = Rs + RT = 1142635 [m] "distance from the center of the earth to the man"
G = universal gravitational constant
M = mass of the earth [kg]
m = mass of the man [kg]
r = distance from the center of the earth to the man [m]
a)
The acceleration he is experimenting is the same acceleration given by the gravity, therefore:
a = g = 9.81[m/s^2]
b)
To find the tangential velocity, we must determinate the force exerted by the earth.
Now we will find the force exerted by the gravity when the man is orbiting the earth at distance r.
And this force will be equal to the following expression:
Answer:
de Broglie wavelength of an electron with speed 0.78 c taking relativistic effects into account is given as:
λ = 1.943 * 10^(-12) m
Explanation:
Given:
v = 0.78 c
we know:
c = speed of light = 3 * 10^8 m/s
mass of electron = m = 9.1 × 10-31 kg
de Broglie wavelength:
In 1924 a French physicist Louis de Broglie assumed that for particles the same relations are valid as for the photon:
(Dual-nature of a particle)
Let the wavelength be = λ
According to de Broglie:
λ = h/p = h/mv
where h is planck's constant = 6.626176 x 10^-34 Js
and p is momentum.
Taking relativistic effects into account, we know that the momentum of the particle changes by a factor 'γ'.
At low speed, γ is almost 1. However, at very high velocity (comparable to light), it has a great effect on momentum.
γ =
γ = 1.6
Now at 0.78 c, considering relativistic effects, we know:
λ = h/γp = h/γ*mv
= (6.62 x 10^(-34))/(1.6*0.78*3*10^(8)*9.1 × 10-31
λ = 1.943 * 10^(-12) m