You might want to go to this website, http://www.indiana.edu/~p1013447/dictionary/mendel.htm
Welcome, And i hope this helps :P
For this, you need the v-squared equation, which is v(final)² = v(initial)² + 2aΔx
The averate acceleration is thus a = (v(final)² - v(initial)²) / 2Δx = (20² - 15²) / 2(50) = 175 / 100 = 1.75 m/s²
So the average acceleration is 1.75 m/s²
Answers:
a) 5400000 J
b) 45.92 m
Explanation:
a) The kinetic energy 
of an object is given by:
Where:
is the mass of the train
is the speed of the train
Solving the equation:
This is the train's kinetic energy at its top speed
b) Now, according to the Conservation of Energy Law, the total initial energy is equal to the total final energy:
Where:
is the train's initial kinetic energy
is the train's initial potential energy
is the train's final kinetic energy
is the train's final potential energy, where 
is the acceleration due gravity and 
is the height.
Rewriting the equation with the given values:
Finding :
Let R be radius of Earth with the amount of 6378 km h = height of satellite above Earth m = mass of satellite v = tangential velocity of satellite
Since gravitational force varies contrariwise with the square of the distance of separation, the value of g at altitude h will be 9.8*{[R/(R+h)]^2} = g'
So now gravity acceleration is g' and gravity is balanced by centripetal force mv^2/(R+h):
m*v^2/(R+h) = m*g' v = sqrt[g'*(R + h)]
Satellite A: h = 542 km so R+h = 6738 km = 6.920 e6 m g' = 9.8*(6378/6920)^2 = 8.32 m/sec^2 so v = sqrt(8.32*6.920e6) = 7587.79 m/s = 7.59 km/sec
Satellite B: h = 838 km so R+h = 7216 km = 7.216 e6 m g' = 9.8*(6378/7216)^2 = 8.66 m/sec^2 so v = sqrt(8.32*7.216e6) = 7748.36 m/s = 7.79 km/sec
Answer:
40.0⁰
Explanation:
The formula for calculating the magnetic flux is expressed as:
where:
is the magnetic flux
B is the magnetic field
A is the cross sectional area
is the angle that the normal to the plane of the loop make with the direction of the magnetic field.
Given
A = 0.250m²
B = 0.020T
= 3.83 × 10⁻³T· m²
3.83 × 10⁻³ = 0.020*0.250cosθ
3.83 × 10⁻³ = 0.005cosθ
cosθ = 0.00383/0.005
cosθ = 0.766
θ = cos⁻¹0.766
θ = 40.0⁰
<em>Hence the angle normal to the plane of the loop make with the direction of the magnetic field is 40.0⁰</em>
