First we need to find the time in the air
d = 1/2gt^2 + vt
Since there is no inital vertical velocity
4.1 = 1/2 (9.8) t^2
8.2/9.8 = t^2
0.8367 = t^2
t = 0.9147s
Now we multiply by speed
7m/s x 0.9147s = 6.4m
To find the acceleration of the car, we can use the equation acelleration=final velocity-initial velocity/elapsed time. Since we know the initial velocity is zero from the start, than we can calculate it as final velocity/elapsed time. If we divide 36 by 10, we get 3.6. Your acceleration is 3.6 km/h.
Hope this helps! :D
Answer:
Yes
Explanation:
Eclipses: Eclipses are also known as game of shadows where one object comes between the star(light source) and another object in a straight line such that the shadow of one object falls on other object. This can occur when the apparent size of the star and the object is almost same.
Talking about the Earth, the geometry is such that the Moon and the Sun are of same apparent size as seen from the Earth. Thus Lunar and Solar eclipse can be seen from the Earth. If we were to go on any other planet the same phenomenon can be seen provided the apparent size of moon and the Sun from that planet is same.
We have seen and recorded many such eclipses on Jupiter. These are from the perspective of Earth. When the moons of Jupiter comes exactly between the Sun and Jupiter the shadow of moon will fall on Jupiter. The places where the shadow falls, one will see a solar eclipse.
Answer:
a) The magnitude of the car's total displacement (T) from the starting point is T = 82.67 Km
b) The angle (θ) from east of the car's total displacement measured from the starting direction is θ = 40.88 °
Explanation:
Attached you can see a diagram of the problem.
a) Find the magnitude of the vector T that goes from point A to point D (see the diagram).
The x and y components of this vector are
The magnitude of the vector is find using the pythagoras theorem:
, being a, b and c the 3 sides of the triagle that forms the vector:
Replacing the values
b) Find the angle θ that forms the vector T and the vector AB (see diagram).
To find this angle you can use the inverse tangent
θ
θ
θ=40.88°