Answer:
Frames of reference.
For measuring the physical quantities you have to rely on something as a reference point or a standard value. In the case of physical quantities that are vectors, and an abstract coordinate system is required to express it clearly.
Let us consider a body located on a ship at some point. We can measure its distance from a ship or from the terrace of the building. Hence, we are assigning a coordinate system with respect to a person standing on the ship and terrace. These coordinate systems are called the frames of reference.
While measuring the distance, both coordinate system gives specific values for the same location with regard to the frames of reference.
When the ship is moving, the person on the terrace observes that his values of the distance change. But the person on the ship did not observe any change with respect to his coordinate system.
Similarly, If a person considers a moving train as his frame of reference where his coordinate system is in motion, he finds the things that are stationary on earth seem to move.
The train is his frame of reference which has a velocity with respect to Earth.
Therefore, anything that is stationary with respect to the ground seems to be in motion with respect to his frame of reference.
20 m/s is the answer this is due to displacement over change in time. Thus 200/10=20
<h2>
The speed of the rock just before it reaches the water 25.0 m below the point where the rock left your hand is 45.06 m/s</h2>
Explanation:
First let us find the initial velocity,
We have after 8 seconds the displacement is zero,
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = ?
Acceleration, a = -9.81 m/s²
Time, t = 8 s
Displacement,s = 0 m
Substituting
s = ut + 0.5 at²
0 = u x 8 + 0.5 x -9.81 x 8²
u = 39.24 m/s
Initial velocity is 39.24 m/s.
Now this case is similar to case where a rock is thrown at 39.24 m/s downward.
We have equation of motion v² = u² + 2as
Initial velocity, u = 39.24 m/s
Acceleration, a = 9.81 m/s²
Final velocity, v = ?
Displacement, s = 25 m
Substituting
v² = u² + 2as
v² = 39.24² + 2 x 9.81 x 25
v = 45.06 m/s
The speed of the rock just before it reaches the water 25.0 m below the point where the rock left your hand is 45.06 m/s
Answer:
0.8582 m/s^2
Explanation:
mass of the bob m = 80 kg
length of the string L= 10.0 m
angle made with the vertical θ= 5.0
let the force exerted by the string = T
from the FBD
T cosθ= mg
T cos 6°= 80×9.81
⇒T=
= 787.79 N
how horizontal component of tension
=787.79 sin 5°= 68.66 N
Now, radial acceleration,
= 68.66/80= 0.8582 m/s^2
<span>Frequency of the pendulum F = 0.15Hz
Acceleration due to gravity g = 9.81 m/s^2
Period of the pendulum is calculated by T = 2 x 3.14 x squareroot of (L / g), L being the length of the pendulam.
We have an equation between T and F => T = 1 / F
T = 1 / 0.15 => T = 20 / 3
Now substituting the values
=> 20 / 3 = 2 x 3.14 x squareroot of (L / 9.81)
=> 3.33 = 3.14 x squareroot of (L / 9.81)
=> squareroot of (L / 9.81) = 3.33 / 3.14
=> L / 9.81 = 1.06^2
=> L = 1.13 x 9.81
Length of the pendulum = 11 m</span>