You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength
1 answer:
You might be interested in
Answer:
The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.
Explanation:
Given:
Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]
Speed of the cart (v) = 3.0 m/s
Thrust force by the rocket engine (F) = 8.5 N
Vertical height of the loop (y) = 20 m
Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.
Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.
So, we will apply kinematics of motion in the two directions separately.
Vertical motion:
Given:
Force acting in the vertical direction is given as:
So, acceleration in the vertical direction is given as:
Acceleration = Force ÷ mass
Vertical displacement of rocket is same as the height of loop. So,
There is no initial velocity in the vertical direction. So,
Now, applying equation of motion in vertical direction. we have:
Now, time taken to reach the loop is 2.15 s.
Horizontal motion:
There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.
Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,
Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.
Answer:
The fraction of kinetic energy to the total energy is .
Explanation:
displacement is one third of the amplitude.
Let the amplitude is A.
x= A/3
The kinetic energy of the body executing SHM is
The total energy is
Divide (1) by (2)
Answer:
0.64 m
Explanation:
The first thing is calculate the center of mass of the system.
now multiplying every coordinate x by the mass of each object (romeo, juliet and the boat) and dividing all by the total mass taking by reference the position of juliet.
X_cm = 1.4589 m
When the forces involved are internals, the center of mass don't change
After the movement the center of mass remains in the same distance from the shore, but change relative to the rear of the boat.
X_cm= 2.10 m
this displacement is how the boat move toward the shore.
2.10-1.46= 0.64 m