Answer:
Explanation:
Given that,
The frequency of vibration is 120Hz
The mass of object attached is 0.5kg.
We want to find the tension In the string
The tension in the string is the weight of the object, it is how much gravity is pulling the object to the centre of the earth
Using newton second law.
F_net = m•a_y
The body is not accelerating the y-direction, then, a_y = 0
F_net = 0
Force acting on the string is the weight of the object and the tension in the string
T - W = 0
T = W
Where weight is mass × gravity
W = mg
Then,
T = W = mg
T = mg = 0.5 × 9.81
T = 4.905 N
The tension in the string is 4.905 N
The so-called "velocity-time" graph is actually a "speed-time" graph. At any point
on it, the 'x'-coordinate is a time, and the 'y'-coordinate is the speed at that time.
'Velocity' is a speed AND a direction. Without a direction, you do not have a velocity,
and these graphs never show the direction of the motion. It seems to me that it would be
pretty tough to draw a graph that shows the direction of motion at every instant of time,
so my take is that you'll never see a true "velocity-time" graph.
At best, it would need a second line on it, whose 'y'-coordinate referred to a second
axis, calibrated in angle and representing the 'bearing' or 'heading' of the motion at
each instant. The graph of uniform circular motion, for example, would have a straight
horizontal line for speed, and a 'sawtooth' wave for direction.
We are asked to solve for the index of refraction and the formula is n = c/v where "n" represents the index of refraction, "c" represents the speed of light in the vacuum while "v" represents the speed of another medium.
In the problem, we have the given values below:
c = 3 x 10^8 m/s
v = 2 x 10^8 m/s
n =?
Solving for n, we have the solution below:
n = 3x10^8 / 2x10^8
n = 1.5
The answer is 1.5 for the index of refraction.
Answer:
vector of zero magnitude
Explanation:
The displacement is a vector magnitude, therefore, in addition to being a module, it has direction and sense.
In this case it moved 350 m and then returned the same 350 m, so the total displacement is zero.
If we draw the vector, one has a directional direction to the right and the other direction to the left, therefore when adding the two vectors gives a vector of zero magnitude
The ground exerts an equal force on the golf ball.
