The time base on the oscilloscope is set for 2 ms/cm and the vertical input has a frequency of 3000 Hz. How many wave cycles wil
2 answers:
Answer:
60 wave cycles
Explanation:
As the horizontal axis in a oscilloscope represents time, the time base is simply the scale, in other words, the amount of time that each division of oscilloscope represents. Therefore, multiplying the width of the screen times the time base will give us the total amount of time graphed on the screen.
The frequency is the amount of oscillations or waves cycles per second. So, in order to find the total amount of oscillations:
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Answer:
The two forces are;
1) Force 1 with magnitude of approximately 183.013 N, acting 30° to the left of the resultant force
2) Force 2 with magnitude of approximately 129.41 N acting at an inclination of 45° to the right of the resultant force
Explanation:
The given parameters are;
The (magnitude) of the resultant of two forces = 250 N
The angle of inclination of the two forces to the resultant = 30° and 45°
Let, F₁ and F₂ represent the two forces, we have;
F₁ is inclined 30° to the left of the resultant force and F₂ is inclined 45° to the right of the resultant force
The components of F₁ are = -F₁ × sin(30°)·i + F₁ × cos(30°)·j
The components of F₂ are = F₂ × sin(45°)·i + F₂ × cos(45°)·j
The sum of the forces = F₂ × sin(45°)·i + F₂ × cos(45°)·j + (-F₁ × sin(30°)·i + F₁ × cos(30°)·j) = 250·j
The resultant force, R = 250·j, which is in the y-direction, therefore, the component of the two forces in the x-direction cancel out
We have;
F₂ × sin(45°)·i = F₁ × sin(30°)·i
F₂ ·√2/2 = F₁/2
∴ F₁ = F₂ ·√2
∴ F₂ × cos(45°)·j + F₁ × cos(30°)·j = 250·j
Which gives;
F₂ × cos(45°)·j + F₂ ·√2 × cos(30°)·j = 250·j
F₂ × ((cos(45°) + √2 × cos(30°))·j = 250·j
F₂ × ((√2)/2 × (1 + √3))·j = 250·j
F₂ × ((√2)/2 × (1 + √3))·j = 250·j
F₂ = 250·j/(((√2)/2 × (1 + √3))·j) ≈ 129.41 N
F₂ ≈ 129.41 N
F₁ = √2 × F₂ = √2 × 129.41 N ≈ 183.013 N
F₁ ≈ 183.013 N
The two forces are;
A force with magnitude of approximately 183.013 N is inclined 30° to the left of the resultant force and a force with magnitude of approximately 129.41 N is inclined 45° to the right of the resultant force.
(a) The value of the gravitational acceleration is given by:
where
is the gravitational constant
M is the Earth's mass
is the Earth's radius at the pole
If we use the value for g given by the problem, , and we rearrange the equation above, we find the value of the Earth's mass:
(b) The value we found for the Earth's mass is , and we see that this value is slightly larger than the accepted value of
.