A voltmeter<span> its </span>instrument<span> used for </span>measuring<span> electrical potential difference between two points in an electric circuit. </span>An ammeter<span> is a </span>measuring device<span> used to</span>measure<span> the electric current in a circuit.
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Answer:
sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j
Explanation:
We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis
So x component of the vector 
y component of the vector 
So vector will be 6.06i+3.5j
Now other vector of length of 7 units and makes an angle of 120° with positive x-axis
So x component of vector 
y component of the vector 
Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j
So, <u>the value of the work is approximately 84.65 J</u>.
<h2>Introduction</h2>
Hi ! Here I will help you to discuss the subject about work that caused by force in amount value of angle. Work is affected by the force and displacement.
- If related to the magnitude of the force, the amount of work will be proportional to the magnitude of the applied force. Thats mean, if the value of the force that applied on it is greater, then the value of the work will be greater.
- If related to the magnitude of shift, the amount of work will be proportional to the magnitude of shift of object. Thats mean, if the value of the shift on it is greater, then the value of the work will be greater.
<h3>Formula Used</h3>
The work done by a moving object can be expressed in the equation:
If the Angle Is Ignored

If the Angle Effect on Work

With the following condition:
- W = work that done by object (J)
- F = force that applied (N)
- s = shift or distance (m)
= angle of elevation (°)
<h3>Solution</h3>
We know that :
- F = force that applied =
N - s = shift or distance = 84.9 m
= angle of elevation = 45°
What was asked ?
- W = work that done by object = ... J
Step by step :






<h3>Conclusion</h3>
So, the value of the work is approximately 84.65 J.
It is fine to use the equation given by Plitter, since we are told that the mass is about the same as it is now, and I seriously doubt the original question wants the student to go into relativistic effects, electron degeneracy pressure and magnetic effects that govern a real white dwarf star.
There is no need to make it unnecessarily complicated, when the question is set at high school level. The question asks, given a particular radius, and a given mass, what will the density be (which in this case will be the average density). To answer the question, one needs to know the mass of the sun (which is about 2×1030 Kg. One needs to convert the diameter to a radius, and then calculate the spherical volume of the white dwarf. Then one can use the formula given above, namely density=mass/volume
Answer:
Explanation:
Assuming that velocities remain negligible,
The work will be equal to the change in potential energy of the center of mass
mass is 30 m(5 kg/m) = 150 kg
W = 150(9.8)(30/2) = 22,050 J ≈ 22 kJ