Consider two resistors connected in parallel and attached to a voltage source ofV. If a thirdresistor is added in parallel to th
2 answers:
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Answer:
At the closest point
Explanation:
We can simply answer this question by applying Kepler's 2nd law of planetary motion.
It states that:
"A line connecting the center of the Sun to any other object orbiting around it (e.g. a comet) sweeps out equal areas in equal time intervals"
In this problem, we have a comet orbiting around the Sun:
- Its closest distance from the Sun is 0.6 AU
- Its farthest distance from the Sun is 35 AU
In order for Kepler's 2nd law to be valid, the line connecting the center of the Sun to the comet must move slower when the comet is farther away (because the area swept out is proportional to the product of the distance and of the velocity: , therefore if r is larger, then v (velocity) must be lower).
On the other hand, when the the comet is closer to the Sun the line must move faster (, if r is smaller, v must be higher). Therefore, the comet's orbital velocity will be the largest at the closest distance to the Sun, 0.6 A.
Answer:
magnitude = 161.3m, ∅ = 32.9°
Explanation:
Vector addition always works the same. Add two vectors by adding their respective components.
vector A:
vector B:
Adding vector A and B:
The magnitude of any vector is given by the Pythagorean theorem:
In the case of the vector A+B:
The angle ∅ of the vector can by found by using trigonometric functions:
For instance, the angle ∅ for a vector is given by the equation:
The direction ∅ can be found by solving the trigonometric function.
In the example of vector A+B:
Solving for ∅:
Answer:
(a) The speed of the lighter block is .
The speed of the heavier block is .
(b) The smaller block goes up to 0.69 m.
Explanation:
We will divide this question into three parts: Part A is for the smaller mass from the top of the incline to the collision. Part B is the collision. And Part C is for the smaller mass from the bottom to the highest point it can achieve.
In order to solve this question, I will assume that the smaller mass is initially at rest.
Part A:
We will use conservation of energy.
This is the speed of the smaller mass just before the collision. The velocity of the mass is directed 30° above horizontal, since the mass is sliding down the incline.
Part B:
Momentum is a vector identity, so the x- and y-components of momentum are to be investigated separately. Since the collision occurs at the horizontal surface, only the x-component of momentum is conserved.
During the collision kinetic energy is also conserved. Since kinetic energy is a scalar quantity, we don't have to separate its components.
The following relation will be used when combining the two equations:
The following equation is useful for combining the two equations:
Therefore from the first equation,
Part C:
We will again use the conservation of energy to find the highest point that the mass can go:
Answer:
Explanation:
Given:
fibers diameter d= 18 μm
distance of the screen D= 30 cm
wavelength λ= 560 nm
we know that fringe width
putting values we get