Answer:
(a) 0.177 m
(b) 16.491 s
(c) 25 cycles
Explanation:
(a)
Distance between the maximum and the minimum of the wave = 2A ............ Equation 1
Where A = amplitude of the wave.
Given: A = 0.0885 m,
Distance between the maximum and the minimum of the wave = (2×0.0885) m
Distance between the maximum and the minimum of the wave = 0.177 m.
(b)
T = 1/f ...................... Equation 2.
Where T = period, f = frequency.
Given: f = 4.31 Hz
T = 1/4.31
T = 0.23 s.
If 1 cycle pass through the stationary observer for 0.23 s.
Then, 71.7 cycles will pass through the stationary observer for (0.23×71.7) s.
= 16.491 s.
(c)
If 1.21 m contains 1 cycle,
Then, 30.7 m will contain (30.7×1)/1.21
= 25.37 cycles
Approximately 25 cycles.
Answer:
A) x4
Explanation:
Magnification is equal to image size divided by the actual size, or M = I/A.
The image size is the student's drawing, which is 28.8 cm, and the actual size is 7.2 cm. Divide them, and cancel out the units, and you should get:
28.8 cm/7.2 cm = 4
If the boat's speed is s, and the river's speed is r, and the boat is traveling east (0 degrees),
(0,r) + (s cos297,s sin297) = (6,0)
now just solve for r and s.
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Answer:
As you know, the denser objects have more weight per unit of volume, this will mean that the force that pulls down these objects is a bit larger.
This will mean that the denser objects will always go to the bottom.
This clearly implies that the red liquid, the one with one of the smaller densities, can not be at the bottom.
There are some cases where a liquid with a small density may become a lot denser as the temperature or pressure changes, and in a case like that, we could see the red liquid at the bottom, but for this case, there is no mention of changes in the temperature nor in the pressure, so this can be discarded.
The only thing that makes sense is that the red part at the bottom is the base of the tube, and has nothing to do with the red liquid.
Answer:
The speed of the heavier fragment is 0.335c.
Explanation:
Given that,
Mass of the lighter fragment 
Mass of the heavier fragment 
Speed of lighter fragment = 0.893c
We need to calculate the speed of the heavier fragment
Let v is the speed of the second fragment after decay
Using conservation of relativistic momentum
Hence, The speed of the heavier fragment is 0.335c.
