Answer:
The answer is true.
Explanation:
A ray coming from the object hits the surface of a mirror and then passes through focus point after reflection. Since the image is formed by the actual meeting of the reflected rays therefore it is called as real image. the direction of arrow will be upward.
Same is true if the object is placed at center of curvature the coming ray along the parallel axis passes through the focus and bends downward. The point where the two reflecting rays will meet is the point of image. And the direction of image arrow will be downward. In this case image is formed between F and C.
.
Answer:
The atoms of the system with higher temperature will gain the heat energy and the vibration of these atoms will increase. The atoms or molecules on the system with lower temperature have lower temperature and are not vibrating. ... So, we can say that heat always flows from higher temperature to a lower temperature.
If it was removed it would not be a predator
Given Information:
Mass of sock = 0.23 kg
Stretched length of sock = x = 2.54 cm = 0.0254 m
Required Information:
Spring constant = k = ?
Answer:
Spring constant = k = 88.82 N/m
Explanation:
We know from the Hook's law that
F = kx
Where k is spring constant, F is the applied force and x is length of sock being stretched.
k = F/x
Where F is given by
F = mg
F = 0.23*9.81
F = 2.256 N
So the spring constant is
k = 2.256/0.0254
k = 88.82 N/m
Therefore, the spring constant of the sock is 88.82 N/m
The answer is <span>higher than.
</span><span>A sound-producing object is moving toward an observer. The sound the observer hears will have a frequency higher than that actually being produced by the object.
Why?
</span>As the source of the waves is moving toward the observer, each of the successive wave crest<span> is emitted from a position closer to the observer than the previous wave.
Thus each wave takes slightly less time to reach the observer than the previous wave. So, the time between the arrival of successive wave crests at the observer is reduced, increasing the frequency. </span>
