#14 isn't really a Physics problem. It's more of just reading a graph.
A). When speed changes, acceleration is
(change in speed) / (time for the change) .
To be correct about it, acceleration can be positive ... when speed
is increasing ... or it can be negative ... when speed is decreasing.
So, on this graph, there are two periods of acceleration:
From zero to 2 seconds, acceleration = (8 m/s) / (4 sec) = 2 m/s² .
From 10 to 12 seconds, acceleration = (-4 m/s) / (2 sec) = -2 m/s² .
B). From 12 to16 seconds, you can read the speed right from
the graph. It's 4 m/s .
C). From 2 to 10 seconds, the objects speed is a steady 8 m/s.
Covering 8 m/s every second for 8 seconds, it covers 64 meters.
Do you remember that distance is the area under the speed/time
graph? You can see that plainly on this graph. From 2 to 10 sec,
there are 16 blocks. Each block is (2 m/s) high and (2 sec) wide,
so its area is (2 m/s) x (2 sec) = 4 meters. The area of 16 blocks
is (16) x (4 meters) = 64 meters.
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#15.
a). constant velocity on a distance graph is a line that slopes up;
constant velocity on a velocity graph is a horizontal line;
b). positive constant acceleration on a distance graph is a
line that curves up;
positive constant acceleration on a velocity graph is a
straight line that slopes up;
c). "uniformly slowing down to a stop" on a distance graph
is a line that's less and less curved as time goes on, and
eventually reaches the x-axis.
"uniformly slowing down to a stop" on a velocity graph is
a straight line that slopes down, and stops when it reaches
the x-axis.
Hi,
Your question is "which graph represents uniform motion", the answer would be option 1
Have a good one!
The sources of the waves must be coherent, which means they emit identical waves with a constant phase difference and the waves should be monochromatic - they should be of a single wavelength.
~Sanjith U.
Answer:
332m/s
Explanation:
We know the formula for velocity is v=d/t, but this particular question is asking about an echo. The problem tells us the distance to the bird from the bat, which is 19.4. To find the echo, you need to find the distance from the bat to the bird and back.
So
19.4 x 2 = 38.8
And then plug into the equation
v = 38.8 / 0.117
= 332m/s
Answer:
Approximately 
.
Explanation:
Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.
Look up the Rydberg constant for hydrogen: 
.
Look up the speed of light in vacuum: 
.
Look up Planck's constant: 
.
Apply the Rydberg formula to find the wavelength 
(in vacuum) of the photon in question:
.
The frequency of that photon would be:
.
Combine this expression with the Rydberg formula to find the frequency of this photon:
.
Apply the Einstein-Planck equation to find the energy of this photon:
.
(Rounded to three significant figures.)
