Length of the copper rod = 2.5 meters
Speed at which sound travels through copper = 3560 meter per second
Let us assume the time taken
by sound to cover the given distance = x seconds
We already know that
Speed = Distance/ Time
Then
Time = Distance/ Speed
x = 2.5/3560 seconds
= 0.0007 seconds.
This can ve done by hitting one end of the rod and then receiving the sound at the other end and using the stop clock to measure the time taken.
I believe the correct answer from the choices listed above is the last option. If the volatility of X is higher than that of Y, then <span>Y’s molecules experience stronger London dispersion forces than X’s molecules. All molecules has london dispersion forces. Also, the stronger the bond, the harder it is to volatilize. Hope this answers the question.</span>
Answer:
1) P₁ = -2 D, 2) P₂ = 6 D
Explanation:
for this exercise in geometric optics let's use the equation of the constructor
where f is the focal length, p and q are the distance to the object and the image, respectively
1) to see a distant object it must be at infinity (p = ∞)
q = f₁
2) for an object located at p = 25 cm
We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm
we substitute in the equations
1) f₁ = -50 cm
2)
= 0.06
f₂ = 16.67 cm
the expression for the power of the lenses is
P = 
where the focal length is in meters
1) P₁ = 1/0.50
P₁ = -2 D
2) P₂ = 1 /0.16667
P₂ = 6 D
Constant velocity means moving in a straight line at a speed that doesn't change. If the object is moving with constant velocity then its acceleration is zero. Acceleration is the rate at which velocity is changing.
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
