Answer:
The tiger would not be able to produce glucose causing it to stop running
Explanation:
Since the mitochondria is in charge of producing ATP the tiger would not be able to use any glucose causing it to not be able to run.
Answer:
6 m/s
Explanation:
12m / 2s = 6 m/s
Hope that's the answer you seek.
Answer:
velocity = 472 m/s
velocity = 52.4 m/s
Explanation:
given data
steady rate = 0.750 m³/s
diameter = 4.50 cm
solution
we use here flow rate formula that is
flow rate = Area × velocity .............1
0.750 =
× (4.50×
)² × velocity
solve it we get
velocity = 472 m/s
and
when it 3 time diameter
put valuer in equation 1
0.750 =
× 3 × (4.50×
)² × velocity
velocity = 52.4 m/s
Answer: a = 0.4m/s^2 - 9.8*c where c is the coefficient of kinetic friction of the surface
Explanation: We know that, by the second Newton's law, a = F/m
where a is the acceleration, F is the net force and m is the mass of the object.
Then, if the surface is frictionless, the total force applied in the object is 10N, and the mass of the object is 25kg, so the acceleration is:
a =10N/25kg = 0.4m/s^2.
But if the surface is frictional, there will be a force of friction applied in the mass (this depends on the coefficient of friction and the weight of the mass), this means that the acceleration will be reduced.
If = -(9.8*25)*c
where c is a number that is bigger than 0 and smaller than 1, is called the coefficient of kinetic friction.
So the total force is now:
F = (10 - 9.8*25*c)
Then, the acceleration in a frictional surface is equal to:
a = (10 - 9.8*25*c)/25 = 0.4m/s^2 - 9.8*c
Answer:
<em>Part A</em><em>:</em>
a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.
<em>Part B</em><em>:</em>
b) If the spacing between the slits is decreased the fringe spacing Δy will increase.
<em>Part C</em><em>:</em>
a) If the distance to the screen is decreased the fringe spacing will decrease.
<em>Part D</em><em>:</em>
The dot in the center of fringe E is
farther from the left slit than from the right slit.
Explanation:
In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.
The position of bright fringes in the screen where the pattern is formed can be calculated with


- m is the order number.
is the wavelength of the monochromatic light.- L is the distance between the screen and the two slits.
- d is the distance between the slits.
- Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light
is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.
- Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
- Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.
- Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be
We simply replace the values in that equation :


The dot in the center of fringe E is
farther from the left slit than from the right slit.