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)
Answer:
a) y = 2.4 x 10⁻³ m = 0.24 cm
b) y = 3.2 x 10⁻³ m = 0.32 cm
Explanation:
The formula of Young's Double Slit experiment will be used here:
where,
y = distance between dark spots = ?
λ = wavelength
L = distance of screen = 2 m
d = slit width = 4 x 10⁻⁴ m
a) FOR λ = 480 nm = 4.8 x 10⁻⁷ m:
<u>y = 2.4 x 10⁻³ m = 0.24 cm</u>
<u></u>
a) FOR λ = 640 nm = 6.4 x 10⁻⁷ m:
<u>y = 3.2 x 10⁻³ m = 0.32 cm</u>
Answer:
4.7 s
Explanation:
The complete question is presented in the attached image to this solution.
v(t) = 61 - 61e⁻⁰•²⁶ᵗ
At what time will v(t) = 43 m/s?
We just substitute 43 m/s into the equation for the velocity of the diver and solve for t.
43 = 61 - 61e⁻⁰•²⁶ᵗ
- 61e⁻⁰•²⁶ᵗ = 43 - 61 = -18
e⁻⁰•²⁶ᵗ = (18/61) = 0.2951
In e⁻⁰•²⁶ᵗ = In 0.2951 = -1.2205
-0.26t = -1.2205
t = (1.2205/0.26) = 4.694 s = 4.7 s to the nearest tenth.
Hope this Helps!!!
Answer:
Explanation:
If a number of less than 1, then the number has a decimal point like
0.085, 0.008 e.t.c.
The zeros before the none zero digit are insignificant. The significant figure is 8 and 5.
But if there a zero between the none zero e.g. 0.0087056
Here the zero between 7 and 5 is significant, then the significant numbers are 8,7,0,5,6
But if the zero is not in between the none zero digit, then the zero is insignificant
E.g 0.05800
The last two zero is insignificant, the significant number is 5 and 8
So, If a positive numbers less than 1, the zeros between the decimal point and a non-zero number are NOT significant.
