Answer:
53.32°C
Explanation:
Length of the aluminium wing = 35 m
Change in length of aluminium wing = 0.03 m
The linear expansion coefficient of aluminium 
We know that change in length is given by 
So 

So final temperature
Answer:
Electric field by charged disk is given as
E = (Charge Density/2u0)*[1 - (z/sqrt(z^2 - R^2))]
R = 9.54cm = 0.0954m, z = 1.01m, Charge density = 4.07 x 10^-6C/m2, e0 = 8.85 x 10^-12F/m.
Substituting all the values in to equation,
E = (2.299 x 10^5) x (8.931 x 10^-3)
E = 2.053 x 10^3N/C
Explanation:
The time elapsed is 9 seconds
Explanation:
The motion of the ball is a uniformly accelerated motion (a motion with constant acceleration), so we can use the following suvat equation:
where
:
v is the final velocity of the ball
u is the initial velocity
a is the acceleration
t is the time elapsed
For the ball in this problem, we have:
u = 3 m/s is the initial velocity
v = 34.5 m/s is the final velocity
is the acceleration
Solving for t, we find the time taken for this change in velocity:

Learn more about acceleration:
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Answer:
a) A = 0.07129 m²
b) A / A ’= 1.77
Explanation:
In this exercise we are asked to find the area in SI units, so let's start by reducing the dimensions to SI units.
width a = 8.5 inch (2.54 cm 1 inch) (1 m / 100 cm)
a = 0.2159 m
length l = 13 inch (2.54 cm / 1 inch) (1 m / 100 cm)
l = 0.3302 m
The area of a rectangle is
A = l a
A = 0.3302 0.2159
A = 0.07129 m²
b) we have a second sheet with reduced dimensions
a ’= 3/4 a
l ’= ¾ l
Let's find the area of this glossy sheet
A ’= l’ a ’
A ’= ¾ l ¾ a
A ’= 9/16 l a
To find the factor we divide the two quantities
A / A ’= l a 16 / (9 l a
A / A ’= 1.77
Your position in meters will, measured relative to the starting point of the car behind you, be
x1(t) = 10 + 23.61 t - 1/2 4.2 t^2
his position will be
x2(t) = 16.67 t
Hence at any time the separation s(t) will be
s(t) = x1(t) - x2(t) = 10 + 6.94 t -2.1 t^2
Now I assume you mean that you will decelerate UNTIl you are driving at the legal speed limit (60 km/h). That will take you:
16.67 m/s = 23.61m/s - 4.2 m/s^2 * t
t = 1.65 seconds
What is the separation at that time? If it is still greater than zero, there will be no collision:
s(1.65) = 10 + 6.94 *1.65 -2.1 (1.65)^2 = 15.73 meters.
Hence you will NOT collide. The 1.65 s you calculated was the time needed to brake to the speed of 60 km/h.