Answer:
The coefficient of thermal expansion of material B is more than A
Explanation:
As we know that when we apply heat to a material it expands. The amount of expansion depends on the nature of the material. The amount of expansion depends on the coefficient of thermal expansion of the material.
If a material has high value of thermal expansion coefficient then it expands more.
here material A expands less and material B expands more because the pin which is made by material A becomes loosen, so the coefficient of thermal expansion of B is more than the coefficient of thermal expansion of A
Answer:
Explanation:
Assuming the squirrel is jumping off the ground, here's what we know but don't really know...
v₀ = 4.0 at 50.0°
So that's not really the velocity we are looking for. We are dealing with a max height problem, which is a y-dimension thing. Therefore, we need the squirrel's upward velocity, which is NOT 4.0 m/s. We find it in the following way:
which gives us that the upward velocity is
v₀ = 3.1 m/s
Moving on here's what we also know:
a = -9.8 m/s/s and
v = 0
Remember that at the very top of the parabolic path, the final velocity is 0. In order to find the max height of the squirrel, we need to know how long it took him to get there. We are using 2 of our 3 one-dimensional equations in this problem. To find time:
v = v₀ + at and filling in:
0 = 3.1 - 9.8t and
-3.1 = -9.8t so
t = .32 seconds.
Now that we know how long it took him to get to the max height, we use that in our next one-dimensional equation:
Δx = 
and filling in:
Δx = 
and using the rules for adding and subtracting sig fig's correctly, we can begin to simplify this:
Δx = .99 - .50 so
Δx = .49 meters
Charge stored in a capacitor given by Q = C*V.
So, C = Q / V.
V=30V, Q=0.003C
C = 0.0001F or 100μF
<span>Charge of the glass bead Q = 8.0 x 10^-9 C
Distance d = 2.0 cm = 0.02 m
Coulombs constant K = 8.99 x 10^9 Nm^2/C^2
Electric Field E = k x Q / d^2 = 8.99 x 10^9 x 8.0 x 10^-9 / (0.02)^2
E = 71.92 / 0.0004 = 17.98 x 10^4
The electric field is 1.8 x 10^5 N/C</span>
Answer:
A) The ball hits the ground 74.45 m far from the hitting position.
B) Maximum height of the ball = 18.57 m
Explanation:
There are two types of motion in this horizontal and vertical motion.
We have velocity = 27 m/s at 45° above the horizontal
Horizontal velocity = 27cos45 = 19.09 m/s
Vertical velocity = 27sin45 = 19.09 m/s
Time to reach maximum height,
v = u + at
0 = 19.09 - 9.81 t
t = 1.95 s
So total time of flight = 2 x 1.95 = 3.90 s
A) So the ball travels at 19.09 m/s for 3.90 seconds.
Horizontal distance traveled = 19.09 x 3.90 = 74.45 m
So the ball hits the ground 74.45 m far from the hitting position.
B) We have vertical displacement
S = ut + 0.5 at²
H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m
Maximum height of the ball = 18.57 m
