To solve this problem we will use the concepts related to thermal expansion in a body for which the initial length, the coefficient of thermal expansion and the temperature change are related:
![\Delta L = L0\alpha\Delta T](https://tex.z-dn.net/?f=%5CDelta%20L%20%3D%20L0%5Calpha%5CDelta%20T)
Where,
= Change in Length
= Coefficient of linear expansion
= Change in temperature
= Initial Length
Our values are:
![L_0 = 3.45m](https://tex.z-dn.net/?f=L_0%20%3D%203.45m)
![\Delta T = 235-20 = 215\°C](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%20235-20%20%3D%20215%5C%C2%B0C)
Replacing we have,
![\Delta L = (3.49) (5.5*10^{-7}) [(215)](https://tex.z-dn.net/?f=%5CDelta%20L%20%3D%20%283.49%29%20%285.5%2A10%5E%7B-7%7D%29%20%5B%28215%29)
![\Delta L = 0.0004126m](https://tex.z-dn.net/?f=%5CDelta%20L%20%3D%200.0004126m)
![\Delta L = 0.4126mm](https://tex.z-dn.net/?f=%5CDelta%20L%20%3D%200.4126mm)
Therefore the change in milimiters was 0.4126mm
Speed and velocity have the same magnitudes. The only difference is that speed is a scalar quantity and velocity is a vector quantity. In other words, speed is just a magnitude, while velocity is a magnitude with direction. They're essentially the same.
Let's convert miles to meters and minutes to seconds
1/4 mile = 402.34 meters ( 1 mile = 1609 m)
13.1 minutes = 786 seconds (1 minute = 60 seconds)
Speed is calculated as distance over time, thus,
Speed = (402.34 meters)*8/786 seconds
a.) Speed = 4.1 m./s
b.) Velocity = 4.1 m/s
Answer:
D) 31 m, 19 Hz
Explanation:
The equation of the wave in the problem is
![y = 0.5 sin (0.20 x+120t)](https://tex.z-dn.net/?f=y%20%3D%200.5%20sin%20%280.20%20x%2B120t%29)
In general, the equation of a travelling wave is written as
![y=Asin(kx+\omega t)](https://tex.z-dn.net/?f=y%3DAsin%28kx%2B%5Comega%20t%29)
where
A is the amplitude
is the wave number, with
being the wavelength of the wave
is the angular frequency and f is the frequency
By comparing the two equations, we see that for this wave:
![k = 0.20 m^{-1}\\\omega = 120 rad/s](https://tex.z-dn.net/?f=k%20%3D%200.20%20m%5E%7B-1%7D%5C%5C%5Comega%20%3D%20120%20rad%2Fs)
So now we can use the two equations for k and
to find the wavelength and the frequency of the wave:
![\lambda=\frac{2\pi}{k}=\frac{2\pi}{0.20}=31 m\\f = \frac{\omega}{2\pi}=\frac{120}{2\pi}=19 Hz](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B2%5Cpi%7D%7Bk%7D%3D%5Cfrac%7B2%5Cpi%7D%7B0.20%7D%3D31%20m%5C%5Cf%20%3D%20%5Cfrac%7B%5Comega%7D%7B2%5Cpi%7D%3D%5Cfrac%7B120%7D%7B2%5Cpi%7D%3D19%20Hz)
Answer:
c. The less massive object will have less momentum if the velocities are the same.
Explanation:
p = mv
If the velocities are the same, p ∝ m, so the less massive object will have less momentum.
a. is wrong. If the masses are equal, p ∝ v, so the object with the higher velocity will have the greater momentum.
b. is wrong. If an object has both more mass and a greater velocity, it will have the greater momentum.
d. is wrong. If the velocities are the same, the more massive object will have more momentum.
Answer:
a) θ = 14.23º, b) θ₂ = 75.77, c) t = 0.6019 s
Explanation:
This is a missile throwing exercise.
a) the reach of the ball is the distance traveled for the same departure height
R =
sin 2θ =
sin 2θ = 7.00 9.8 / 12.0²
2θ = sin⁻¹ (0.476389) = 28.45º
θ = 14.23º
the complementary angle that gives the same range is the angle after 45 that the same value is missing to reach 90º
θ ’= 90 -14.23
θ’= 75.77º
b) the two angles that give the same range are
θ₁ = 14.23
θ₂ = 75.77
the greater angle has a much greater height so the time of the movement is greater and has a greater chance of being intercepted by the other team.
C) the time of the pass can be calculated with the expression
x = v₀ₓ t
t = x / v₀ₓ
t = 7 / 11.63
t = 0.6019 s