Answer:
A = 2 cm
, λ = 8 cm
Explanation:
The amplitude of a wave is the maximum height it has, in this case the height is measured by the vertical ruler,
We are told the balance point is in the reading of 5 cm, that the maximum reading is 3 cm and the Minimum reading is 7 cm. Therefore, the distance from the ends of the ridge to the point of equilibrium is
d = 7-5 = 2 cm
d = 5-3 = 2 cm
A = 2 cm
The wavelength is the minimum horizontal distance for which the wave is repeated, that is measured by the horizontal ruler.
The initial reading for 4 cm and the final reading for 8 cm, this distance corresponds to a crest of the wave, the complete wave is formed by two crests whereby the wavelength is twice this value
Δx = 8-4 = 4 cm
λ = 2 Δx
λ = 8 cm
To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as
![m_1v_1+m_2v_2 = (m_1+m_2)v_f](https://tex.z-dn.net/?f=m_1v_1%2Bm_2v_2%20%3D%20%28m_1%2Bm_2%29v_f)
Where,
= Mass of each object
= Initial velocity of each object
= Final Velocity
Since the receiver's body is static for the initial velocity we have that the equation would become
![m_2v_2 = (m_1+m_2)v_f](https://tex.z-dn.net/?f=m_2v_2%20%3D%20%28m_1%2Bm_2%29v_f)
![(0.42)(21) = (90+0.42)v_f](https://tex.z-dn.net/?f=%280.42%29%2821%29%20%3D%20%2890%2B0.42%29v_f)
![v_f = 0.0975m/s](https://tex.z-dn.net/?f=v_f%20%3D%200.0975m%2Fs)
Therefore the velocity right after catching the ball is 0.0975m/s
Answer:
88 m/s
Explanation:
To solve the problem, we can use the following SUVAT equation:
![v^2-u^2=2ad](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2ad)
where
v is the final velocity
u is the initial velocity
a is the acceleration
d is the distance covered
For the car in this problem, we have
d = 484 m is the stopping distance
v = 0 is the final velocity
is the acceleration
Solving for u, we find the initial velocity:
![u=\sqrt{v^2-2ad}=\sqrt{-2(8.0)(484)}=88 m/s](https://tex.z-dn.net/?f=u%3D%5Csqrt%7Bv%5E2-2ad%7D%3D%5Csqrt%7B-2%288.0%29%28484%29%7D%3D88%20m%2Fs)
Answer:
In the table, 1=46.7 °C, 1=165 J, 2=819 J, 3=1510 J, and 4=2830 J.
Other experiments determine that the material has a temperature of fusion of
fusion =235 °C and a temperature of vaporization of vapor=481 °C.
If the sample of material has a mass of =8.60 g, calculate the specific heat when this material is a solid, and when it is liquid, l