The scientists that study the forces and shape of the earth is called geologist.
Answer:
900 cm/s or 9 m/s.
Explanation:
Data obtained from the question include the following:
Length (L) = 30 cm
frequency (f) = 60 Hz
Velocity (v) =.?
Next, we shall determine the wavelength (λ).
This is illustrated below:
Since the wave have 4 node, the wavelength of the wave will be:
λ = 2L/4
Length (L) = 30 cm
wavelength (λ) =.?
λ = 2L/4
λ = 2×30/4
λ = 60/4
λ = 15 cm
Therefore, the wavelength (λ) is 15 cm
Now, we can obtain the speed of the wave as follow:
wavelength (λ) = 15 cm
frequency (f) = 60 Hz
Velocity (v) =.?
v = λf
v = 15 × 60
v = 900 cm/s
Thus, converting 900 cm/s to m/s
We have:
100 cm/s = 1 m/s
900 cm/s = 900/100 = 9 m/s
Therefore, the speed of the wave is 900 cm/s or 9 m/s.
Answer:
Explanation:
Given
Mass of object (m)=6 kg
falling height(h)=10 m
mass of water(
)=600 gm
temperature of water =15
specific heat of water ![=4.186 j/g-^{\circ}C](https://tex.z-dn.net/?f=%3D4.186%20j%2Fg-%5E%7B%5Ccirc%7DC)
Let T be the Final Temperature of water
Here Object Potential Energy is converted into Heat energy which will be absorbed by water
Potential Energy(P.E.)![=mgh=6\times 9.81\times 10=588.6 J](https://tex.z-dn.net/?f=%3Dmgh%3D6%5Ctimes%209.81%5Ctimes%2010%3D588.6%20J)
Heat supplied![=m_wc(\Delta T)](https://tex.z-dn.net/?f=%3Dm_wc%28%5CDelta%20T%29)
![H.E.=600\times 4.186\times (T-16)](https://tex.z-dn.net/?f=H.E.%3D600%5Ctimes%204.186%5Ctimes%20%28T-16%29%20)
![588.6=2511.6\times (T-16)](https://tex.z-dn.net/?f=588.6%3D2511.6%5Ctimes%20%28T-16%29)
T-16=0.234
![T=16.234^{\circ}C](https://tex.z-dn.net/?f=T%3D16.234%5E%7B%5Ccirc%7DC)
This is not an efficient way of heating water as there is only
increase in temperature.
Answer:
The increase in potential energy of the ball is 115.82 J
Explanation:
Conceptual analysis
Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:
U = m × g × h
U: Potential Energy in Joules (J)
m: mass in kg
g: acceleration due to gravity in m/s²
h: height in m
Equivalences
1 kg = 1000 g
1 ft = 0.3048 m
1 N = 1 (kg×m)/s²
1 J = N × m
Known data
![h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m](https://tex.z-dn.net/?f=h_2%20%3D%20265ft%20%2A%20%5Cfrac%7B0.3048m%7D%7Bft%7D%20%3D%2080.77m)
![h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m](https://tex.z-dn.net/?f=h_1%20%3D%203ft%20%2A%20%5Cfrac%7B0.3048m%7D%7Bft%7D%20%3D%200.914m)
![m = 148g*\frac{1kg}{1000g} = 0.148kg](https://tex.z-dn.net/?f=m%20%3D%20148g%2A%5Cfrac%7B1kg%7D%7B1000g%7D%20%3D%200.148kg)
![g = 9.8 \frac{m}{s^2}](https://tex.z-dn.net/?f=g%20%3D%209.8%20%5Cfrac%7Bm%7D%7Bs%5E2%7D)
Problem development
ΔU: Potential energy change
ΔU = U₂ - U₁
U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁
U₂ - U₁ = mₓg(h₂ - h₁)
![U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J](https://tex.z-dn.net/?f=U_2%20-%20U_1%20%3D%200.148kg%20%2A%209.8%20%5Cfrac%7Bm%7D%7Bs%5E2%7D%2A%2880.77m%20-%200.914m%29%20%3D%20115.82%20N%20%2A%20m%20%3D%20115.82J)
The increase in potential energy of the ball is 115.82 J