Answer:
changes electrical energy into mechanical energy
<em>The gravitational force between two objects is inversely proportional to the square of the distance between the two objects.</em>
The gravitational force between two objects is proportional to the product of the masses of the two objects.
The gravitational force between two objects is proportional to the square of the distance between the two objects. <em> no</em>
The gravitational force between two objects is inversely proportional to the distance between the two objects. <em> no</em>
The gravitational force between two objects is proportional to the distance between the two objects. <em> no</em>
The gravitational force between two objects is inversely proportional to the product of the masses of the two objects. <em> no</em>
Answer:
the answer is b
Explanation:
Second and third class levers are differentiated by <u>the location of the </u><u>load.</u>
<em>Hope</em><em> </em><em>this</em><em> </em><em>help</em><em> </em><em>you</em><em> </em><em>out </em><em>and have</em><em> </em><em>a </em><em>nice</em><em> </em><em>day </em><em>=</em><em>)</em>
Answer:
-The speed of sound at 33°C is 362.8 m/s.
-The wavelength at a frequency at 5 kHz is 0.07256 m .
Explanation:
let v = 343 m/s be the speed of sound.
let T be the temperature.
then the speed of sound V, at 33°C is given by:
V = v + 0.6×T
= 343 + 0.6×33
= 362.8 m/s
Therefore, the speed of sound at 33°C is 362.8 m/s.
the wavelength at a frequency of f = 5kHz = 5000 Hz is given by:
λ = V/f
= (362.8)/(5000)
= 0.07256 m
Therefore, the wavelength at a frequency at 5 kHz is 0.07256 m .
Answer:
Recall the Diffraction grating formula for constructive interference of a light
y = nDλ/w Eqn 1
Where;
w = width of slit = 1/15000in =6.67x10⁻⁵in =
6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m
D = distance to screen
λ = wavelength of light
n = order number = 1
Given
y1 = ? from 1st order max to the central
D = 2.66 m
λ = 633 x 10-9 m
and n = 1
y₁ = 0.994m
Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 1) = 0.994m
Q b. How far (m) from the central maximum (m = 0) is the second-order maximum (m = 2) observed?
w = width of slit = 1/15000in =6.67x10⁻⁵in =
6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m
D = distance to screen
λ = wavelength of light
n = order number = 1
Given
y1 = ? from 1st order max to the central
D = 2.66 m
λ = 633 x 10⁻⁹ m
and n = 2
y₂ = 0.994m
Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 2) =1.99m
