Answer:
0.47 m
Explanation:
= Number of vibrations = 37
= total time taken = 33 s
= time period of each vibration
frequency of vibration is given as
Hz
= distance traveled along the rope = 421 cm = 4.21 m
= time taken to travel the distance = 8 s
= speed of the wave
Speed of the wave is given as
= wavelength of the harmonic wave
wavelength of the harmonic wave is given as
Answer:
The magnitude of the force required to bring the mass to rest is 15 N.
Explanation:
Given;
mass, m = 3 .00 kg
initial speed of the mass, u = 25 m/s
distance traveled by the mass, d = 62.5 m
The acceleration of the mass is given as;
v² = u² + 2ad
at the maximum distance of 62.5 m, the final velocity of the mass = 0
0 = u² + 2ad
-2ad = u²
-a = u²/2d
-a = (25)² / (2 x 62.5)
-a = 5
a = -5 m/s²
the magnitude of the acceleration = 5 m/s²
Apply Newton's second law of motion;
F = ma
F = 3 x 5
F = 15 N
Therefore, the magnitude of the force required to bring the mass to rest is 15 N.
If it's Kepler's law of equal areas you're talking about,
then the first of the four statements is true.
Answer:
The answer to the question is
The distance d, which locates the point where the light strikes the bottom is 29.345 m from the spotlight.
Explanation:
To solve the question we note that Snell's law states that
The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction
n₁sinθ₁ = n₂sinθ₂
y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and
θ₁ = 14.511 °
n₁ = 1.0003 = refractive index of air
n₂ = 1.33 = refractive index of water
Therefore sinθ₂ = 
= 
= 0.1885 and θ₂ = 10.86 °
Since the water depth is 4.0 m we have tanθ₂ = 
or x₂ = 
=
= 20.845 m
d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.
100 ml
100 ml of the stock solution is required to prepare the order.
We know that C1V1 = C2V2
where C1= 2%
V1 = 500ml
C2= 10%
V2 = ?
V2 = C1V1 / C2
= 500 * 2% / 10%
=100
V2 = 100 ml
<h3>What is meant by stock solution?</h3>
- A stock solution is a sizable amount of a typical reagent in a standardized concentration, like sodium hydroxide or hydrochloric acid.
- This phrase is frequently used in analytical chemistry while doing operations like titrations where it's crucial to employ precise solution concentrations.
<h3>What distinguishes a standard solution from a stock solution?</h3>
- The main distinction between stock solution and standard solution is that the former is a highly concentrated solution while the later is a solution whose concentration is precisely known.
- Because standard solutions frequently arrive as stock solutions, the phrases "stock solution" and "standard solution" are connected.
To learn more about stock solution preparation visit:
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