The goal of Science is to expand knowledge.
Answer:
C) 20 m/s
Explanation:
Wave: A wave is a disturbance that travels through a medium and transfers energy from one point to another, without causing any permanent displacement of the medium itself. Examples of wave are, water wave, sound wave, light rays, radio waves. etc.
The velocity of a moving wave is
v = λf ............................ Equation 1
Where v = speed of the wave, λ = wave length, f = frequency of the wave.
Given: f = 2 Hz (two complete cycles in one seconds), λ = 10 meters
Substituting these values into equation 1
v = 2×10
v = 20 m/s.
Thus the speed of the wave = 20 m/s
The right option is C) 20 m/s
Answer:
(a) Eₐ = 6.36 J/s
(b) Eₐ = 4.64 J/s
Explanation:
Stefan-Boltzmann law: States that the total energy per second radiated or absorbed by a black body is directly proportional to the absolute temperature.
Using, Stefan-Boltzmann equation
Eₐ =eσAT⁴ ................ Equation 1
where Eₐ = Radiant energy absorbed per seconds, e = emissivity, σ = stefan - boltzman constant, A = Surface area. and T = temperature in kelvin
(a) Where e = 0.89, σ = 5.67 ×10⁻⁸ watt/m²/K⁴, A = 140 cm² = 140 cm²(m²/10000cm²) = 0.014 m², T = 35 °C = (35 + 273) K = 308 K.
Applying these values in equation 1 above,
Eₐ = 0.89 × 5.67 ×10⁻⁸ × 0.014 × (308)⁴
Eₐ =6.36 J/s
(b) when e = 0.65,
∴ Eₐ = 0.65 × 5.67 × 10⁻⁸ × 0.014 × (308)⁴
Eₐ = 4.64 J/s
Answer:
22.7 meters
Explanation:
Let's remind the difference between distance and displacement:
- distance: the total distance travelled by an object in all its paths
- displacement: the different between the final and initial position of the object
In this case, the problem asks to find the distance covered by the ball. This will be the sum of the distances covered by the ball in each part of its motion, therefore:
(instead, the displacement will be the difference between the final and initial position of the ball, therefore:
)
Answer:
x = 1.63 m
Explanation:
mass (m) = 10 kg
μk = 0.3
velocity (v) = 3.1 m/s
Assuming that most of the computers weight is applied on the belt instantaneously, we can apply the constant acceleration equation below
x = 
where a = μk.g , therefore
x = 
/2μk.g
x = (3.1 x 3.1)/(2 x 0.3 x 9.8)
x = 1.63 m
