Explanation:
The total distance in a path is called distance.
The shortest distance between two points is called displacement.
a) Here, the distance travelled between the park to your friend's house and back is
Distance between park to friends house + Distance from friend's house to your house.
b) Displacement would be the shortest distance between the park and your house.
a) Distance walked between your house to library and back is
Distance between your house and library + Distance between your house and library
b) Displacement would be zero (0) as the distance between you initial point and final point is zero. Here, the initial and final points are the same
Answer:
121.0 W
Explanation:
We use the equation for rate of heat transfer during radiation.
Q/t = σεA(T₂⁴ - T₁⁴)
Since temperature of surroundings = T₁ = -20.0°C = 273 +(-20) = 253 K, and temperature of skier's clothes = T₂ = 5.50°C = 273 + 5.50 = 278.5 K.
Surface area of skier , A = 1.60 m², emissivity of skier's clothes, ε = 0.70 and σ = 5.67 × 10⁻⁸ W/m²K⁴
.
Therefore, the rate of heat transfer by radiation Q/t is
Q/t = σεA(T₂⁴ - T₁⁴) = (5.67 × 10⁻⁸ W/m²K⁴
) × 0.70 × 1.60 m² × (278.5⁴ - 253⁴) = 6.3054 × (1918750544.0625) × 10⁻⁸ W = 1.2098 × 10² W = 120.98 W ≅ 121.0 W
I know that 4. Is wave 2 and 5. Is wave 3.
Answer:
The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)
Explanation:
Fundamental frequency = wave velocity/2L
where;
L is the length of the stretched rubber
Wave velocity = 
Frequency (F₁) = 
To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.
Given:
L₂ =2L₁ = 2L
T₂ = 2T₁ = 2T
(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)
F₂ = ![\frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B%5Cfrac%7B2T%7D%7B0.5%28%5Cfrac%7BM%7D%7BL%7D%29%7D%7D%7D%7B4%2AL%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B4%28%5Cfrac%7BT%7D%7B%5Cfrac%7BM%7D%7BL%7D%7D%7D%29%7D%7B4%2AL%7D%20%3D%20%5Cfrac%7B2%7D%7B2%7D%20%5B%5Cfrac%7B%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cfrac%7BM%7D%7BL%7D%7D%7D%7D%7B2%2AL%7D%5D%20%3D%20F_1)
Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).
