To solve this problem, it will be necessary to apply the concepts related to the fundamental resonance frequency in a closed organ pipe.
This is mathematically given as
For fundamental frequency n is 0, then,
When,
v = Velocity of sound
L = Length,
Rearranging to find the velocity,
Therefore the speed of sound in this gas is 416m/s
Answer:
Explanation:
The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:
Here d is the distance at which the earthquake take place and 
is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:
The velocity of the canoe is 1.7 m/s.
<h3>What is momentum?</h3>
Momentum in physics is the products of mass and velocity. Now we have to find momentum with the formula; p = mv
a) Initial momentum = (15)8 m/s + 135 = 255 Kgms-1
b) Since momentum is conserved, the total momentum after throwing the anchor is still 255 Kgms-1
c) The final velocity of the boat is obtained from;
255 Kgms-1 = (15Kg + 135 Kg) v
v = 255 Kgms-1/(15Kg + 135 Kg)
v = 1.7 m/s
Learn more about momentum: brainly.com/question/904448
The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g
<u>Explanation:</u>
Given,
Temperature, T = 0°C
Initial mass, Mi = 62kg
Speed, s = 5.48m/s
Distance, x = 26.8m
Friction is present.
Mass of ice melted = ?
We know,
The amount of energy required for the melting of ice is exactly equal to the initial kinetic energy of the block of ice
and
Therefore, 
KE = 930.94 Joules
Ice melting lateral heat is 334 kJ/kg = 334000 J/kg.
Therefore, the melted mass of the ice = 930.94 / 334000 = 0.00278 kg = 2.78 g.
Thus, The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g
Answer:
W = 1080.914 J
Explanation:
f(x) = 1100xe⁻ˣ
Work done by a variable force moving through a particular distance
W = ∫ f(x) dx (with the integral evaluated between the interval that the force moves through)
W = ∫⁶₀ 1100xe⁻ˣ dx
W = 1100 ∫⁶₀ xe⁻ˣ dx
But the integral can only be evaluated using integration by parts.
∫ xe⁻ˣ dx
∫ vdu = uv - ∫udv
v = x
(dv/dx) = 1
dv = dx
du = e⁻ˣ dx
∫ du = ∫ e⁻ˣ dx
u = -e⁻ˣ
∫ vdu = uv - ∫udv
∫ xe⁻ˣ dx = (-e⁻ˣ)(x) - ∫ (-e⁻ˣ)(dx)
= -xe⁻ˣ - e⁻ˣ = -e⁻ˣ (x + 1)
∫ xe⁻ˣ dx = -e⁻ˣ (x + 1) + C (where c = constant of integration)
W = 1100 ∫⁶₀ xe⁻ˣ dx
W = 1100 [-e⁻ˣ (x + 1)]⁶₀
W = 1100 [-e⁻⁶ (6 + 1)] - [-e⁰ (0 + 1)]
W = 1100 [-0.0173512652 + 1]
W = 1100 × (0.9826487348)
W = 1080.914 J
Hope this Helps!!!
