The answer for this change in the magnitude of momentum is the same for both because momentum is always conserved so both vehicles have the identical change.
So for determining who has the greater change in kinetic energy, momentum (P) = mv so P^2 = m^2 v^2 P^2 / 2m = 1/2 m v^2 = energy So the weightier the mass the smaller the energy change for the same momentum change so in here, the car has a greater change in kinetic energy.
Answer:
The starting position of the runner.
Explanation:
When you look at the graph, you can see that the first point on the graph is twenty on the y-axis.
The runner starts at twenty, and ends at thirty.
Therefore, the runner starts at twenty on the y-axis, so it's the starting position of the runner.
Answer:
A) T1 = 566 k = 293°C
B) T2 = 1132 k = 859°C
Explanation:
A)
The average kinetic energy of the molecules of an ideal gas is givwn by the formula:
K.E = (3/2)KT
where,
K.E = Average Kinetic Energy
K = Boltzman Constant
T = Absolute Temperature
At 10°C:
K.E = K10
T = 10°C + 273 = 283 K
Therefore,
K10 = (3/2)(K)(283)
FOR TWICE VALUE OF K10:
T = T1
Therefore,
2 K10 = (3/2)(K)(T1)
using the value of K10:
2(3/2)(K)(283) = (3/2)(K)(T1)
<u>T1 = 566 k = 293°C</u>
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B)
The average kinetic energy of the molecules of an ideal gas is given by the formula:
K.E = (3/2)KT
but K.E is also given by:
K.E = (1/2)(m)(vrms)²
Therefore,
(3/2)KT = (1/2)(m)(vrms)²
vrms = √(3KT/m)
where,
vrms = Root Mean Square Velocity of Molecule
K = Boltzman Constant
T = Absolute Temperature
m = mass
At
T = 10°C + 273 = 283 K
vrms = √[3K(283)/m]
FOR TWICE VALUE OF vrms:
T = T2
Therefore,
2 vrms = √(3KT2/m)
using the value of vrms:
2√[3K(283)/m] = √(3KT2/m)
2√283 = √T2
Squaring on both sides:
(4)(283) = T2
<u>T2 = 1132 k = 859°C</u>
Answer:
A) 12.08 m/s
B) 19.39 m/s
Explanation:
A) Down the hill, we will apply Newton’s second law of motion in the downward direction to get:
mg(sinθ) – F_k = ma
Where; F_k is frictional force due to kinetic friction given by the formula;
F_k = (μ_k) × F_n
F_n is normal force given by mgcosθ
Thus;
F_k = μ_k(mg cosθ)
We now have;
mg(sinθ) – μ_k(mg cosθ) = ma
Dividing through by m to get;
g(sinθ) – μ_k(g cosθ) = a
a = 9.8(sin 12.03) - 0.6(9.8 × cos 12.03)
a = -3.71 m/s²
We are told that distance d = 24.0 m and v_o = 18 m/s
Using newton's 3rd equation of motion, we have;
v = √(v_o² + 2ad)
v = √(18² + (2 × -3.71 × 24))
v = 12.08 m/s
B) Now, μ_k = 0.10
Thus;
a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)
a = 1.08 m/s²
Using newton's 3rd equation of motion, we have;
v = √(v_o + 2ad)
v = √(18² + (2 × 1.08 × 24))
v = 19.39 m/s
