The calculated coefficient of kinetic friction is 0.33125.'
The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
given mass of the block=10 kg
spring constant k= 2250 Nm
now according to principal of conservation of energy we observe,
the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.
mgh= μ (mgl) +1/2 kx²
10 x 10 x 3= μ(600) +(1125) (0.09)
μ(600) =300 - 101.25
μ = 198.75÷600
μ =0.33125
The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)
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Answer:
15.3 s and 332 m
Explanation:
With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon
gm = 1/6 ge
gm = 1/6 9.8 m/s² = 1.63 m/s²
We calculate the range
R = Vo² sin 2θ / g
R = 25² sin (2 30) / 1.63
R= 332 m
We will calculate the time of flight,
Y = Voy t – ½ g t2
Voy = Vo sin θ
When the ball reaches the end point has the same initial height Y=0
0 = Vo sin t – ½ g t2
0 = 25 sin (30) t – ½ 1.63 t2
0= 12.5 t – 0.815 t2
We solve the equation
0= t ( 12.5 -0.815 t)
t=0 s
t= 15.3 s
The value of zero corresponds to the departure point and the flight time is 15.3 s
Let's calculate the reach on earth
R2 = 25² sin (2 30) / 9.8
R2 = 55.2 m
R/R2 = 332/55.2
R/R2 = 6
Therefore the ball travels a distance six times greater on the moon than on Earth
Answer:
E_particle = 1,129 10⁻²⁰ J / particle
T= 817.5 K
Explanation:
Energy is a scalar quantity so it is additive, let's look for the total energy of each gas
Gas a
E_a = 2 5000 = 10000 J
Gas b
E_b = 3 8000 = 24000 J
When the total system energy is mixed it is
E_total = E_a + E_b
E_total = 10000 + 24000 = 34000
The total mass is
M = m_a + m_b
M = 2 +3 = 5
The average energy among the entire mass is
E_averge = E_total / M
E_averago = 34000/5
E_average = 6800 J
One mole of matter has Avogadro's number of atoms 6,022 10²³ particles
Therefore, each particle has an energy of
E_particle = E_averag / 6.022 10²³ = 6800 /6.022 10²³
E_particle = 1,129 10⁻²⁰ J / particle
For find the temperature let's use equation
E = kT
T = E / k
T = 1,129 10⁻²⁰ / 1,381 10⁻²³
T = 8.175 102 K
T= 817.5 K
Answer:
A
Explanation:
if the man doubles his force to 40 and the box was yet to move that means acceleration also doubled so your answer would be A
Either one is fun and great to play!
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