Answer:
The final components of velocity of the composite object is 3.33 m/s.
Explanation:
Given;
mass of the first object, m₁ = 3.35 kg
initial velocity of the first object, u₁ = 4.90 m/s in positive x-direction
mass of the second object, m₂ = 1.88 kg
initial velocity of the second object, u₂ = 3.12 m/s in negative y-direction
initial momentum of the first object, P₁ = 3.35 x 4.9 = 16.415 kgm/s
initial momentum of the second object, P₂ = 1.88 x 3.12 = 5.8656 kgm/s
The resultant velocity of the two objects is given by;
R² = 16.415² + 5.8656²
R² = 303.858
R = √303.858
R = 17.432 kgm/s
Apply the principle of conservation of linear momentum for inelastic collision;
total initial momentum before = total final momentum after collision
P₁(x) + P₂(y) = Pf
R = Pf
R = v(m₁ + m₂)
17.432 = v(m₁ + m₂)
where;
v is the final components of velocity of the composite object
Therefore, the final components of velocity of the composite object is 3.33 m/s.
Answer:
Explanation:
When an electromagnetic wave passes through the interface between two mediums, it undergoes refraction, which means that it bents and its speed and its wavelength change.
In particular, the wavelength of an electromagnetic wave in a certain medium is related to the index of refraction of the medium by:
where
is the wavelength in a vacuum (air is a good approximation of vacuum)
n is the refractive index of the medium
In this problem:
is the original wavelength of the wave
n = 1.47 is the index of refraction of corn oil
Therefore, the wavelength of the electromagnetic wave in corn oil is:
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)
Learn more about kinetic friction here-
brainly.com/question/13754413
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m = mass of the penny
r = distance of the penny from the center of the turntable or axis of rotation
w = angular speed of rotation of turntable
F = centripetal force experienced by the penny
centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as
F = m r w²
in the above equation , mass of penny "m" and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .
hence greater the distance from center , greater will be the centripetal force to remain in place.
So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.
Answer:
Earth attract the Moon with a force that is greater.
Explanation:
According to the law of gravitation, the gravitational force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Mathematically, F1 = Gm1m2/r²... 1
Let m1 be the mass of the earth and m2 be that of the moon
If the Earth is much more massive than is the Moon, the new force of attraction between them will become;
F2= G(2m1)m2/r²
F2 = 2Gm1m2/r² ... (2)
Dividing eqn 1 by 2 we have;
F1/F2 = (Gm1m2/r²)÷(2Gm1m2/r²)
F1/F2 = Gm1m2/r²×r²/2Gm1m2
F1/F2 = 1/2
F2=2F1
This shows that that the earth will attract the moon by a force 2times the initial force of the masses(i.e a much greater force)
