Three small objects are arranged along a uniform rod of mass m and length L. one of mass m at the left end, one of mass m at thecenter, and one of mass 2m at the right end. How far to the left or right of the rod's center should you place a support so that the rod
with the attached objects will balance there?
Answers
Answer:
See answer below
Explanation:
Hi there,
To get started, recall the Center of Mass formula for two masses:
where m is mass and x is displacement from the center of the shape.
Since masses at the center of a geometric shape have a displacement (x) value of 0, as the mass is already of the center, and does not affect Xcm. So, we can disregard the central mass, hence we use the above formula for two masses.
We can arbitrarily define left to be a negative (-) displacement, and vice versa for right direction. We proceed with the formula:
Since we defined left (-) and right (+), we notice the center of mass is (+) value. This makes sense, as there is slightly more mass on the right side. Hence, you should place a support 1/6 of the rod's length away from the rod's center.
Study well and persevere.
thanks,
Final answer:
To balance the rod with the attached objects, place a support at a distance of L/3 from the left end of the rod.
Explanation:
To balance the rod with the attached objects, you need to place a support at a distance of L/3 from the left end of the rod. This is because the center of gravity of the system should be directly above the support.
The center of gravity is given by the equation xcg = (m*0 + m*L/2 + 2m*L)/ (m + m + 2m). Solving this equation, we get xcg = 2L/3. Therefore, the support should be placed at a distance of L/3 from the left end of the rod.
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A 0.26-kg stone is held 1.2 m above the top edge of a water well and then dropped into it. The well has a depth of 4.5 m. (a) Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone−Earth system before the stone is released?
Answers
Answer:
Explanation:
The volume rate of flow = a x v where a is cross sectional area of pipe and v is velocity of flow
putting the values
π x .2945² x 12.1
= 3.3 m³ /s
To know the pipe's diameter at the refinery we shall apply the following formula
a₁ v₁ = a₂ v₂
a₁ v₁ and a₂ v₂ are volume rate of flow of liquid which will be constant .
3.3 = a₂ x 6.29
a₂ = .5246 m³
π x r² = .5246
r = .4087 m
= 40.87 cm
diameter
= 81.74 cm
Answers
Answer:
Explanation:
designer
illusionist
engineer
entrepreneur
salesperson
human
inventor
fromthe cliff. Determine how fast the vehicle was pushed off
thecliff.
Answers
Answer:
v = a/√(2h/g) m/s
Explanation:
Lets say the distance away from the cliff is a.
then, a = v t
where v is velocity with which it was thrown and t is time taken to fall.
Using equations of motion, we can also say that
h=1/2gt^2
where h is the height of the cliff
Thus, t^2 = 2h/g and t = √(2h/g)
Thus, v = a/√(2h/g).
the vehicle was pushed off the cliff with the velocity , v = a/√(2h/g). m/s
Answers
Answer:
Explanation:
solution below
Final answer:
The quantum of energy for one atomic oscillator in tungsten, given the effective interatomic spring stiffness of 360 N/m, the mass of one tungsten atom as 3.074 x 10^-25 kg, and the reduced Planck's constant of 1.0546 x 10^-34 J · s, can be calculated to be approximately 1.33 x 10^-21 J.
Explanation:
To calculate the quantum of energy for one atomic oscillator in tungsten, we will consider the model of an atom being connected to two springs, both having an effective interatomic spring stiffness of four times the given value (90 N/m). This value thus becomes 360 N/m.
One mole of tungsten has a mass of 0.185 kg, thus the mass of one atom can be determined by dividing this value by Avogadro's number (6.0221 x 10^23 molecules/mole), which gives approximately 3.074 x 10^-25 kg.
The quantum of energy, or the energy of one quantum (the smallest possible energy increment), is given by the formula E = ħω, where ħ is the reduced Planck's constant (1.0546 x 10^-34 J · s) and ω is the angular frequency, given by sqrt(k/m), where k is the spring constant and m is the mass.
Substituting the known values into these equations gives ω= sqrt((360)/(3.074 x 10^-25)) and E= (1.0546 x 10^-34) x sqrt((360)/(3.074 x 10^-25)), which results in a quantum of energy of approximately 1.33 x 10^-21 J.
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a) 2v
b) 1/2v
c) v
d) 4v
e) 1/4v
Answers
Answer:
Explanation:
Given that,
Initial speed of the cart, u = 0
Let F force is applied to the cart for time after which the car has speed v. The force on an object is given by :
F = ma
m is the mass of the cart
We need to find the speed of second cart, if the same force is applied for the same time to a second cart with twice the mass. Force becomes,
So, the speed of second cart is half of the initial speed of first cart. So, the correct option is (b).
Answers
Answer:
ωB = 300 rad/s
ωC = 600 rad/s
Explanation:
The linear velocity of the belt is the same at pulley A as it is at pulley D.
vA = vD
ωA rA = ωD rD
ωD = (rA / rD) ωA
Pulley B has the same angular velocity as pulley D.
ωB = ωD
The linear velocity of the belt is the same at pulley B as it is at pulley C.
vB = vC
ωB rB = ωC rC
ωC = (rB / rC) ωB
Given:
ω₀A = 40 rad/s
αA = 20 rad/s²
t = 3 s
Find: ωA
ω = αt + ω₀
ωA = (20 rad/s²) (3 s) + 40 rad/s
ωA = 100 rad/s
ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s
ωB = ωD = 300 rad/s
ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s