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Tanya [424]
2 years ago
10

An object of mass M is placed on a disk that can rotate. One end of a string is tied to the object while the other end of the st

ring is held by a pole that is located at the center of the disk. Initially, the object is spun in a horizontal circle of radius R at a constant tangential speed of v0, as shown in the figure, such that the tension in the string is T. At a later time, the disk is spun such that the tangential speed of the object is increased to 2v0 . At a later time, the tangential speed of the object is increased to 3v0 . Which of the following diagrams could represent the forces exerted on the object at one of the given speeds? Assume that the force of static friction is negligible. Select two answers.
2T
3T
4T
9T
Physics
2 answers:
Mars2501 [29]2 years ago
7 0

Answer:

Graphic labeled 4t and graphic labeled 9t

polet [3.4K]2 years ago
6 0

Answer: 4t and 9t

Explanation:

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From a height of 40.0 m, a 1.00 kg bird dives (from rest) into a small fish tank containing 50.5 kg of water. Part A What is the
Rom4ik [11]

Answer:

0.00185 °C

Explanation:

From the question,

The potential energy of the bird = heat gained by the water in the fish tank.

mgh = cm'(Δt)................... Equation 1

Where m = mass of the bird, g = acceleration due to gravity, h = height, c = specific heat capacity of water, m' = mass of water, Δt = rise in temperature of water.

make Δt the subject of the equation

Δt = mgh/cm'............... Equation 2

Given: m = 1 kg, h = 40 m, m' = 50.5 kg

constant: g = 9.8 m/s², c = 4200 J/kg.K

Substitute into equation 2

Δt = 1(40)(9.8)/(50.5×4200)

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3 years ago
A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc
Harman [31]

Answer:

E=88200\ J

Explanation:

Given:

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  • coefficient of kinetic friction, \mu_k=0.4

<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

f=\mu_k.N

where N = normal reaction by ground on the car

f=0.4\ties 1500\times 9.8

f=5880\ N

<em>Now from the work-energy equivalence:</em>

E=f.d

E=5880\times 15

E=88200\ J is the dissipated energy.

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3 years ago
What is the frequency of an x- ray if the wavelength is 4.5 E - 10m?
AnnZ [28]
V (speed) = F (frequency) x Wavelength
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F (frequency) = Speed ÷ Wavelength
F = 300,000 m\s x 4.5 e -10m
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AnnyKZ [126]

Answer:

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Explanation:

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solniwko [45]
The answer would be 48
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