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ale4655 [162]
3 years ago
6

small ball is attached to one end of a spring that has an unstrained length of 0.245 m. The spring is held by the other end, and

the ball is whirled around in a horizontal circle at a speed of 3.55 m/s. The spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.0194 m. By how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless? Number Enter your answer in accordance to the question statement Units Choose the answer from the menu in accordance to the question statement
Physics
1 answer:
Helga [31]3 years ago
6 0

Answer:

x = 4 \times 10^{-3} m

Explanation:

As we know that the force required to move the mass in circle with uniform speed is known as centripetal force

This centripetal force is given as

F_c = \frac{mv^2}{R}

while mass is revolving in horizontal circle the force is due to spring so it is given as

F_c = kx

kx = \frac{mv^2}{R}

k(0.0194) = \frac{m\times 3.55^2}{(0.245 + 0.0194)}

k(0.0194) = 47.66 m

now we have

\frac{k}{m} = 2457

now when mass is suspended at the end of spring then we have

mg = kx

x = \frac{mg}{k}

x = \frac{9.81}{2457}

x = 4 \times 10^{-3} m

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3 years ago
In a particular case of an object in front of a spherical mirror with a focal length of +12.0 cm, the magnification is +4.00.(a)
salantis [7]

Answer:

9 cm

-36 cm

Explanation:

u = Object distance

v = Image distance

f = Focal length = 12

m = Magnification = 4

m=-\frac{v}{u}\\\Rightarrow 4=-\frac{v}{u}\\\Rightarrow v=-4u

Lens equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{12}=\frac{1}{u}+\frac{1}{-4u}\\\Rightarrow \frac{1}{12}=\frac{3}{4u}\\\Rightarrow u=9\ cm

Object distance is 9 cm

v=-4\times 9=-36\ cm

Image distance is -36 cm (other side of object)

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3 years ago
When measuring espresso for a drink, which instrument would give the<br> greatest precision?
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How many mL is an espresso?

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5 0
2 years ago
Help!
dmitriy555 [2]

Answer:

1.It's the world's most famous equation, but what does it really mean? "Energy equals mass times the speed of light squared." On the most basic level, the equation says that energy and mass (matter) are interchangeable; they are different forms of the same thing.

2.The process releases energy because the total mass of the resulting single nucleus is less than the mass of the two original nuclei.

3.In nuclear reactions, mass is never conserved—some mass is exchanged for energy and energy for mass. Nuclear reactions take place in an atom's nucleus. In a spontaneous nuclear reaction, such as radioactive decay, mass is "lost" and appears as energy in the form of particles or gamma rays.

4.In a nuclear reaction, mass decreases and energy increases. The sum of mass and energy is always conserved in a nuclear reaction.

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Explanation:

hope it helps

7 0
2 years ago
A rope of length L has circular cross-sectional area A and density rho = m/V , where m is the mass of the rope and V = A · L is
hram777 [196]

Answer: µ = ρ¹ * A¹

Where x=1 and y=1

Explanation: According to the question, the mass per unit length (µ) is related to the density (ρ) and area A are related by the formulae below

µ = ρ * A

The dimension for each of these quantities is given below

Since µ is mass per unit length, unit is Kg/m and the dimension is ML^-1

ρ is density with unit kg/m³ and the dimension is ML^3

A is area with unit m², thus the dimension is M^2

Note that using dimensional analysis means we will be using the 3 fundamental quantities (mass, length and time) in our analysis.

Their dimensions below

Mass = M

Length = L

Time = T

Since the mass per unit length is related to density and area, we have a mathematical equation to provide a solution as shown below

µ = ρ^x * A^y.

By getting the power of x and y we will be able to get the formula that relates the quantities.

This is done by slotting in the dimensions of the respective quantities.

ML^-1 = (ML^-3)^x * (L²) ^y

By using law of indices on the right hand side of the equation, we have that

ML^-1 = (M^x * L^-3x) * (L^2y)

Also applying law of indices on the right hand side, we have that

ML^-1 = (M^x) * (L^-3x +2y)

The next step is to relate equal variables on both sides

For the M variable

M¹ = M^x which results to

x = 1

For the L variable

L^-1 = L^-3x+2y which results to

-1 = - 3x +2y

But x = 1

We have that

-1 = - 3(1) +2y

-1 = - 3 + 2y

-1 +3= 2y

2 = 2y

y = 1

Thus x=1 and y=1 and the formulae that relates the quantities is

µ = ρ¹ * A¹

3 0
3 years ago
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