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kaheart [24]
3 years ago
9

Imagine a simple pendulum swinging in an elevator. If the cable holding the elevator up was to snap, allowing the elevator to go

into free fall, what would happen to the frequency of oscillation of the pendulum? Justify your answer.
Physics
1 answer:
Lady_Fox [76]3 years ago
5 0

One of the components that affect the period is gravity (the other is length). This gravity is basically the value of the effective acceleration that acts on the body due to gravity. When the elevator is over free fall, the effective gravity becomes zero. Mathematically this can be visualized as,

T = 2\pi \sqrt{\frac{l}{\dot{g_{eff}}}}

Since this value is zero, the period would tend to be infinite,

T \rightarrow \infty

Therefore the frequency that is inversely proportional to the period would be defined as

f = \frac{1}{T}

f = \frac{1}{\infty}

f \approx 0

In this way there is no frequency on the body which will not generate any oscillation on the body

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Why should a physical quantity be measured?​
Natasha_Volkova [10]

Answer:

It allows us to understand nature much more deeply than does qualitative description alone.

Explanation:

Without explaining the measurements, a quantity cannot always be measured.

Hope this helped, and please mark as Brainliest :)

8 0
2 years ago
Will convection always work faster and more efficiently than conduction?
Stells [14]

Answer:

Yes convection will always work faster and more efficiently.

Explanation:

When a gas or a liquid is heated, hot areas of the material flow and mix with the cool areas. ... Convection transfers heat over a distance faster than conduction. But ultimately conduction must transfer the heat from the gas to the other object, though molecular contact.

5 0
3 years ago
0. A 3.00-kg block is dropped from rest on a vertical spring whose spring constant is 750 N/m. The block hits the spring, compre
Dmitry_Shevchenko [17]

Answer:

the spring compressed is 0.1878 m

Explanation:

Given data

mass = 3 kg

spring constant k = 750 N/m

vertical distance h = 0.45

to find out

How far is the spring compressed

solution

we will apply here law of mass of conservation

i.e

gravitational potential energy loss = gain of eastic potential energy of spring

so we say m×g×h = 1/2× k × e²

so e² = 2×m×g×h / k

so

we put all value here

e² = 2×m×g×h / k

e² = 2×3×9.81×0.45 / 750

e²  = 0.0353

e = 0.1878 m

so the spring compressed is 0.1878 m

5 0
3 years ago
Speed is the ratio of the distance an object moves to
guajiro [1.7K]
A - the amount of time needed to travel the distance

The formula for speed is distance/time and the unit is m/s. Therefore you divide the distance travelled by the time it took to travel it.
6 0
3 years ago
Read 2 more answers
A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 50.9 J and a maximum d
Llana [10]

(a) 2446 N/m

When the spring is at its maximum displacement, the elastic potential energy of the system is equal to the total mechanical energy:

E=U=\frac{1}{2}kA^2

where

U is the elastic potential energy

k is the spring constant

A is the maximum displacement (the amplitude)

Here we have

U = E = 50.9 J

A = 0.204 m

Substituting and solving the formula for k,

k=\frac{2E}{A^2}=\frac{2(50.9)}{(0.204)^2}=2446 N/m

(b) 50.9 J

The total mechanical energy of the system at any time during the motion is given by:

E = K + U

where

K is the kinetic energy

U is the elastic potential energy

We know that the total mechanical energy is constant: E = 50.9 J

We also know that at the equilibrium point, the elastic potential energy is zero:

U=\frac{1}{2}kx^2=0 because x (the displacement) is zero

Therefore the kinetic energy at the equilibrium point is simply equal to the total mechanical energy:

K=E=50.9 J

(c) 8.55 kg

The maximum speed of the block is v = 3.45 m/s, and it occurs when the kinetic energy is maximum, so when

K = 50.9 J (at the equilibrium position)

Kinetic energy can be written as

K=\frac{1}{2}mv^2

where m is the mass

Solving the equation for m, we find the mass:

m=\frac{2K}{v^2}=\frac{2(50.9)}{(3.45)^2}=8.55 kg

(d) 2.14 m/s

When the displacement is

x = 0.160 m

The elastic potential energy is

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

So the kinetic energy is

K=E-U=50.9 J-31.3 J=19.6 J

And so we can find the speed through the formula of the kinetic energy:

K=\frac{1}{2}mv^2 \rightarrow v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(19.6)}{8.55}}=2.14 m/s

(e) 19.6 J

The elastic potential energy when the displacement is x = 0.160 m is given by

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

And since the total mechanical energy E is constant:

E = 50.9 J

the kinetic energy of the block at this point is

K=E-U=50.9 J-31.3 J=19.6 J

(f) 31.3 J

The elastic potential energy stored in the spring at any time is

U=\frac{1}{2}kx^2

where

k = 2446 N/m is the spring constant

x is the displacement

Substituting

x = 0.160 m

we find the elastic potential energy:

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

(g) x = 0

The postion at that instant is x = 0, since it is given that at that instant  the system passes the equilibrium position, which is zero.

4 0
3 years ago
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