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3 years ago

Why can an object still be seen when it is at absolute zero?

1 answer:

6 0

<span>As the temperature goes down, the chaotic motion (velocity) of atoms start decreasing. If the temperature hits the absolute zero (which, in reality, is impossible to achieve), the atoms of the body would freeze, making the body still and stiff. One thing to note here is that the atoms do not get destroyed when the temperature reaches the absolute zero. That is the reason why the object can still be seen when it is at absolute zero.</span>

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A circular conducting loop with a radius of 0.10 m and a small gap filled with a 10.0 ȍresistor is oriented in the xy-plane. If

lidiya [134]

Answer:

The magnitude of the current is 5.45 mA.

Explanation:

Given that,

Resistance = 10.0 ohm

Radius = 0.10 m

Magnetic field = 1.0 T

Angle = 30°

Increase magnetic field = 7.0 T

Time t = 3.0 s

Number of turns = 1

We need to calculate the initial flux

Using formula of flux

$\phi=NB_{1}A\cos\theta$

Put the value into the formula

$\phi=1\times1.0\times\pi\times(0.10)^2\times\cos30^{\circ}$

$\phi=1\times1.0\times\pi\times(0.10)^2\times\dfrac{\sqrt{3}}{2}$

$\phi=0.027\ wb$

We need to calculate the final flux

$\phi=1\times7.0\times\pi\times(0.10)^2\times\dfrac{\sqrt{3}}{2}$

$\phi=0.1904\ wb$

We need to calculate the induced emf

Using formula of emf

$\epsilon=\dfrac{\phi_{f}-\phi_{i}}{t}$

Put the value into the formula

$\epsilon=\dfrac{0.1904-0.027}{3.0}$

$\epsilon=0.0545\ V$

We need to calculate the current

Using formula of current

$I=\dfrac{\epsilon}{R}$

Put the value into the formula

$I=\dfrac{0.0545}{10.0}$

$I=5.45\ mA$

Hence, The magnitude of the current is 5.45 mA.

4 0

3 years ago

When you blink your eye, the upper lid goes from rest with your eye open to completely covering your eye in a time of 0.024 s.

dusya [7]

a) Distance moved by the top lid during a blink: 1 cm (estimate)

b) The acceleration is $34.7 m/s^2$

c) The final speed is 0.83 m/s

Explanation:

For the purpose of this problem, we can estimate the size of the eye from the top lid to the bottom lid to be 1 cm, so this is the distance moved by the top lid during a blink.

Assuming the motion of the eyelid to be at constant acceleration, we can find the acceleration by using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the eyelid, we have:

u = 0 (it starts from rest)

s = 1 cm = 0.01 m is the distance covered

t = 0.024 s is the time taken

Solving for a, we find the acceleration:

$a=\frac{2s}{t^2}=\frac{2(0.01)}{0.024^2}=34.7 m/s^2$

The final speed of the upper eyelid can be found by using another suvat equation:

$v=u+at$

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

For the eyelid in this problem, we have

u = 0

$a=34.7 m/s^2$

t = 0.024 s

Substituting, we find the final speed:

$v=0+(34.7)(0.024)=0.83 m/s$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

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brainly.com/question/2562700

#LearnwithBrainly

4 0

3 years ago

Two forces P and Q act on an object of mass 15.0 kg with Q being the larger of the two forces. When both forces are directed to

Rom4ik [11]

Answer:

P=6.25N and Q=16.25N

Explanation:

In order to solve this problem we must first draw a free body diagram for both situation, (see attached picture).

Now, we need to analyze the two free body diagrams. So let's analyze the first diagram. Since the body is accelerated, then the sum of forces is equal to mass times acceleration, so we get:

$\Sigma F=ma$

We can assume there will be only the two mentioned forces P and Q, so

the sum of forces will be:

P+Q=ma

$P+Q=(15kg)(1.50m/s^{2})$

P+Q=22.5N

We can do the same analysis for the second free body diagram:

$\Sigma F=ma$

$Q-P=(15kg)(0.7m/s^{2})$

Q-P=10.5N

so now we have a system of equations we can solve by elimination:

Q+P=22.5N

Q-P=10.5N

Now, we can add the two equations together so the P force is eliminated, so we get:

2Q=32.5N

now we can solve for Q:

$Q=\frac{32.5N}{2}$

Q=16.25N

Now we can use any of the equations to find P.

Q+P=22.5N

P=22.5N-Q

when substituting for Q we get:

P=22.5N-16.25N

P=6.25N

5 0

4 years ago

DESPERATE!! WILL GIVE BARINLIST AND THANKS

Finger [1]

Answer:

it grabs unto soil and keeps it clamped together it prevents this makes the soil harder to wash away

3 0

3 years ago

Read 2 more answers

A runner achieves a velocity of 11.1 m/s, 9 sec after he begins. What is his

Elena-2011 [213]

Answer:

1.23 m/s²

Explanation:

Given:

v₀ = 0 m/s

v = 11.1 m/s

t = 9 s

Find: a

Equation:

v = at + v₀

Plug in:

11.1 m/s = a (9 s) + 0 m/s

a = 1.23 m/s²

The runner's acceleration is 1.23 m/s².

7 0

3 years ago