Can a material have negative permittivity?

Which force is reasonable for making fusion possible in the sun?

andreev551 [17]

The force is reasonable for making fusion possible in the Sun is heat energy.

<h3>What is nuclear fission and fusion?</h3>

When the slow moving neutrons are bombarded with the heavy radioactive nuclei, the product is the more number of neutrons are produced with the large amount of energy. This multiplying process is called nuclear fusion.

The amount of energy produced in such a reaction can be calculated using the equivalence of mass and energy relationship.

E = mc²

The same happens in nuclear fusion where large amount of energy is needed to make more heavy nuclei.

Thus, fusion requires heat energy to continue the reaction.

Learn more about nuclear fusion and fission.

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8 0

2 years ago

Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s starting from rest. Assuming that they have constan

elena55 [62]

Answer:

$10.07m/s^2$

Explanation:

Information we have:

velocities:

initial velocity: $v_{i}=0mph$ (starts from rest)

final velocity: $v_{f}=50mph$

time: $t=2.22s$

Since we need the answer in $m/s^2$ , we nees to convert the speed to meters per second:

$v_{f}=\frac{50miles}{1hour}(\frac{1hour}{3,600s} )(\frac{1609.34meters}{1mile} ) \\\\v_{f}=\frac{50*1609.34}{3600} m/s\\\\v_{f}=22.35m/s$

We find the acceleration with the following formula:

$a=\frac{v_{f}-v_{i}}{t}$

substituting the known values:

$a=\frac{22.35m/s-0m/s}{2.22s}\\ \\a=10.07m/s^2$

the acceleration is 10.07 $m/s^2$

8 0

3 years ago

Read 2 more answers

You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

$t=3.41 s$ This is the time the ball is in air

5 0

3 years ago

Does the earths magnetic field change with time?

ZanzabumX [31]

Answer:

As per the fossil fuel records, magnetic field reversal does not impact living beings. It will take almost a century for the poles to complete the shift. Meanwhile, the earth is left with almost zero magnetic field.

6 0

3 years ago

Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power P needed for it to ope

Mandarinka [93]

Answer:

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s.

the rate at which they raise the temperature of the room.

2.1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

Explanation:

A refrigerator is a device that uses work to remove heat energy from a cold reservoir and deposit it into a hot reservoir. .A good refrigerator (with a large performance coefficient) will remove a large amount of heat energy from the cold reservoir for a small amount of work input

The performance coefficient of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir to the work input to the refrigerator:

k=QC/W

power is defined as work per unit time

1.k=1500/750=2

2. 1200/400=3

3.2000/500=4

4.1000/250=4

5.1500/500=3

6.3000/1000=3

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s

2, Rate at which they raise the temperature of the room.

rate at which temperature rises in the inner chamber of the refrigerator is proportional to the rate of energy used to dispel heat from the refrigerator

1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

5 0

3 years ago