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3 years ago

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Relativistic velocity is of the order of _____ of velocity of light A- 1/15th of the velocity of light B-1/20th of the velocity

of light C-1/10th of the velocity of light

1 answer:

ratelena [41]3 years ago

4 0

Answer:

Relativistic velocity is of the order of 1/10th of the velocity of light

Explanation:

We define relativistic speed (or velocity) as a speed that is a significant fraction of the speed of light: c = 3*10^8 m/s

Such that for these speeds, the special relativity theory starts to apply (the relativity effects starts to apply).

Usually, we define relativistic speeds as those that are of the order (or larger) of c/10, which is one-tenth of the speed of light.

Then the correct option is C:

Relativistic velocity is of the order of 1/10th of the velocity of light

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One of the characteristic of sound enogy Similar<br>to light energy is?

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Answer:

They both produce heat energy.

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3 years ago

A coin and feather are dropped in a moon. what will fall earlier on ground.give reasons.if they are dropped in the earth,which o

Sergeeva-Olga [200]

Answer:

on the moon, they will fall at the time

on earth, the coin will fall faster to the ground

Explanation:

A coin and feather dropped in a moon experience the same acceleration due to gravity as small as 1.625 m/s², and because of the absence of air resistance both will fall at the same rate to the ground.

If the same coin and feather are dropped in the earth, they will experience the same acceleration due to gravity of 9.81 m/s² and because of the presence of air resistance, the heavier object (coin) will be pulled faster to the ground by gravity than the lighter object (feather).

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3 years ago

A 100 mH inductor whose windings have a resistance of 6.0 Ω is connected across a 12 Vbattery having an internal resistance of 3

Alexus [3.1K]

Answer:

The store energy in the inductor is 0.088 J

Explanation:

Given that,

Inductor = 100 mH

Resistance = 6.0 Ω

Voltage = 12 V

Internal resistance = 3.0 Ω

We need to calculate the current

Using ohm's law

$V = IR$

$I=\dfrac{V}{R+r}$

Put the value into the formula

$I=\dfrac{12}{6.0+3.0}$

$I=1.33\ A$

We need to calculate the store energy in the inductor

$U=\dfrac{1}{2}LI^2$

$U=\dfrac{1}{2}\times100\times10^{-3}\times(1.33)^2$

$U=0.088\ J$

Hence, The store energy in the inductor is 0.088 J

7 0

3 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

3 years ago

An electric dipole consisting of charges of magnitude 1.70 nC separated by 6.80 μm is in an electric field of strength 1160 N/C.

bazaltina [42]

Answer:

p = 1.16 10⁻¹⁴ C m and ΔU = 2.7 10 -11 J

Explanation:

The dipole moment of a dipole is the product of charges by distance

p = 2 a q

With 2a the distance between the charges and the magnitude of the charges

p = 1.7 10⁻⁹ 6.8 10⁻⁶

p = 1.16 10⁻¹⁴ C m

The potential energie dipole is described by the expression

U = - p E cos θ

Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line

Orientation parallel to the field

θ = 0º

U = 1.16 10⁻¹⁴ 1160 cos 0

U1 = 1.35 10⁻¹¹ J

Antiparallel orientation

θ = 180º

cos 180 = -1

U2 = -1.35 10⁻¹¹ J

The difference in energy between these two configurations is the subtraction of the energies

ΔU = | U1 -U2 |

ΔU = 1.35 10-11 - (-1.35 10-11)

ΔU = 2.7 10 -11 J

6 0

3 years ago