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3 years ago

Which is an example of nuclear energy being converted into heat and light energy?

Physics

2 answers:

likoan [24]3 years ago

6 0

The answer to the question is A

Akimi4 [234]3 years ago

3 0

<span>Which is an example of nuclear energy being converted into heat and light energy?

a.the sun</span>

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calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the s

IceJOKER [234]

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m * 1609.344 m

= 149.668 * 10^8 m

next we will express the distance between the earth and the sun

$r = \frac{a(1-E^2)}{1+Ecos\beta }$ --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

$\beta$ = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

$v^2 = \frac{4\pi^2 }{r_{c} }$ ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

( $V^2 = (\frac{4\pi^{2} }{149.626*10^9})$

therefore : V = 1.624* 10^-5 m/s

4 0

3 years ago

What is the difference between radial acceleration and tangential acceleration and how do you calculate both of these accelerati

sergey [27]

Answer:

Tangential acceleration is in the direction of velocity - along the circumference of a circle if the object is undergoing circular motion

a = (V2 - V1) / T

Radial acceleration is perpendicular to the direction of motion if the object is not moving in a straight line (perhaps along the circumference of a circle)

a = m V^2 / R = m ω^2 R where R is the radius vector of the velocity - note that the Radius vector is directed from the center of motion to the object and for circular motion would be constant in magnitude but not in direction

8 0

1 year ago

Plzz answer the question.

Aleksandr [31]

Ur answer is 3 and i'm sure of it

8 0

2 years ago

A capacitor consists of two square plates, 8.7 cm on a side, separated by a 2.0 mm air gap. How much energy would be stored in t

slavikrds [6]

Answer:

122.84 J

Explanation:

Since plate is square, area, A is given by $(8.7/100)^{2}=0.007569m^{2}$

The distance between plates, d, is given in the question as 2mm=0.002m

Charge on plate, Q, as given in the question is 240 $\mu c$

Assuming mica dielectric constant, k of 7

Capacitance, C is given by

C= $\frac {k\epsilon_{o}A}{d}=\frac {(7)(8.85*10^{-12})(0.007569)}{0.002}=2.34*10*^{-10}F$

Stored energy, E is given by

E= $\frac {Q^{2}}{2C}=\frac {(240*10^{-6})^{2}}{2*(2.34*10^{-10})}=122.84J$

Therefore, the stored energy is 122.84 J

5 0

2 years ago

The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.

S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

$\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }$

Where:

$\omega$ - Angular frequency, measured in radians per second.

$m$ - Mass of the physical pendulum, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$d$ - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

$I_{O}$ - Moment of inertia with respect to pivot point, measured in $kg\cdot m^{2}$ .

In addition, frequency and angular frequency are both related by the following formula:

$\omega =2\pi\cdot f$

Where:

$f$ - Frequency, measured in hertz.

If $f = 0.658\,hz$ , then angular frequency of the physical pendulum is:

$\omega = 2\pi \cdot (0.658\,hz)$

$\omega = 4.134\,\frac{rad}{s}$

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

$\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}$

$I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}$

Given that $m = 1.15\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $d = 0.425\,m$ and $\omega = 4.134\,\frac{rad}{s}$ , the moment of inertia associated with the physical pendulum is:

$I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}$

$I_{o} = 0.280\,kg\cdot m^{2}$

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

8 0

3 years ago