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3 years ago

Can energy be converted into other forms?

2 answers:

6 0

Energy can exist in many forms within a system and may be converted from one form to another within the constraint of the conservation law. These different forms include gravitational, kinetic, thermal, elastic, electrical, chemical, radiant, nuclear, and mass energy.

DanielleElmas [232]3 years ago

3 0

Kinetic and potential energy

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Vector A⃗ points in the positive y direction and has a magnitude of 12 m. Vector B⃗ has a magnitude of 33 m and points in the ne

gavmur [86]

vector A has magnitude 12 m and direction +y

so we can say

$\vec A = 12 \hat j$

vector B has magnitude 33 m and direction - x

$\vec B = -33 \hat i$

Now the resultant of vector A and B is given as

$\vec A + \vec B = 12 \hat j - 33 \hat i$

now for direction of the two vectors resultant will be given as

$\theta = tan^{-1}\frac{12}{-33}$

$\theta = 160 degree$

so it is inclined at 160 degree counterclockwise from + x axis

magnitude of A and B will be

$R = \sqrt{A^2 + B^2}$

$R = \sqrt{12^2 + 33^2} = 35.11 m$

so magnitude will be 35.11 m

6 0

3 years ago

300 grams of ethanol is heated with 14640 Joules of energy to reach a final temperature of 30 °C. What was the initial temperatu

levacccp [35]

Answer:

10 °C

Explanation:

Applying

q = cm(t₂-t₁)............... Equation 2

Where q = heat energy, c = specific heat of ethanol, m = mass of ethanol, t₁ = initial temperature, t₂ = Final temperature.

Given: c = 2.44 J/g.°C, m = 300 g, q = 14640 J, t₂ = 30°C

Substitute into equation 2 and solve for t₁

14640 = 2.44×300(30-t₁)

14640 = 732(30-t₁)

732(30-t₁) = 14640

(30-t₁) = 14640/732

(30-t₁) = 20

t₁ = 30-20

t₁ = 10 °C

6 0

3 years ago

Read 2 more answers

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

Explanation:

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

3 years ago

As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horiz

Eddi Din [679]

a)E= U + K = $\frac{1}{2}$ kx² + $\frac{1}{2}$ mv²

The total energy of the system at any point in the motion is equal to the sum of the elastic potential energy of the spring, U, and of the kinetic energy of the mass, K:

E= U + K = $\frac{1}{2}$ kx² + $\frac{1}{2}$ mv²

where

'k' represents the spring constant

'x' is the compression/stretching of the spring with respect to its equilibrium position

'm' is the mass of the block attached to the spring

and 'v' is the speed of the block

b) A= $\sqrt{\frac{2E}{k}}$

The amplitude of the motion compares to the most extreme displacement of the mass-spring system. The displacement of the system, x(t), at time t, for a simple harmonic oscillator is given by,

x= Asin(ωt+∅)

where

amplitude is 'A'

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency of the motion

t is the time

$\phi$ is the phase (we can take $\phi$ =0 )

The amplitude of the motion occurs when the displacement of the motion is maximum: x=A. Regarding energy, the mass-spring system is at its maximum displacement (x=A) when all the mechanical energy of the framework is elastic potential energy, so when the kinetic energy is zero:

K= $\frac{1}{2}mv^2$ =0

E= $\frac{1}{2}kA^2\\$ -->(1)

A= $\sqrt{\frac{2E}{k}}$

c) $v_{max}=\omega A$

When the elastic potential energy is zero, the maximum speed of the system occurs i.e U=0 and the kinetic energy is maximum, so:

U=0

E= $\frac{1}{2}mv_{max}^2$

According to the law of conservation of the mechanical energy, this energy must be equal to the energy of the system at its maximum displacement (1), so we can write

$\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2$