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4 years ago

6

How much work is required to turn an electric dipole 180° in a uniform electric field of magnitude E = 46.0 N/C if the dipole mo

ment has a magnitude of p = 3.02 × 10−25 C·m and the initial angle is 64°?

1 answer:

chubhunter [2.5K]4 years ago

3 0

Answer:

$W=1.22*10^{-23}J$

Explanation:

Torque and energy of an electric dipole in an electric field we find:

$W=U(\alpha_{o}+\pi )-U(\alpha_{o} )=-pE(cos(\alpha_{o}+\pi )-cos(\alpha_{o} ))\\W=2pECos\alpha_{o}\\ W=2(3.02*10^{-25}C.m )(46.oN/C)Cos64\\W=1.22*10^{-23}J$

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3 years ago

A weight of 6 kg increases the speed of its force from 2 m / s to 4 m / s. By how many joules does the kinetic energy of the bod

Leviafan [203]

Answer:

36 J

Explanation:

Let's start off with the Kinetic Energy formula: 1/2m $v^{2}$

So to find the <em>change</em> in Kinetic Energy, you would first have to find the Kinetic Energy in the beginning and in the end.

*Note: Mass: you incorrectly wrote weight = 6 kg- because weight is a force you have to write 6 Newtons. If you are talking about the <em>mass</em>, you would write 6 kg. If you are talking about weight, you would write 6 Newtons. The difference will make significant changes to the answer, so I will give you the answer for both.

If 6 kg is the MASS:

Beginning: find the kinetic energy.

Plug in all parts of the formula: 1/2mv^2 = $\frac{1}{2}$ (6)(2 squared)= $\frac{1}{2}$ (6)(4)= 12 J

End: find the kinetic energy.

Plug in all parts of the formula: 1/2mv^2= $\frac{1}{2}$ (6)(4 squared)= $\frac{1}{2}$ (6)(16)= 48 J

Answer:

36 J

If 6 kg is the WEIGHT:

We know that the weight formula is mg, or mass times acceleration due to gravity (which is always 9.8 m/s^2). Plug in the numbers:

Weight = mass x acceleration due to gravity

6 = mass x 9.8

6 = 9.8m

*Divide both sides*

mass = 0.6 kg

Now, we can use the mass to find the kinetic energy.

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Plug in all parts of the formula: 1/2mv^2 = $\frac{1}{2}$ (0.6)(2 squared)= $\frac{1}{2}$ (0.6)(4)= 1.2 J

End: find the Kinetic Energy

Plug in all parts of the formula: 1/2mv^2= $\frac{1}{2}$ (0.6)(4 squared)= $\frac{1}{2}$ (0.6)(16)= 4.8 J

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3.6 J

(this answer is not very feasible, so 36 J is the way to go. But just remember, don't mix up weight and mass again- as you can see, they lead to different answers!)

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3 years ago

Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t

Readme [11.4K]

Answer: $3.66(10)^{33}kg$

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity $\omega$ of the planet P1 with a period $T=750years=2.36(10)^{10}s$ :

$\omega=\frac{2\pi}{T}=\frac{V_{1}}{R}$ (1)

Where:

$V_{1}=40.2km/s=40200m/s$ is the velocity of planet P1

$R$ is the radius of the orbit of planet P1

Finding $R$ :

$R=\frac{V_{1}}{2\pi}T$ (2)

$R=\frac{40200m/s}{2\pi}2.36(10)^{10}s$ (3)

$R=1.5132(10)^{14}m$ (4)

On the other hand, we know the gravitational force $F$ between the star S with mass $M$ and the planet P1 with mass $m$ is:

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Where $G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

In addition, the centripetal force $F_{c}$ exerted on the planet is:

$F_{c}=\frac{m{V_{1}}^{2}}{R^{2}}$ (6)

Assuming this system is in equilibrium:

$F=F_{c}$ (7)

Substituting (5) and (6) in (7):

$G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}}$ (8)

Finding $M$ :

$M=\frac{V^{2}R}{G}$ (9)

$M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}$ (10)

Finally:

$M=3.66(10)^{33}kg$ (11) This is the mass of the star S

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