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Misha Larkins [42]
3 years ago
6

when people walked on the moon, they found that they could jump higher than theu could back on earth. why is this true?

Physics
1 answer:
Leni [432]3 years ago
4 0

Answer:

There is less gravity on the moon

hope this helps

have a good day :)

Explanation:

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Suppose that the dipole moment associated with an iron atom of an iron bar is 2.8 × 10-23 J/T. Assume that all the atoms in the
masya89 [10]

To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.

The diple moment associated with an iron bar is given by,

\mu = \alpha *N

Where,

\alpha = Dipole momento associated with an Atom

N = Number of atoms

\alpha y previously given in the problem and its value is 2.8*10^{-23}J/T

L = 5.8cm = 5.8*10^{-2}m

A = 1.5cm^2 = 1.5*10^{-4}m^2

The number of the atoms N, can be calculated as,

N = \frac{\rho AL}{M_{mass}}*A_n

Where

\rho = Density

M_{mass} = Molar Mass

A = Area

L = Length

A_n =Avogadro number

N = \frac{(7.9g/cm^3)(1.5cm)(5.8cm^2)}{55.9g/mol}(6.022*10^{23}atoms/mol)

N = 7.4041*10^{23}atoms

Then applying the equation about the dipole moment associated with an iron bar we have,

\mu = \alpha *N

\mu = (2.8*10^{-23})*(7.4041*10^{23})

\mu = 20.72Am^2

PART B) With the dipole moment we can now calculate the Torque in the system, which is

\tau = \mu B sin(90)

\tau = (20.72)(2.2)

\tau = 45.584N.m

<em>Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.</em>

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3 years ago
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.8 1010 m (inside the orbit
creativ13 [48]

Answer:

Explanation:

From the given information:

Distance d_i = 4.8 \times 10^{10} \ m

Speed of the comet V_i = 9.1 \times 10^{4} \ m/s

At distance d_2 = 6 \times 10^{12} \ m

where;

mass of the sun = 1.98 \times 10^{30}

G = 6.67 \times 10^{-11}

To find the speed V_f:

Using the formula:

E_f = E_i + W \\ \\  where; \  \  W = 0  \ \  \text{since work done by surrounding is zero (0)}

E_f = E_i + 0 \\ \\  K_f + U_f = K_i + U_i  \\ \\ = \dfrac{1}{2}mV_f^2 +  \dfrac{-GMm}{d^2} =  \dfrac{1}{2}mV_i^2+ \dfrac{-GMm}{d_i} \\ \\ V_f = \sqrt{V_i^2 + 2 GM \Big [  \dfrac{1}{d_2}- \dfrac{1}{d_i}\Big ]}

V_f = \sqrt{(9.1 \times 10^{4})^2 + 2 (6.67\times 10^{-11}) *(1.98 * 10^{30} ) \Big [  \dfrac{1}{6*10^{12}}- \dfrac{1}{4.8*10^{10}}\Big ]}

\mathbf{V_f =53.125 \times 10^4 \ m/s}

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2 years ago
An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch
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Answer:

Explanation:

Solution is in the picture attached

8 0
3 years ago
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Power = 1800W (or 1.8KW by dividing by 1000)

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x3

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(5.4KJ or 5400J used)

$0.15 Kw/h

$0.15 X 5.4 = 0.81

Thus, cost $0.81

Hope this helps!

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Can you help me with this paper please I will give you 20 points!
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3) magnetic pole

4)magnet

5) magnetic force

6) magnetism

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2 years ago
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