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r-ruslan [8.4K]
3 years ago
6

FIND THE WEIGHT OF A 80 KG MAN ON THE SURFACE OF MOON? WHAT SHOULD BE HISS MASS ON THE EARTH AND ON THE MOON? (ge = 9.8 m/s2 ; g

m = 1.63 m/s2)
pls answer fast
Physics
1 answer:
elena55 [62]3 years ago
8 0

Weight = (mass) x (acceleration of gravity at the place where the mass is) .

Man's mass = 80 kg

His weight on Earth = (80 kg) x (9.8 m/s²) = 784 newtons (about 176 pounds)

His weight on the Moon = (80 kg) x (1.63 m/s²) = <em>130.4 newtons</em> (about  29.2 pounds)

His mass is <em>80 kg</em>.  Mass is the thing about him that doesn't change. 
He has the same mass on the Earth, on the Moon, or anywhere.

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What is the term for substances that have several unpaired electrons and are strongly magnetic
Murljashka [212]

Answer:

Scientific definitions for ferromagnetic

The property of being strongly attracted to either pole of a magnet. Ferromagnetic materials, such as iron, contain unpaired electrons, each with a small magnetic field of its own, that align readily with each other in response to an external magnetic field.

Explanation:

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Answer question 83 and 84
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A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis
bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

B=\frac{\mu_o I_r}{2\pi r}     (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

I_r=JA_r

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by  A_r=\pi(r^2-R_1^2)

The current density is given by the total area and the total current:

J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current  = 4 A

Then, the current in the wire for a distance r is:

I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}  (2)

You replace the last result of equation (2) into the equation (1):

B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})

Finally. you replace the values of all parameters:

B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0
3 years ago
Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The mo
ddd [48]

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

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From the question,

r = i + j + k

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Therefore,

M₀ = (i + j + k) x (1i + 0j +  5k)

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M₀ = 5i - 4j - k

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