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geniusboy [140]
4 years ago
8

The weight of a 72.0 kg astronaut on the Moon, where g = 1.63 m/s2 is (5 points) Select one: a. 112 N b. 117 N c. 135 N d. 156 N

e. 171 N
Physics
2 answers:
Hunter-Best [27]4 years ago
5 0

Answer: The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

Explanation:

Mass of the astronaut on the moon , m= 72 kg

Acceleration due to gravity on moon,g  = 1.63 m/s^2

According to Newton second law of motion: F = ma

This will changes to = Weight = mass × g

Weight=72 kg\times 1.63m/s^2=117.36 N

The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

lesya692 [45]4 years ago
3 0
B. 117 N because you multiply 72 Kg with 1.63 m/s^2 and Kgm/s^2 is considered to be newtons
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The graph between the strength of the magnet(number of paper clips picked) and battery is approximately a straight line.

For 25 coil, with increase of 1.5 V battery voltage, the electromagnet picks about 5 more clips. So, for a 7.5 V battery, it would pick about 30 paper clips.

For 50 coil, with increase of 1.5 V battery voltage, the electromagnet picks about 15 more clips. So, for a 7.5 V battery, it would pick about 30 paper clips.

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4 years ago
What did the scientist say to the hydrogen atom that claimed it lost an electron
lys-0071 [83]

Answer:

"Are you positive?"

5 0
4 years ago
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The gas in a balloon has T=280 K and V=0.0279 m^3. if the temperature increases to 320 K at constant pressure, what is the new v
victus00 [196]

Answer:

0.0319 m³

Explanation:

Use ideal gas law:

PV = nRT

where P is pressure, V is volume, n is amount of gas, R is the gas constant, and T is temperature.

Since P, n, and R are held constant:

n₁ R / P₁ = n₂ R₂ / P₂

Which means:

V₁ / T₁ = V₂ / T₂

Plugging in:

0.0279 m³ / 280 K = V / 320 K

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8 0
3 years ago
A length of copper wire carries a current of 11 A, uniformly distributed through its cross section. Calculate the energy density
NISA [10]

Answer:

a)0.983 \frac{J}{m^3}

b)u_E =7.329x10^-3 \frac{J}{m^3}

Explanation:

The energy density is "the energy per unit volume, in the electric field.  The energy stored between the plates of the capacitor equals the energy per unit volume stored in the electric field times the volume between the plates".

A magnetic field is a "vector field that describes the magnetic influence of electric charges in relative motion and magnetized materials".

Part a

For this case we can assume use the equation for the magnetic field in terms of the energy per unit of volume.

B=\sqrt{2\mu_o u}

Where μ0 represent the permeability constant, also known as the magnetic constant. If we solve for u we got:

u=\frac{B^2}{2\mu_o}

We also know that the magnetic field can be expressed in terms of the current and the radius of action R like this:

B=\frac{\mu_o i}{2\pi R}

Replacing this on the formula for u we have:

u=\frac{1}{2\mu_o}(\frac{\mu_o i}{2\pi R})^2

And simplyfing we got:

u=\frac{\mu_o i^2}{8\pi^2 R^2}

Replacing the values given we have:

u=\frac{(4\pix10^{-7} \frac{H}{m} (11A)^2}{8\pi^2 (0.0014m)^2} =0.983 \frac{J}{m^3}

Part b

The density current is given by this formula J=i/A and the resistance by R=\frac{\rho l}{A}

If we use the equation for the energy density we have this:

u_E =\frac{1}{2}\varepsilon_o E^2 =\frac{\varepsilon}{2}(\rho J)^2=\frac{\varepsilon}{2}(\frac{iR}{l})^2

And replacing the values given we have:

u_E =\frac{8.85x10^{-12}\frac{F}{m}}{2}(\frac{11A(3700\frac{\Omega}{m})}{l})^2 =7.329x10^-3 \frac{J}{m^3}

4 0
4 years ago
Can someone help me rq?
Nana76 [90]
How are we suppose to answer this tho?
6 0
3 years ago
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