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ivanzaharov [21]
3 years ago
11

What is the difference in Neil Armstrong’s weight on the moon and on earth? Neils mass is 160kg including his spacesuit and back

pack. The moon’s gravitational pull is only 17% of the Earths gravitational pull. Assume the Earth’s gravitational pull is 10N/kg (if you can show working out.)
Physics
1 answer:
Len [333]3 years ago
3 0

Explanation:

Given parameters:

Mass of Neil Armstrong = 160kg

Gravitational pull of earth = 10N/kg

Moon's pull = 17% of the earth's pull

Unknown:

Difference between Armstrong's weight on moon and on earth.

Solution:

To find the weight,

   Weight = mass x acceleration due to gravity = mg

Moon's gravitational pull = 17% of the earth's pull = 17% x 10 = 1.7N/kg

Weight on moon = 160 x 1.7 = 272N

Weight on earth = 160 x 10 = 1600N

The difference in weight = 1600 - 272 = 1328N

The weight of Armstrong on earth is 1328N more than on the moon.

Learn more:

Weight and mass brainly.com/question/5956881

#learnwithBrainly

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Answer:

You could try finding a familiar peer to join the activity with your child. Or ask your child who their friends are at school, or what they look for in a friend at school.

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Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
storchak [24]

Answer:

\large \boxed{42\, \mu \text{C}}$

Explanation:

The formula for the force exerted between two charges is

F=k \dfrac{ q_1q_2}{r^2}

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

F=k\dfrac{q^{2}}{r^2}

\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

7 0
3 years ago
2.5 g of helium at an initial temperature of 300 K interacts thermally with 9.0 g of oxygen at an initial temperature of 620 K .
muminat

Answer:

Explanation:

2.5 g of He = 2.5 / 4  mole

= .625 moles

9 g of oxygen = 9/32

= .28 mole of oxygen

C_p of He = 3/2 R

C_p of O₂ = 5/2 R

A ) Initial thermal energy of He = 3/2 n R T

= 1.5 x .625 x 8.32 x 300

= 2340 J

Initial thermal energy of O₂ = 5/2 n R T

= 2.5 x .28 x 8.32 x 620

= 3610.88 J

B ) If T be the equilibrium temperature after mixing

gain of heat by helium

= n C_p Δ T

= .625 x 3/2 R x ( T - 300 )

Loss of heat by oxygen

n C_p Δ T

= .28 x 5/2 R x ( 620 - T )

Loss of heat = gain of heat

.625 x 3/2 R x ( T - 300 ) = .28 x 5/2 R x ( 620 - T )

1.875 T- 562.5 = 868- 1.4 T

3.275 T = 1430,5

T = 436.8 K

Thermal energy of He

= 1.5 x .625 x 8.32 x 436.8

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= 2.5 x .28  x 8.32 x 436.8

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C )

Heat energy transferred

=  .28 x 5/2 R x ( 620 - T )

=  .28 x 5/2 x  8.32 x ( 620 - 436.8 )

1066.95 J

Heat will flow from O₂ to He

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The energy of a photon was found to be
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Answer:

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b.6.5025 J

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Spring constant,k=1800 N/m

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a.Potential energy stored in the compressed spring  is given by

P.E=\frac{1}{2}kx^2

P.E=\frac{1}{2}(1800)(8.5\times 10^{-2})^2

P.E=6.5025 J

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P.E of spring=K.E of the pellet

K.E of the pellet=6.5025 J

6 0
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