Does anyone know this?

An object weighs 100 newtons on earth.What is its weight on the moon?

Pachacha [2.7K]

Answer:

a) The gravitational acceleration at the surface of the Moon is g moon=1.67 m/s

2

The ratio of weights (for a given mass ) is the ratio of g-values, so

W

moon

=(100N)(1.67/9.8)=17N.

(b) For the force on that object caused by Earth's gravity to equal 17 N, then the free fall acceleration at its location must be

ag

=1.67m/s

2

. Thus , .

ag

= r 2

Gm

E

⇒

a

g

Gm

=1.5×10

7

m

So the object would need to be a distance of r/R

E

=2.4 "radii" from Earth's center.

4 0

3 years ago

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Earth rotates on is axis approximately once every 24 hours, causing the day to night cycle and?

katovenus [111]

Yes this is true this causes day and night

6 0

3 years ago

Consider the elementary gas-phase reversible reaction A 3C Pure A enters at a temperature of 400 K and a pressure of 10 atm. At

xxMikexx [17]

Answer:

39%

Explanation:

The equilibrium equation is:

$A\rightleftharpoons 3C$

The initial concentration of A can be calculated from the ideal gas equation:

$pV=nRT\\\\n/v=p/(RT)\\\\C_A=\dfrac{10atm}{0.08206(atm-dm^3/K-mol)/times 400K}\\\\C_A=0.304mol/liter$

Determine the conversion of substance A to substance C using an ICE table and the Kc constant:

A ⇄ 3C

I 0.304 0

C - x + 3x

E 0.304 - x 3x

$K_c=0.25=\dfrac{(3x)^3}{(0.304-x)}$

Solve for x:

You need to use a graphing calculator:

108x³ = 0.304 - x

108x³ + x - 0.304 = 0

x ≈ 0.1195 mol/liter

Then:

$C_A=0.304mol/liter-0.1195mol/liter=0.1845mol/liter\\\\C_C=3\times 0.304mol/liter=0.912mol/liter$

The equilbrium conversion is:

$\% = [0.304mol/liter-0.1845mol/liter]/(0.304mol/liter)\times 100$

$\% \approx 39\%$

3 0

3 years ago

Rachel has an unknown sample of a radioisotope listed in the table. Using a special technique, she is able to measure the mass o

mart [117]

<span>Barium-139 is the correct answer.</span>

3 0

3 years ago

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A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 15-m-long cable. You are (un

Charra [1.4K]

Answer:

1.9 s

Explanation:

We are given that

Length of cable=l=15 m

We have to find the time you have to move out of the way.

We know that

Time period,T= $2\pi\sqrt{\frac{l}{g}}$

Where g= $9.8m/s^2$

By using the formula

$T=2\pi\sqrt{\frac{15}{9.8}}$

$T=2\times 3.14\times \sqrt{\frac{15}{9.8}}=7.77 s$

Time you have to move out

$t=\frac{T}{4}=\frac{7.77}{4}=1.9 s$

Hence,time you have to move out of the way=7.77 s

6 0

4 years ago