What is the equation to find the volume of a hemisphere? I know the equation to get the volume of a sphere. It's 4/3 pi R cubed?

The Global Positioning System (GPS) is a constellation of about 24 artificial satellites. The GPS satellites are uniformly distr

Liono4ka [1.6K]

Answer:

b) 3.72m/s²

c) 9.33*10^5

d) 9.33*10^5

e) 11.85 hrs

Explanation:

a) to confirm that gEarth is about 98 m/s².

Let's use the formula:

$gEarth= \frac{G*M}{R^2}$

$= \frac{6.67*10^-^1^1*5.972*10^2^4}{(6378*10^3)^2}$

= 9.78 m/s²

=> 9.8m/s²

b) Given:

$m = 6.417*10^2^3$

r = 2106 miles

$g_Mars = \frac{G*M}{R^2}$

$= \frac{6.67*10^-^1^1*6.417*10^2^3}{(2106*1.61*10^3)^2}$

=3.72 m/s²

c) we use:

$F = \frac{G*M*m}{R^2}$

$=\frac{6.67*10^-^1^1*5.972*10^2^4*1630*10^3}{((20000+6378)*10^3)^2}$

$= 9.33*10^5 N$

d) Let's take the force of gravitybon earth due to satellite as our answer in (c) because the Earth's gravitational force on a GPS satellite and the force of gravity on a GPS satellite on earth are equal and opposite (two mutual forces).

$F = 9.33*10^5 N$

e) In a circular motion,

Gravitional force = Centripetal force.

$\frac{GM*m}{R^2}=\frac{m*v^2}{R}$

$\frac{GM}{R}= v^2$

Solving for v, we have

$v= \sqrt{\frac{6*67*10^-^1^1*5.972*10^2^4}{(20000+6278)*10^3}}$

v = 3886m/s

Therefore,

v = 2πR/T

$3886 = \frac{2*pi*(20000+6378)*10^3}{T}$

Solving for T, we have:

T = 42650seconds

Convert T to hours

T = 42650/60*60

T = 11.86hrs

6 0

4 years ago

A train travels 600 km in one hour. What is the trains velocity in meters/second

eimsori [14]

v= s/t

600 ÷ 60

10 m/s

4 0

4 years ago

Read 2 more answers

Small number written below a symbol in a chemical formula

pochemuha

Subscript... shows how many atoms of an element is present in a substance

3 0

4 years ago

Can someone help me please? I’m so confused

Svetradugi [14.3K]

Answer:

what do you need help with? Like which type of print each image is?

Explanation:

5 0

3 years ago

Radon-222 ( 222/86 Rn) is a radioactive gas with a half-life of 3.82 days. A gas sample contains 4.1 e 8 radon atoms initially.

kow [346]

Answer :

(a) The number of radon atoms will remain after 12 days is, $4.67\times 10^7$

(b) The number of radon nuclei have decayed by this time will be, $3.6\times 10^8$

Explanation :

Half-life = 3.82 days

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{3.82\text{ days}}$

$k=1.81\times 10^{-1}\text{ days}^{-1}$

Now we have to calculate the number of radon atoms will remain after 12 days.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.81\times 10^{-1}\text{ days}^{-1}$

t = time passed by the sample = 12 days

a = initially number of radon atoms = $4.1\times 10^8$

a - x = number of radon atoms left = ?

Now put all the given values in above equation, we get

$12=\frac{2.303}{1.81\times 10^{-1}}\log\frac{4.1\times 10^8}{a-x}$

$a-x=4.67\times 10^7$

Thus, the number of radon atoms will remain after 12 days is, $4.67\times 10^7$

Now we have to calculate the number of radon nuclei will have decayed by this time.

The number of radon nuclei have decayed = Initial number of radon atoms - Number of radon atoms left

The number of radon nuclei have decayed = $(4.1\times 10^8)-(4.67\times 10^7)$

The number of radon nuclei have decayed = $3.6\times 10^8$

Thus, the number of radon nuclei have decayed by this time will be, $3.6\times 10^8$

5 0

3 years ago