1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
timofeeve [1]
3 years ago
7

On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole

and g = ago at the equator (with 0 < a < 1). find g(9), the freefall acceleration at colatitude 9 as a function of 9.

Physics
1 answer:
ohaa [14]3 years ago
5 0
The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. \alpha is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\&#10;g_0=g'\\&#10;g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\&#10;ag_0=g_0-w^2r\\&#10;w^2r=g_0(a-1)&#10;
Our final equation is:
g=g_0-g_0(a-1)cos^2(\alpha)
Colatitude is:
\alpha_c=90^\circ-\alpha
The answer is:
g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)

You might be interested in
On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0
2 years ago
A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)
bezimeni [28]

Answer:

205 V

V_{R} = 2.05 V

Explanation:

L = Inductance in Henries, (H)  = 0.500 H

resistor is of 93 Ω so R = 93 Ω

The voltage across the inductor is

V_{L} = - IwLsin(wt)

w = 500 rad/s

IwL = 11.0 V

Current:

I = 11.0 V / wL

 = 11.0 V / 500 rad/s (0.500 H)

 = 11.0 / 250

I = 0.044 A

Now

V_{R} = IR

    = (0.044 A) (93 Ω)

V_{R} = 4.092 V

Deriving formula for voltage across the resistor

The derivative of sin is cos

V_{R} = V_{R} cos (wt)

Putting V_{R} = 4.092 V and w = 500 rad/s

V_{R} = V_{R} cos (wt)

    = (4.092 V) (cos(500 rad/s )t)

So the voltage across the resistor at 2.09 x 10-3 s is which means

t = 2.09 x 10⁻³

V_{R} = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))

    =  (4.092 V) (cos (500 rads/s)(0.00209))

    = (4.092 V) (cos(1.045))

    = (4.092 V)(0.501902)

    = 2.053783

V_{R} = 2.05 V

8 0
3 years ago
A radio telescope has a circular collecting dish of diameter 5.0 m. It is used to observe two distant
Likurg_2 [28]
Skskdidododododoorrororo
3 0
2 years ago
In a chemical equation, the chemicals that react are considered . In a chemical equation, the chemicals that are produced are co
vovangra [49]

Answer

Hi,

In a chemical equation, chemicals that react are the reactants, while chemicals that are produced are the products/by products. Both sides of the equation must be balanced.

Explanation

When writing a chemical equation, reactants reacts to produce products. For example in the equation for formation of water, hydrogen combines with oxygen as 2H₂ +O₂→2H₂O where the first part before the arrow represent the reactants and the next part after the arrow are the products. Reactants are on the left where as products are on the right.Coefficient 2, in this cases is used for balancing the equation.

Good luck!

4 0
3 years ago
Read 2 more answers
What does the term Hubble time mean in cosmology, and what is the current best calculation for the Hubble time?
balu736 [363]

Hubble time in cosmology means the estimated age of the universe and the best calculation for it is T=1/H, where H is the Hubble constant
4 0
3 years ago
Other questions:
  • Consider a civilization broadcasting a signal with a power of 1.2×104 watts. The Arecibo radio telescope, which is about 300 met
    5·1 answer
  • If you are traveling in a primary channel and you encounter a red and green marker (with the green band on top) when proceeding
    9·1 answer
  • PLS I NEED HELP ASAP!!!
    10·1 answer
  • Brianna weighs 425 N. She climbs a flight of stairs to a height of 8 m. It takes her 6 seconds.
    6·1 answer
  • Electronegativity is<br>____ at the top of the periodic table.​
    5·2 answers
  • In a chemical reaction blank are the substances left over
    7·1 answer
  • Create a multimedia presentation about your favorite element.
    12·2 answers
  • A 10kg crate sits on a distant planet. The planet has the same mass as Earth, but 80% of the radius of Earth (Earth radius = 637
    6·1 answer
  • What is the difference between thrust and applied force?
    9·1 answer
  • An eraser is tied to a string swung in a horizontal circle. Identify the type of force which causes this object to travel along
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!