Find the period of revolution for the planet Mercury, whose average distance from the Sun is 5.79 x 1010 m.

Answers

Answer 1

Answer:

T = 7.61*10^6 s

Explanation:

In order to calculate the Mercury's period. in its orbit around the sun, you take into account on the Kepler's law. You use the following formula:

[tex]T=\sqrt{\frac{4\pi^2r^3}{GM_s}}[/tex]         (1)

T: period of Mercury

r: distance between Mercury and Sun

Ms: mass of the sun = 1.98*10^30 kg

G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2

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

[tex]T=\sqrt{\frac{4\pi^2(5.79*10^{10}m)^3}{(6.674*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}}\\\\T=7.61*10^6s*\frac{1h}{3600s}*\frac{1d}{24h}=88.13\ days[/tex]

The period of Mercury is 7.61*10^6 s, which is approximately 88.13 Earth's days


Related Questions

Three rocks of equal mass are thrown with identical speeds from the top of the same building (identical height). Rock X is thrown vertically downward, rock Y is thrown vertically upward, and rock Z is thrown horizontally.Required:Which rock has the greatest speed just before it hits the ground? Assume air resistance is negligible.

Answers

Answer:Three rocks of equal mass are thrown

Explanation:

The phenomena: Three rocks of equal mass are thrown with identical speeds from the top of the same building (identical height). Rock X is thrown vertically downward, rock Y is thrown vertically upward, and rock Z is thrown horizontally.

In an undergraduate physics lab, a simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min. What are the period and length of the pendulum

Answers

Explanation:

We have

A simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min.

The frequency of a pendulum is equal to the no of oscillation per unit time. so,

[tex]f=\dfrac{N}{t}\\\\f=\dfrac{71}{1.8\times 60}\\\\f=0.65\ Hz[/tex]

Tim period is reciprocal of frequency. So,

[tex]T=\dfrac{1}{0.65}\\\\T=1.53\ s[/tex]

The time period of a pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

l is length of pendulum

[tex]l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(1.53)^2\times 9.8}{4\pi ^2}\\\\l=0.58\ m[/tex]

So, the period and length of the pendulum are 1.53 s and 0.58 m respectively.

B. Write short notes on:
1. Horticulture
2. Pisciculture
3. Aviculture
4. Veterinary science
5. Intensive farming.

Answers

1. Horticulture is the agriculture of plants, mainly for food, materials, comfort and beauty for decoration.

2.Pisciculture also known as fish farming is the rearing of fish for food in enclosures such as fish ponds or tanks.

3.Aviculture is the practice of keeping and breeding birds, especially of wild birds in captivity. Aviculture is generally focused on not only the raising and breeding of birds, but also on preserving avian habitat, and public awareness campaigns.

4. Veterinary medicine is the branch of medicine that deals with the prevention, control, diagnosis, and treatment of disease, disorder, and injury in animals. Along with this, it also deals with animal rearing, husbandry, breeding, research on nutrition and product development.

5. Intensive agriculture, also known as intensive farming and industrial agriculture, is a type of agriculture, both of crop plants and of animals, with higher levels of input and output per cubic unit of agricultural land area.

Hope this helps.

A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amount that the column has shrunk.

Answers

Answer:

0.4757 mm

Explanation:

Given that:

Load P = 223,000 N

the length of the height of the aluminium column = 1.22 m

the diameter of the aluminum column = 10.2 cm = 0.102 m

The amount that the column has shrunk ΔL can be determined by using the formula:

[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]

where;

A = πr²

2r = D

r = D/2

r = 0.102/2

r = 0.051

A = π(0.051)²

A = 0.00817

Also; the young modulus of aluminium [tex]E_{Al}[/tex] is:

[tex]E_{Al}= 7*10^{10} \Nm^{-2}[/tex]

[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]

[tex]\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}[/tex]

ΔL = 4.757 × 10⁻⁴ m

ΔL =  0.4757 mm

Hence; the amount that the column has shrunk is 0.4757 mm

The upward velocity of a 2540kg rocket is v(t)=At + Bt2. At t=0 a=1.50m/s2. The rocket takes off and one second afterwards v=2.00m/s. Determine the constants A and B with units.

Answers

Answer:

The value of A is 1.5m/s^2 and B is 0.5m/s^³

Explanation:

The mass of the rocket = 2540 kg.

Given velocity, v(t)=At + Bt^2

Given t =0  

a= 1.50 m/s^2

Now, velocity V(t) = A*t + B*t²

If,  V(0) = 0, V(1) = 2

a(t) = dV/dt = A+2B × t  

a(0) = 1.5m/s^²  

1.5m/s^²  =  A + 2B ×  0  

A = 1.5m/s^2

now,

V(1) = 2 = A× 1 + B× 1^²  

1.5× 1 +B× 1 = 2m/s

B = 2-1.5  

B = 0.5m/s^³

Now Check V(t) = A× t + B × t^²

So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² ×  1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s  

Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)

Which describes any compound that has at least one element from group 17? Halide;noble gas; metalliod; transition metal

Answers

Answer:

Halide

Explanation:

It has at least one element from the halogen group (17)

Halide describes any compound that has at least one element from group 17, therefore the correct option is option A.

What are halides?

When the elements belonging to group 17 of the periodic table form ionic compounds with other electropositive elements, then these compounds are known as halides.

These elements from group 17 are also known as halogens. Generally, these halides have very high electronegativity as they reside on the right side of the periodic table.

Generally, the valency of the halogens element involved in the halide compound is one and they form ionic compounds with the alkali and alkaline earth metals.

Thus, halides are compounds that have at least one element from group 17.

Learn more about halides from here

https://brainly.com/question/5504882

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Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/kg. The final specific volume is 0.027 m^3/kg. Find the specific work in the process.

Answers

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

[tex]W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.[/tex]

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

An alarm clock is plugged into a 120 volt outlet and has a resistance of 15,000 ohms. How much power does it use?

Answers

Answer:

The power used is 0.96 watts.

Explanation:

Recall the formula for electric power (P) as the product of the voltage applied  times the circulating current:

[tex]P=V\,\,I[/tex]

and recall as well that the circulating current can be obtained via Ohm's Law as the quotient of the voltage applied divided the resistance:

[tex]V=I\,\,R\\I=\frac{V}{R}[/tex]

Then we can re-write the power expression as:

[tex]P=V\,\,I=V\,\,\frac{V}{R} =\frac{V^2}{R}[/tex]

which in our case becomes:

[tex]P=\frac{V^2}{R}=\frac{120^2}{15000} =0.96\,\,watts[/tex]

How much work is needed to move an object from one position to another when both positions are located the same distance from the center of the earth

Answers

Answer:

The product of the object's weight and the horizontal distance between the two positions.

Explanation:

Work is the product of force and the distance through which this force is moved. The distance moved can be vertical, or horizontal. For two bodies located the same distance from the center of the earth, the work done will be the product of the weight of the product and the horizontal distance between the two positions. If the vertical work is needed, then the work is zero, since there is no height gradient between them.

The magnetic field strength at the north pole of a 2.0-cm-diameter, 8-cm-long Alnico magnet is 0.10 T. To produce the same field with a solenoid of the same size, carrying a current of 1.8 A , how many turns of wire would you need

Answers

Answer:

The number of turns of the solenoid is 3536 turns

Explanation:

Given;

magnetic field of the solenoid, B = 0.1 T

current in the solenoid, I = 1.8 A

length of the solenoid, L = 8cm = 0.08m

The magnetic field near the center of the solenoid is given by;

B = μ₀nI

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

n is number of turns per length

I is the current in the coil

The number of turns per length is calculated as;

n = B / μ₀I

n = (0.1 ) / (4π x 10⁻⁷ x 1.8)

n = 44203.95 turns/m

The number of turns is calculated as;

N = nL

N = (44203.95)(0.08)

N = 3536 turns

Therefore, the number of turns of the solenoid is 3536 turns

Suppose that the voltage of the battery in the circuit is 3.9 V, the magnitude of the magnetic field (directed perpendicularly into the plane of the paper) is 1.1 T, and the length of the rod between the rails is 0.22 m. Assuming that the rails are very long and have negligible resistance, find the maximum speed attained by the rod after the switch is closed.

Answers

Answer:

v = 16.11 m/s

Explanation:

In order to calculate the maximum speed of the rod, you use the following formula:

[tex]\epsilon=vBLsin\theta[/tex]        (1)

ε = voltage of the circuit = 3.9V

v: maximum speed of the rod = ?

B: magnitude of the magnetic field = 1.1T

L: length of the rod = 0.22m

θ: angle between the direction of motion of the rod and the direction of the magnetic field = 90°

You solve the equation (1) for v and replace the values of the other parameters:

[tex]v=\frac{\epsilon}{BLsin\theta}=\frac{3.9V}{(1.1T)(0.22m)sin90\°}\\\\v=16.11\frac{m}{s}[/tex]

The maximum speed of the rod is 16.11 m/s

A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediately after the impact, a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction relative to the +x-axis does the third piece move?
(a) 39.8º from the +x-axis
(b) 36.9° from the +x-axis
(c) 39.9° from the +x-axis
(d) 216.9° from the +x-axis
(e) 219.8° from the +X-axis

Answers

Answer:

M1 Vx1 + M2 Vx2 + M3 Vx3 = 0     conservation of momentum in x direction

Vx3 = -(M1 Vx1 + M2 Vx2 ) / M3

Vx3 = - 320 * 2 / 100 = -6.4 m/s      M2 has no x-component of momentum

Likewise:

Vy3 = -(M1 Vy1 + M2 Vy2 ) / M3

Vy3 = - 355 * 1.5 / 100 = -5.33 m/s

tan theta = -5.33 / -6.4 = .833    where theta is in the third quadrant and measured from the negative x-axis

theta = 39.8 deg

180 + 39.8 = 219.8     from the positive x-axis

Which scientist's work led to our understanding of how planets move around
the Sun?
A. Albert Einstein
B. Lord Kelvin
C. Johannes Kepler
D. Edwin Hubble

Answers

Answer:

Johannes Kepler

Explanation:

He made rules about planetary motion.The scientist Johannes Kepler was a German astronomer.He found out that the planets evolved around the Sun.He also made the laws of planetary motion.

Hope this helped,

Kavitha

It was Johanne keplers work

A 32-cm-long solenoid, 1.8 cm in diameter, is to produce a 0.30-T magnetic field at its center. If the maximum current is 4.5 A, how many turns must the solenoid have?

Answers

Answer:

16,931 turns

Explanation:

The magnetic field produced is expressed using the formula

[tex]B = \frac{\mu_0NI}{L}[/tex]

B is the magnetic field = 0.30T

I is the current produced in the coil = 4.5A

[tex]\mu_0[/tex] is the magnetic permittivity in vacuum = 1.26*10^-6Tm/A

L is the length of the solenoid = 32 cm = 0.32 m

N is the number of turns in the solenoid.

Making N the subject of the formula from the equation above;

[tex]B = \frac{\mu_0NI}{L}\\\\BL = \mu_0NI\\\\Dividing\ both\ sides \ by \ \mu_0I\\\\\frac{BL}{\mu_0I} =\frac{\mu_oNI}{\mu_0I} \\\\[/tex]

[tex]N = \frac{BL}{\mu_0I}[/tex]

Substituting the give values to get N;

[tex]N = \frac{0.3*0.32}{1.26*10^{-6} * 4.5}\\\\N = \frac{0.096}{0.00000567} \\\\N = 16,931.21[/tex]

The number of turns the solenoid must have is approximately 16,931 turns

An 75-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical elevation increases 540 m. Determine the change in gravitational potential energy of the climber-Earth system.

Answers

Answer:

The change in gravitational potential energy of the climber-Earth system is  [tex]\Delta PE = 396900 \ J[/tex]

Explanation:

From the question we are told that

    The mass of the hiker is  [tex]m = 75 \ kg[/tex]

    The time  taken is  [tex]T = 2 \ hr = 2 * 3600 = 7200 \ s[/tex]

    The  vertical elevation after time  T is  [tex]H = 540 \ m[/tex]

   

The  change  in gravitational potential is  mathematically represented as

         [tex]\Delta PE = mgH[/tex]

here g is the acceleration due to gravity with value  [tex]g = 9.8 \ m/s^2[/tex]  

     substituting values  

        [tex]\Delta PE = 75 * 9.8 * 540[/tex]

       [tex]\Delta PE = 396900 \ J[/tex]

A conventional current of 8 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius 0.078 m. Another circular loop of wire lies in the same plane, with its center at the origin and with radius 0.03 m. How much conventional current must run counterclockwise in this smaller loop in order for the magnetic field at the origin to be zero

Answers

Answer:

I2 = 3.076 A

Explanation:

In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:

[tex]B=\frac{\mu_oI}{2R}[/tex]        (1)

I: current in the wire

R: radius of the wire

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:

[tex]B_T=\frac{\mu_oI_1}{2R_1}-\frac{\mu_oI_2}{2R_2}[/tex]         (2)

I1: current of the first ring = 8A

R1: radius of the first ring = 0.078m

I2: current of the second ring = ?

R2: radius of the first second = 0.03m

To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:

[tex]\frac{\mu_oI_2}{2R_2}=\frac{\mu_oI_1}{2R_1}\\\\I_2=I_1\frac{R_2}{R_1}=(8A)\frac{0.03m}{0.078m}=3.076A[/tex]

The current of the second ring is 3.076A and makes that the magntiude of the total magnetic field generated for both rings is equal to zero.

A circuit element maintains a constant resistance. If the current through the circuit element is doubled, what is the effect on the power dissipated by the circuit element

Answers

Answer:

This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.

Explanation:

The formula for calculating the power expended in a circuit is P =  I²R... 1

i is the current (in amperes)

R is the resistance (in ohms)

If  a circuit element maintains a constant resistance and the current through the circuit element is doubled, then new current I₂ = 2I

New power dissipated P₂ = (I₂)²R

P₂ = (2I)²R

P₂ = 4I²R ... 2

Dividing equation 2 by 1 will give;

P₂/P = 4I²R/I²R

P₂/P = 4

P₂ = 4P

This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.

A rigid massless rod is rotated about one end in a horizontal circle. There is a particle of mass m1 attached to the center of the rod and a particle of mass m2 attached to the outer end of the rod. The inner section of the rod sustains a tension that is 5 times as great as the tension that the outer section sustains. Find the ratio m1/m2

Answers

Answer:

m₁/m₂ =8

Explanation:

Writing the centripetal force equation: where T₁ is inner tension and T₂ is outer tension

T₂ = m₂ × ω²R....equation 1

T₁-T₂ = m₁ × ω²R/2

Also T₁=5T₂

5T₂-T₂ = m₁ × ω²R/2

4 × 2T₂ = m₁ × ω²R

8T₂ = m₁ × ω²R

dividing it by equation 1

8 = m₁/m₂

Hence m₁/m₂ =8

5. Two men, Joel and Jerry, push against a car that has stalled, trying unsuccessfully to get it moving. Jerry stops after 10 min, while Joel is able to push for 5.0 min longer. Compare the work they do on the car

Answers

Answer:

The work done by both Joel and Jerry is equal to 0 J.

Explanation:

The work done on a body by an external agency is the product of the force applied on the body and the distance through which the body moves. Therefore,

W = F.d

where,

W = Work Done on the Body

F = Force Applied on the Body

d = displacement covered by the body

In the given case of both Joel and Jerry, they are unable to move the car. It means that the displacement covered by the car is zero. Hence,

W = F(0)

W = 0 J (For both Joel and Jerry)

According to the law of conservation of energy, if the ocean water cools, then something else should warm. What is it that warms?

Answers

Answer:

The answer is air

Explanation:

The combustion of propane (C3H8) in the presence of excess oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g) When 2.0 mol of O2 are consumed in this reaction, ________ mol of CO2 are produced.

Answers

Answer:

1.2

Explanation:

2.0 mol O₂ × (3 mol CO₂ / 5 mol O₂) = 1.2 mol CO₂

A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.

Answers

Complete question:

Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. In contrast,  a passenger not wearing a seat belt or using an air bag decelerates over a distance of 5mm.

(a) A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.

Answer:

The net force on the person as the air bad deploys is -6750 N backwards

Explanation:

Given;

mass of the passenger, m = 60 kg

velocity of the car at impact, u = 15 m/s

final velocity of the car after impact, v = 0

distance moved as the front of the car crumples, s = 1 m

First, calculate the acceleration of the car at impact;

v² = u² + 2as

0² = 15² + (2 x 1)a

0 = 225 + 2a

2a = -225

a = -225 / 2

a = -112.5 m/s²

The net force on the person;

F = ma

F = 60 (-112.5)

F = -6750 N backwards

Therefore, the net force on the person as the air bad deploys is -6750 N backwards

the time required for one cycle, a complete motion that returns to its starting point, it called the_____. period medium frequency periodic motion

Answers

Answer:

The time required for one cycle, a complete motion that returns to its starting point,it is called periodic motion

Explanation:

I hope this will help you:)

During a move, Jonas and Matías carry a 115kg safe to the third floor of a building, covering a height of 6.6m.

1) = what work do they do?
2) = what power do they develop if the work is done in 5.5 minutes?

Answers

Answer:

work is =7590joules

power = 23watts

Answer:

1) 7590 Joules

2) 23 Watts

Explanation:

1) Work = force × distance

W = mgh

W = (115 kg) (10 m/s²) (6.6 m)

W = 7590 J

2) Power = work / time

P = W / t

P = (7590 J) / (330 s)

P = 23 W

A wheel has a rotational inertia of 16 kgm2. Over an interval of 2.0 s its angular velocity increases from 7.0 rad/s to 9.0 rad/s. What is the average power done by the torque

Answers

Answer:

128.61 Watts

Explanation:

Average power done by the torque is expressed as the ratio of the workdone by the toque to time.

Power = Workdone by torque/time

Workdone by the torque = [tex]\tau \theta[/tex] = [tex]I\alpha * \theta[/tex]

I is the rotational inertia = 16kgm²

[tex]\theta = angular\ displacement[/tex]

[tex]\theta = 2 rev = 12.56 rad[/tex]

[tex]\alpha \ is \ the\ angular\ acceleration[/tex]

To get the angular acceleration, we will use the formula;

[tex]\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}[/tex]

[tex]\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}[/tex]

Workdone by the torque = 16 * 1.28 * 12.56

Workdone by the torque = 257.23 Joules

Average power done by the torque = Workdone by torque/time

=  257.23/2.0

= 128.61 Watts

The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 m, and a baseball has a mass of 145 g.(a) What force did he produce on the ball during this record-setting pitch? (b) Draw free-body diagrams of the ball during the pitch and just after it left the pitcherâs hand.

Answers

Answer:

Explanation:

F ×1 = 0.5×0.145×47×47

F = 160.15 N

What is the equivalent temperature in kelvin if you have a metal at 50°F?

Answers

Answer:

The required temperature is 283 K.

Explanation:

[tex]T\:=\:\left(50-32\right)\times \frac{5}{9}+273\\\\T=283\:K[/tex]

Best Regards!

What is the equivalent resistance between the points A and B of the network?​

Answers

Explanation:

First, simplify the circuit. Then calculate the parallel and consecutive resistances to find the answer.

HELP ILL MARK BRAINLIEST PLS!!!!

A patch of mud has stuck to the surface of a bicycle tire as shown. The stickiness of
the mud is the centripetal or tension force that keeps the mud on the tire as it spins.
Has work been done on the mud as the tire makes one revolution, if the mud stays
on the tire? Explain.

Answers

Answer:

Yes, work has been done on the mud.

Explanation:

Work is done on a body, when a force is applied on the body to move it through a certain distance. In the case of the mud, the tire exerts a centripetal force on the mud. The centripetal force moves the mud along a path that follows the circle formed by the tire in one revolution of the tire. The total distance traveled is the circumference of the circle formed. The work done on the mud is therefore the product of the centripetal force on the mud from the tire, and the circumference of the circle formed by the tire, usually expressed in radian.

Suppose that 300 keV X-ray photons are aimed at a zinc cube (Zinc, Z = 30). According to the chart below, what effect will predominate when the X-rays hit the metal?
a) Photoelectric Effect 3
b) Compton Effect 3
c) Pair Production

Answers

Answer:

the answer is option A = photoelectric effect

Explanation:

If the threshold frequency of a metal is lower than the energy of X-rays, then photoelectric effect will happen.

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