The distance to the other planets in the Solar System from the Earth depends on where they are in their orbit. The closest that Mars gets to the Earth is 0.5 AU. At this time it is observed to have an angular diameter of 18 arcseconds. The furthest it gets from Earth is 2.5 AU. What would its angular diameter be for this observation

Answers

Answer 1

Answer:

3.6 arcsec

Explanation:

 angular diameter = diameter / distance

diameter is constant

so angular diameter ∝ 1 / distance

angular diameter = k / distance

For first case ,

18 = k / .5

for second case let angular diameter be D .

D = k / 2.5

dividing ,

D / 18 = .5 / 2.5 = 1 / 5

D = 18 / 5 = 3.6 arcsec

3.6 arcsec is the answer .


Related Questions

The equilibrant is the equal to the resultant magnitude but opposite in direction.


True

False


Answer and I will give you brainiliest

Answers

Answer:

The answer is False......

Answer:

true

it is equal but opposite

A scientist reports a measurement of the temperature of the surface of a newly discovered planet as negative 20 Kelvin. What conclusion can you draw from this report

Answers

Answer:

The temperature of this newly discovered planet violates the third law of thermodynamic, there is a mistake in this value.

Explanation:

The third law of the Thermodynamic says:

At zero kelvin all molecular movement stops, which means that the entropy will be zero at this temperature.

So we can say there is no thermodynamic system that has temperature values less than 0 K.

The conclusion of the report will be.

The temperature of this new planet violates the third law of thermodynamic, there is a mistake in this value.

I hope it helps you!

dose contact or noncontact force weaken with distance

Answers

Answer:

The more massive an object is, the greater the gravitational force. Since gravitational force is inversely proportional to the distance between two interacting objects, more separation distance will result in weaker gravitational forces.

Explanation:

hope this help:)

Non-contact force is a force that can exert a push, or pull an object without actually having any physical contact with that object. This happens when force is applied on some object by another object without any interaction or contact between those two objects. In a non-contact force, the force is transmitted over distance. Some refer to these types of forces as “action-at-a-distance forces.” Examples of non-contact forces include magnetic, gravitational and electrostatic force.

A cylinder of gas is at room temperature (20°C). The air conditioner breaks down, and the temperature rises
to 46°C. What is the new pressure of the gas relative to its initial pressure (Pi)?

Answers

Answer: we divide this equation by pV and use {C}_{p}={C}_{V}+ . ... The temperature, pressure, and volume of the resulting gas-air mixture

Explanation:

The new pressure of the gas relative to its initial pressure P₁ is P₁ × 1.09

From the question given above, the following data were obtained:

Initial temperature (T₁) = 20 °C = 20 + 273 = 293 K

Initial pressure (P₁) = P₁

New temperature (T₂) = 46 °C = 46 + 273 = 319 K

New pressure (P₂) =?

The new pressure of the gas can be obtained as follow:

P₁ / T₁ = P₂ / T₂

P₁ / 293 = P₂ / 319

Cross multiply

293 × P₂ = P₁ × 319

Divide both side by 293

P₂ = (P₁ × 319) / 293

P₂ = P₁ × 1.09

Thus, the new pressure relative to it's initial pressure is P₁ × 1.09

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5. Friction has two components __________ and ___________.

Answers

Answer:

static friction and kinetic friction.

Explanation:

what is the frequency of a wave having a period equal to 18 seconds

Answers

Explanation:

The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.

The relation between wave period and frequency is as follows.

T = \frac{1}{f}T=

f

1

where, T = time period

f = frequency

It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.

T = \frac{1}{f}T=

f

1

or, f = \frac{1}{T}f=

T

1

= \frac{1}{18 sec}

18sec

1

= 0.055 per second (1cycle per second = 1 Hertz)

or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz

Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz5.5×10 −2 hertz .

You want to paddle a boat straight across the Mississippi River. You are able to paddle the boat with a speed of 2.2 m/s relative to the water, and the current flows at a speed of 1.1 m/s. In what direction (angle), relative to the river bank, do you need to point the boat to cross the river in the shortest time? The width of the river is 95.0 m.
A) 60 degrees
B) 80 degrees
C) 90 degrees
D) 70 degrees
E) 50 degrees

Answers

Answer:

C) 90 degrees

Explanation:

If we could have an imaginative illustration about what the question is all about.

We could think of the velocity of the river (V_r cosθ) as a horizontal component and the velocity of the boat (V_b sinθ) as the vertical component.

So the velocity of the boat V_b = 2.2 m/s; &

the velocity of the river V_r = 1.1 m/s

The minimum time (which is the shortest time) required for the boat to cross the river is θ = 90°

the time require is:

[tex]t = \dfrac{d}{v_b \ sin \theta}[/tex]

[tex]t = \dfrac{95}{2.2}[/tex]

t = 43.18 sec

There are two metallic spheres and a positive-charge generator. Throughout the entire problem, one sphere is always closest to the generator (call it C), and one is always farther away (call it F). Both spheres start electrically neutral and are in contact with each other. These spheres are now brought near the positivecharge generator (call it G), but do not touch it. The two metallic spheres C and F are then separated from each other, and then individually moved away from the positive-charge generator.

Immediately after the two spheres are moved away from the positive-charge generator, the sphere farther away from the charge generator is charged_______
a. positive
b. negative
c. neutral
d. cannot be determined

Answers

Answer:

The correct answer is C

Explanation:

In this problem we must remember that charges of the same sign repel and of the same sign attract.

Generator G induces a charge in the closest spheres C, otherwise the charge is negative, therefore in the furthest sphere F, with the ac that is in contact, a positive charge is formed, they do not indicate that the spheres It is in contact with Earth, so the charges remain on them.

As the spheres move away from the generator, the load is redistributed, they return to their initial load, that is, they remain Neutral since no load left or entered the spheres, there was only a redistribution due to the proximity of the generator.

The correct answer is C

PLEASEEEE HELP MEEEEE!!!!

Answers

Answer:

D. Yes

m

m mb

m

m

m

n

nn

n

n

n

j

j

jj

Two masses are suspended by cord that passes over a pulley with negligible mass. The cord also has negligible mass. One of the masses, m1, has a mass of 7.0 kg and the other mass, m2, has a mass of 3.0 kg. The pulley turns on a shaft through the center of the pulley and supports the pulley and all the masses. The vertical force of the shaft on the pulley that supports the whole system is

Answers

Answer: F = 98N

Explanation: The shaft have to sustain the pulley, the cord and the two masses. The pulley and the cord have negligible masses, so, they have negligible weight.

The two masses have two vertical forces acting on them: force of traction because of the cord and force due to gravitational force, also known as weight.

So, the vertical force the shaft has to support is the sum of the weight of each mass:

[tex]F_{net}=F_{g}_{1}+F_{g}_{2}[/tex]

[tex]F_{net}=m_{1}.g+m_{2}.g[/tex]

[tex]F_{net}=g(m_{1}+m_{2})[/tex]

[tex]F_{net}=9.8(7+3)[/tex]

[tex]F_{net}=[/tex] 98

The vertical force that supports the whole system is 98 N.

A particle has 37.5 J of kinetic energy and 12.5 J of gravitational potential energy at one point during its fall from a tree to the ground. An instant before striking the ground, how much mechanical energy-rounded to the nearest Joule-does the particle have? Ignore air resistance.​

Answers

Answer: 50J

Explanation:

Mechanical energy follows the same principles of kinetic energy and potential energy, it is conserved. So Ei = Ef.

Mechanical energy is the sum of ALL energy's. There is no friction, so its just kinetic plus potential.

37.5 + 12.5 = 50J

Since the particle has not touched the ground, it has not transferred any energy to the ground yet, therefore the mechanical energy must still be 50J; mostly in kinetic energy with a very small amount of potential because of the low height relative to the ground.

A lake has a surface area of 410 m2 and a volume of 1140 m3 . Suppose that during the day, sunlight with a power averaging 820 W/m2 shines on the lake, and that about 10% of this power is absorbed in the lake, producing heat. Assume that the temperature of the lake water stays constant because the absorbed solar power is exactly balanced by heat lost due to evaporation of water from the lake surface. What is the evaporation rate, in g/s

Answers

Answer:

Explanation:

Total solar energy falling on total surface per second

= 410 x  820 W

= 336200 W

10 % of 336200 = 33620 J is converted into heat which is absorbed by lake water . But its temperature does not rise because heat is used up in evaporating water in the form of vapor .

Total heat released during evaporation = 33620 J

Let evaporation rate be m gram /s

heat absorbed by m gram water = m x latent heat of evaporation

= m x 2260 J .

Given ,

m x 2260 = 33620

m = 14.87 g /s .

Solar flares are disturbances on the Sun’s surface that send large amounts of energy away from the Sun’s surface. These disturbances can often interfere with technology here on Earth. What statement BEST describes the cause of solar flares?

Answers

Answer:

with that  boom and that boom glare boom that boom glare with that boom the tha ba glare boom that boom glare with that.

Explanation:

What is the shortest time that a jet pilot starting from rest can take to reach Mach-3.60 (3.60 times the speed of sound) without graying out? (Use 331 m/s for the speed of sound in cold air.)

Answers

Answer:

30.4 s

Explanation:

A pilot , with plane accelerated at 4 g starts greying out . In the problem , the acceleration of jet is 4 g

a = 4 x 9.8 = 39.2 m /s²

initial velocity u = 0

Final velocity = 3.60 times speed of sound

= 3.6 x 331 = 1191.6 m /s

v = u + at

Putting the values

1191.6 = 0 + 39.2 t

t = 30.4 s .

You are working in a shoe test laboratory measuring the coefficients of friction for running shoes on a variety of surfaces. The shoes are pushed against the surface with a downward force of 400 N, and a sample of the surface material is then pulled out from under the shoe by a machine. The machine pulls with a force of 300 N before the material begins to slide. When the material is sliding, the machine only has to pull with a force of 200 N to keep the material moving. What is the coefficient of static friction between the shoe and the material

Answers

Answer:

0.75

Explanation:

Since the static frictional force is the maximum force applied just before sliding, our frictional force, F is 300 N.

Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).

So, μ = F/N

= 300 N/400 N

= 3/4

= 0.75

So, the  coefficient of static friction μ = 0.75

The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 520 lines/mm , and the light is observed on a screen 1.4 m behind the grating.

Required:
What is the distance between the first-order red and blue fringes?

Answers

Answer:

0.143 m

Explanation:

Since

d = 1/N = 1/520 = 1.92 * 10^-3 mm

For red light;

θ = sin^-1  (1 * λred/d) =  sin^-1  (1 * 656 * 10^-9/1.92 * 10^-6) = 19.98°

L = 1.4 * (tan 19.98) = 0.509 m

For blue light;

θ = sin^-1  (1 * λblue/d) =  sin^-1  (1 * 486 * 10^-9/1.92 * 10^-6) = 14.66°

L = 1.4 * (tan 14.66°) = 0.366 m

Distance between the first-order red and blue fringes= 0.509 m - 0.366 m = 0.143 m

A locomotive is accelerating at 1.60 m/s2. It passes through a 20.0-m-wide crossing in a time of 2.40 s. After the locomotive leaves the crossing, how much time is required until its speed reaches 31.6 m/s

Answers

Answer:

The time to reach 31.6 m/s is 14.54 s.

Explanation:

Given;

acceleration of the locomotive, a = 1.60 m/s²

distance traveled by the locomotive, s = 20 m

time of motion, t = 2.4 s

final velocity of the locomotive, v = 31.6 m/s

The speed of the of the locomotive as it leaves the 20 m wide crossing;

[tex]v = \frac{d}{t} \\\\v = \frac{20}{2.4} \\\\v = 8.333 \ m/s[/tex]

The time to reach 31.6 m/s is calculated from the equation below;

v = u + at

31.6 = 8.333 + 1.6t

1.6t = 31.6 - 8.333

1.6t = 23.267

t =  23.267 / 1.6

t = 14.54 s

Therefore, the time to reach 31.6 m/s is 14.54 s.

Four point charges each having charge Q are located at the corners of a square having sides of length a. Find symbolic expressions for the work required to bring a fifth charge q from infinity to the center of the square.

Answers

Answer:

Explanation:

We shall first calculate electric potential at the centre of square due to four charges of Q .

potential = k Q / r , k = 9 x 10⁹ , Q is charge which is placed at distance r from the point where potential is measured .

Distance of corner of square to centre = √ 2 a / 2 = a / √2 = .707 a

potential due to one charge

= k x Q / .707 a

= 9 x 10⁹ x Q / .707a

= 12.73 x 10⁹ Q/a

Potential due to 4  charges

= 4 x 12.73 x 10⁹ Q/a

= 50.92 x 10⁹ Q/a .

Potential at infinity = 0

Work done in carrying charge q from infinity to centre of square

= potential difference x charge

= ( 50.92 x 10⁹ Q/a - 0 ) x q

= 50.92 x 10⁹ x Qq / a  J .

if you make a sound by tapping on a glass of water what is the order of vibration
A. glass - water - air
C. air - water - glass
B. water - glass - air

Answers

Answer:

A.) glass water air

Explanation:

hope this helps :) have a great day!!

Answer:

A.) glass - water - air

good luck, i hope this helps :)

Which of the following statements is true? A. Both warming up and cooling down or important. B. It is more important warm up then it is to cool down. C. Is more important to cool down then is to warm up. D. Both warming up and cooling down are not important

Answers

Answer:

No. A is correct because both warming up and cooling down are important

Both warming up and cooling down are important.

This is based on aerobics and human body balance regulation as regards exercising.

Warming up and cooling down in exercising are just based on the level of intensity at which the exercise is carried out.

Now, warming up when exercising involves activities like jogging. Warming up is a very vital part of exercising as it helps to get a person's cardiovascular system ready for the subsequent exercises and physical activities to be engaged. This will help in making sure there is enough blood flowinh to your muscles as well as  increasing your body temperature.

In another sense, cooling down is also very vital in activity because after the blood pressure and heart rates have been raised after exercising, they will need to be restored to their normal levels at which they were before commencement of the exercise. It also helps to regulate the blood flow.

Thus, Both warming up and cooling down are important.

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How much horizontal force FFF must a sprinter of mass 48 kgkg exert on the starting blocks to produce this acceleration

Answers

Answer:

720N

Explanation:

According to Newtons second law;

Force = Mass* Acceleration

Let the acceleration given be 15m/s²

Mass = 48kg

Substitute

Force = 48 * 15\

Force = 720N

Hence the required horizontal force is 720N

Note that the value of the acceleration was assumed

A car of mass 150 kg is riding down at constant velocity. What is the normal force?

The formula is ( Fg=mg ; Fnet =ma ; Fx = Fcos theta ; Fy = Fsin theta)





Answer and I will give you brainiliest

Answers

Answer:

1500N

Explanation:

Normal force = mg - F sin theta

constant velocity means acceleration = 0

F= ma = 150× 0 = 0N

thus;

normal force = mg = 150 × 10 = 1500N

Answer:

864N

Explanation:

A 1 800-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.60 m before coming into contact with the top of the beam, and it drives the beam 13.6 cm farther into the ground before coming to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.

Answers

Answer:

F = 614913.88 N

Explanation:

We are given;

Mass of pile driver; m = 1800 kg

Height of fall of pole driver; h = 4.6 m

Depth driven into beam; d = 13.6 cm = 0.136 m

Now, from energy equations and applying to this question, we can write that;

Workdone = Change in potential energy

Formula for workdone is; W = F × d

While the average potential energy here is; W = mg(h + d)

Thus;

Fd = mg(h + d)

Where F is the average force exerted by the beam on the pile driver while in bringing it to rest.

Making F the subject, we have;

F = mg(h + d)/d

F = 1800 × 9.81 × (4.6 + 0.136)/0.136

F = 614913.88 N

Earth has seasons because _____.

it rotates on its axis as it moves around the sun

the temperature of the sun changes

its axis is tilted

the distance between Earth and the sun changes

Answers

Answer:

c, its axis is tilted

maybe

As it works its way around the sun, its tilted axis exposes different parts of earth.

C would be it because the roation of Earth on its axis doesn't have anything to do with the exposer of the revolution on its axis

It’s c .its axis is tilted

4.The equation for Kinetic Energy is 1/2 x m x v2. Calculate the KE when the mass is 5 kg and the velocity is 4 m/s.

a.2 joules
b.10 joules
c.20 joules
d.40 joules​

Answers

Explanation:

KE = ½mv²

= ½ × 5 × 4²

= ½ × 5 × 16

= ½ × 80

= 40 J

A 12 V battery is connected across a device with variable resistance. As the resistance of the device increases, determine whether the following quantities increase, decrease, or remain unchanged. HINT (a) The current through the device. increases decreases remains unchanged (b) The voltage across the device. increases decreases remains unchanged (c) The power consumed by the device. increases decreases remains unchanged

Answers

Answer:

a) DECREASE, b) INCREASE, c)  power remains constant

Explanation:

The resistance and the battery are connected in series.

a) How the current changes when the resistance changes

               V = I R

               I = V / R

               I = 12 / R

if the resinification increases, the current must DECREASE

b) when you check the expression

                V= I R

, if R increases the voltage of the INCREASE

c) the power in an electric circuit is

               P = I V

               

Let's analyze this expression, the voltage increases linearly with the increase in resistance and the current decreases linearly with the increase in r, for which the two effects are compensated and the dissipated power remains constant

A roller coaster moving along its track rolls into a circular loop of radius r. In the loop, it is only affected by its initial velocity, gravity, and the shape of the track. Let v denote the instantaneous speed and a denote the magnitude of the instantaneous acceleration of the roller coaster in the loop. Which of the following is true in the loop?
a. The roller coaster is not in uniform circular motion, but we still have a=v^2/r everywhere on the loop
b. The roller coaster is not in uniform circular motion, but the tangential acceleration is so small that we can approximate a by v^2/r everywhere on the loop
c. The roller coaster is in uniform circular motion
d. The roller coaster is not in uniform circular motion, and a=v^2/r is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes

Answers

Answer:

c. The roller coaster is in uniform circular motion

Explanation:

Since the loop is circular with radius r, and its instantaneous speed, v is always constant, and also, its centripetal acceleration, a' = v²/r.

Since the angular speed, ω = v/r does not change, the magnitude of its  tangential acceleration is zero although there is a change in its direction because the direction of its initial velocity changes. That is a" = rα and α = Δω/Δt since Δω = 0, α = 0 and a" = r(0) = 0

So, there is no tangential acceleration. Since there is no tangential acceleration, our instantaneous acceleration which is the vector sum of our centripetal acceleration and tangential acceleration is a = √(a'² + a"²) =  √(a'² + 0²) = √a'² = a' = v²/r

So, a is always v²/r.

Since the instantaneous acceleration is always (a = v²/r) constant, the motion is uniform. So, it is uniform circular motion.

The roller coaster is not in uniform circular motion, and [tex]a = \frac{v^2}{r}[/tex] is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes.

The given parameters;

radius of the circular path, = r instantaneous speed = v instantaneous acceleration = a

The motion tension on the loop at the lowest point in the circular motion is given as;

[tex]T = mg + \frac{mv^2}{r}[/tex]

The motion tension on the loop at the highest point in the circular motion is given as;

[tex]T = \frac{mv^2}{r} - mg[/tex]

This shows that circular motion is affected by;

acceleration due to gravity, gradius of the circular path, rspeed of the motion, vmass of the object, m

Thus, we can that the roller coaster is not in uniform circular motion, and [tex]a = \frac{v^2}{r}[/tex] is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes.

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An aircraft flies 800km due East and then 600km due north. Determine the magnitude of it's displacement. please it's urgent​

Answers

Answer:

Magnitude of it's displacement = 1,000 km

Explanation:

Given:

Distance towards east = 800 km

Distance towards north = 600 km

Find:

Magnitude of it's displacement

Computation:

Magnitude of it's displacement = √800² + 600²

Magnitude of it's displacement = √640000 + 360000

Magnitude of it's displacement = √10000000

Magnitude of it's displacement = 1,000 km

before:
2m
What is the change in momentum if the v before = 1m/s and the velocity after = 1m/s?
(Change in momentum - momentum after-
after:
2m
momentum before) p - mv

Answers

Answer:

22

Explanation:

before:

2m

What is the change in momentum if the v before = 1m/s and the velocity after = 1m/s?

(Change in momentum - momentum after-

after:

2m

momentum before) p - mv

What must the charge (sign and magnitude) of a particle of mass 1.43 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 700 N/C ? Use 9.80 m/s2 for the magnitude of the free-fall acceleration.

Answers

Answer:

the sign and magnitude of the charge is - 2 x 10⁻⁵ C.

Explanation:

Given;

mass of the particle, m = 1.43 g = 0.00143 kg

electric field experienced by the particle, E = 700 N/C

The force experienced by the particle is calculated as;

F = mg = EQ

Where;

Q is the magnitude of the charge

[tex]Q = \frac{mg}{E} \\\\Q = \frac{0.00143 \times 9.8}{700} \\\\Q = 2\times 10^{-5} \ C[/tex]

The force must be upward in opposite direction to the electric field. Since the force and the electric field are in opposite direction, the charge must be negative.

Therefore, the sign and magnitude of the charge is - 2 x 10⁻⁵ C.

The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed.

The charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.

What is electric charge?

The electric force experienced by the body when placed it into the electromagnetic field is called electric charge.

Given information-

The mass of the particle is 1.43 g or 0.00143 kg.

The magnitude of the downward-directed electric field is 700 N/C.

The magnitude of the free-fall acceleration is 9.80 meter per second squared.

The electric field is defined as the electric force per unit charge. It can be given as,

[tex]E=\dfrac{F}{q}[/tex]

Rewrite the equation for the charge,

[tex]q=\dfrac{F}{E}[/tex]

Force experienced by the particle is equal to the product of mass and free fall acceleration (gravity). Thus,

[tex]q=\dfrac{0.00143\times9.8}{700}\\q=2\times10^{-5}[/tex]

Thus the magnitude of the charge is [tex]2\times10^{-5}[/tex] C.

The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed. For the opposite direction the sign of the charge should be negative.

Thus the charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.

Learn more about the electric charge here;

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