Scientists are constantly exploring the universe, looking for new planets that support life similar to the life on
Earth. A new planet that supports life would have all of the following characteristics except -
A. a gaseous atmosphere.
B. an orbiting moon.
C. liquid water.
D. protection from radiation.

Answers

Answer 1
The answer would be “B” because humans would need water, protection from radiation so we don’t melt or burn to death lol, and a gaseous atmosphere because we would need oxygen.
Answer 2

A new planet that supports life would have all the following characteristics except an orbiting moon. Hence, option B is correct.

What is a Planet?

An enormous, spherical celestial object known as a planet is neither a star nor its remains. The nebular hypothesis, which states how an interstellar cloud falls out of a nebula to produce a young protostar encircled by a protoplanetary disk, is now the best explanation for planet formation.

By gradually accumulating material under the influence of gravity, or accretion, planets develop in this disk.

The rocky planets Mercury, Venus, Earth, and Mars, as well as the giant planets Jupiter, Saturn, Uranus, and Neptune, make up the Solar System's minimum number of eight planets. These planets all revolve around axes that are inclined relative to their respective polar axes.

To know more about Planet:

https://brainly.com/question/14581221

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Related Questions

Momentum
Project: Egg Drop
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Answers

Answer:

get egg and try to make in not crack when it falls by exerting the momentum of the fall into something other than the egg ex. make a box full of bubble wrap and put your egg in it

Explanation:

Materials For Project:Four sheets of 8 1/2 x 11 inch paper + One meter of masking tape + One resealable plastic bag (sized for an egg) + One meter of string + Three Straws + Eggs + White school glue. Draw Three Sketches: of your ideas for the egg do drop and NOT crack. Then, claim the advantages and disadvantages for each sketch. (If you don't know how to draw, try your best or notify a teacher.) FINAL SKETCH: After drawling the three sketching pick the best one for your final work. Explain why you chose that sketch.

-Email your teacher if you are still confused.

A running Marites launched the egg she
stole as she was about to be caught with
a velocity of 25 m/s in a direction making
an angle of 20° upward with the
horizontal
a) What is the maximum height reached by
the egg?
b) What is the total flight time (between
launch and touching the ground) of the
egg?
c) What is the horizontal range (maximum
* above ground) of the egg?
d) What is the magnitude of the velocity
of the egg just before it hits the ground?

Answers

Answer:

a)   y = 3.73 m,  b)  t = 1.74 s,  c)  R = 40.99 m,

d) vₓ  = 23.49  m/s,   v_y = -8.5 m / s

Explanation:

This is a projectile launching exercise, we start by breaking down the initial velocity

          sin θ = v_{oy} / v₀

          cos θ = v₀ₓ / v₀

          v_{oy} = v₀ sin θ

          v₀ₓ = v₀ cos θ

          v_{oy} = 25 sin 20 = 8.55 m / s

          v₀ₓ = 25 cos 20 = 23.49 m / s

a) when the egg reaches the maximum height its vertical speed is zero

          v_y² = v_{oy}² - 2 g y

          0 = v_[oy}² - 2g y

           y = v_{oy}² / 2g

          y = [tex]\frac{8.55^2}{2 \ 9.8 }[/tex]

          y = 3.73 m

b) flight time

          y = v_{oy} t - ½ g t²

the time of flight occurs when the body reaches the ground y = 0

          0 = (v_{oy} - ½ g t) t

         

The results are

          t₁ = 0s        this time is for using the body star

          v_{oy} - ½ g t = 0

           t = [tex]\frac{2v_{oy}^2}{g}[/tex]

           t = 2 8.55 / 9.8

           t = 1.74 s

c) the range

           R = v₀² sin 2θ / g

           R = 25² sin (2 20) / 9.8

           R = 40.99 m

d) speed at the point of arrival

horizontal speed is constant

           vₓ = v₀ₓ = 23.49  m/s

vertical speed is

           v_y = Iv_{oy} - g t

           v_y = 8.55 - 9.8  1.74

           v_y = -8.5 m / s

What is the period of a wave with a speed of 20.0 m/s and a frequency of 10.0 Hz?

Answers

im confused hold on imma send you a link to the answerExplanation:

A winch is capable of hauling a ton of bricks vertically two stories (6.35 m ) in 24.5 s .
If the winch’s motor is rated at 5.80 hp , determine its efficiency during raising the load.

Answers

Answer: 84 %

Explanation:

(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is 3.85 x 108 m. Find the time it took for his voice to reach earth via radio waves. (b) Someday a person will walk on Mars, which is 5.6 x 1010 m from earth at the point of closest approach. Determine the minimum time that will be required for that person's voice to reach earth.

Answers

Answer:

a). 1.28333 seconds

b). 186.66 seconds

Explanation:

a). Given :

Distance between the earth and the moon, d = [tex]$3.85 \times 10^8$[/tex] m

Speed of the radio waves, c = [tex]$3 \times 10^8$[/tex] m/s

Therefore the time required for the voice of Neil Armstrong to reach the earth via radio waves is given by :

[tex]$t=\frac{d}{c}$[/tex]

 [tex]$=\frac{3.85 \times 10^8}{3 \times 10^8}$[/tex]

 = 1.28333 seconds

b). Distance between Mars and the earth, d = [tex]$5.6 \times 10^{10}$[/tex] m

   Speed of the radio waves, c = [tex]$3 \times 10^8$[/tex] m/s

So, the time required for his voice to reach earth is :

[tex]$t=\frac{d}{c}$[/tex]

 [tex]$=\frac{5.6 \times 10^{10}}{3 \times 10^8}$[/tex]

 = 186.66 seconds

10 POINTS! SPACE QUESTION!!

Answers

Answer; they are larger and made of rocky material

what is friction and the types withe examples.​

Answers

Explanation:

The answer is In the picture. Thanks.

You are locked inside the train car and want to get it moving to draw attention to your plight. There is effectively no friction between the axle and the car, and the train is on horizontal tracks. To try and get the car moving with respect to the ground, you run and slam with all your force against the wall at the front. What happens with the car after you slammed against the wall of the car

Answers

Answer:

the car movves briefly as you ran, however, it stops again after you ran in to the wall

Explanation:

Our net total linear momentum was zero at the time both the train and the boy was at rest. As there is no pressure, the train and boy system's linear momentum can be conserved provided no external forces are working on it that might shift its momentum.

If the boy runs inside of the train with a velocity V1 in the forward direction, the train would have a velocity V2 in the reverse direction to V1 to conserve the system's momentum, resulting in Final momentum.

i.e

[tex]m \times V_1 + M \times V_2 = 0[/tex]  --- (1)

here;

m = mass of the boy

M = mass of the train

Thus;

[tex]m \times V_1 =- M \times V_2[/tex] --- (2)

As the boy crashes into the train's wall, a pair of equal and opposing force F intervene on both the train and the boy. This force F causes the boy traveling with momentum m× V1 to come to a halt; its momentum remains zero. As the train moves with about the same momentum as the boy as seen in (2) and is subjected to the same force F, its momentum will be diminished to zero, and it will come to a halt.

Light with a wavelength of 700 nm (7×〖10〗^(-7) m) is incident upon a double slit with a separation of 0.30 mm (3 x 10-4 m). A screen is located 1.5 m from the double slit. At what distance from the screen will the first bright fringe beyond the center fringe appear?

Answers

Answer:

[tex]0.0035\ \text{m}[/tex]

Explanation:

y = Distance from the center point

d = Separation between slits = 0.3 mm

D = Distance between slit and screen = 1.5 m

[tex]\lambda[/tex] = Wavelength = 700 nm

m = Order = 1

We have the relation

[tex]d\dfrac{y}{D}=m\lambda\\\Rightarrow y=\dfrac{Dm\lambda}{d}\\\Rightarrow y=\dfrac{1.5\times 1\times 700\times 10^{-9}}{0.3\times 10^{-3}}\\\Rightarrow y=0.0035\ \text{m}[/tex]

The distance from the screen at which the first bright fringe beyond the center fringe appear is [tex]0.0035\ \text{m}[/tex].

The half-life for a 400-gram sample of radioactive element X is 3 days. How much of element X remains after 15 days have passed?

A.
12.5 g

B.
25 g

C.
50 g

D.
100 g

Answers

The answer to your question is B!!

Which one of the following statements concerning the magnetic field inside (far from the surface) a long, current-carrying solenoid is true?
1) The magnetic field is zero.
2) The magnetic field is independent of the number of windings.
3) The magnetic field varies as 1/r as measured from the solenoid axis.
4) The magnetic field is independent of the current in the solenoid.
5) The magnetic field is non-zero and nearly uniform.

Answers

Correct answe: 5 (the magnetic field is non-zero and nearly uniform)

Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human is taken to be 90.0 years, a spaceship would need to achieve some minimum speed vmin to deliver a living human being to this galaxy. How close to the speed of light would this minimum speed be? Express your answer as the difference between vmin and the speed of light c.

Answers

Answer:

[tex]0.0018833\ \text{m/s}[/tex]

Explanation:

[tex]d[/tex] = Distance of Andromeda Galaxy from Earth = [tex]2.54\times 10^7\ \text{ly}[/tex]

[tex]t[/tex] = Time taken = [tex]90\ \text{years}[/tex]

[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]

We have the relation

[tex]t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}[/tex]

[tex]c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}[/tex]

The required answer is [tex]0.0018833\ \text{m/s}[/tex].

Suppose a 65 kg person stands at the edge of a 6.5 m diameter merry-go-round turntable that is mounted on frictionless bearings and has a moment of inertia of 1850 kgm2. The turntable is at rest initially, but when the person begins running at a speed of 3.8 m/s (with respect to the turntable) around its edge, the turntable begins to rotate in the opposite direction. Calculate the angular velocity of the turntable. (Hint: use what you know about relative velocity to help solve the problem

Answers

Answer:

[tex]0.3165\ \text{rad/s}[/tex]

Explanation:

m = Mass of person = 65 kg

d = Diameter of round table = 6.5 m

r = Radius = [tex]\dfrac{d}{2}=3.25\ \text{m}[/tex]

v = Velocity of person running = 3.8 m/s

[tex]I_t[/tex] = Moment of inertia of turntable = [tex]1850\ \text{kg m}^2[/tex]

Moment of inertia of the system is

[tex]I=I_t+mr^2\\\Rightarrow I=1850+65\times 3.25^2\\\Rightarrow I=2536.5625\ \text{kg m}^2[/tex]

As the angular momentum of the system is conserved we have

[tex]L_i=L_f\\\Rightarrow mvr=I\omega_f\\\Rightarrow \omega_f=\dfrac{mvr}{I}\\\Rightarrow \omega_f=\dfrac{65\times 3.8\times 3.25}{2536.5625}\\\Rightarrow \omega_f=0.3165\ \text{rad/s}[/tex]

The angular velocity of the turntable is [tex]0.3165\ \text{rad/s}[/tex].

Philosophy: The Big Picture Unit 8

How does pragmatism differ from the utilitarianism of the previous era?

A. Utilitarianism did not consider rights and pragmatism did.
B. Pragmatism is concerned with function, utilitarianism with happiness.
C. Utilitarianism was created in England and pragmatism came from Germany.
D. Pragmatism applies to everyone, but utilitarianism is concerned with the upper class.

Answers


D. Pragmatism applies to everyone, but utilitarianism is concerned with the upper class.

1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.
How far will it go before it is moving at 20 m/s?
A) 83
B) 43
C) 39
D) 94
E) 20

Answers

Answer:

It will go up to 93.75 m before it is moving at 20 m/s

Explanation:

As we know that

[tex]v^2 - u^2 = 2aS[/tex]

here v is the final speed i.e 20 m/s

u is the initial speed i.e 5 m/s

a is the acceleration due to gravity i.e 2 m/s^2

Substituting the given values in above equation, we get -

[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters

A 4 kg box is at rest on a table. The static friction coefficient u, between the box and table is 0.30, and
the kinetic friction coefficient Hi is 0.10. Then, a 10 N horizontal force is applied to the box.

Answers

Answer:

The box will not move from its position.

Explanation:

First, we will calculate the static frictional force that is stopping the box to move from its position:

[tex]f = \mu R = \mu W=\mu mg[/tex]

where,

f = static frictional force = ?

μ = coefficient of static friction = 0.3

m = mass of box = 4 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]f = (0.3)(4\ kg)(9.81\ m/s^2)\\f=11.77\ N[/tex]

Since the frictional force (11.77 N) is greater than the applied force (10 N).

Therefore, the box will not move from its position.

1.) Calculate the mass of a solid gold rectangular bar that has dimensions lwh = 4.30 cm ✕ 14.0 cm ✕ 27.0 cm. (The density of gold is 19.3 ✕ 103 kg/m3.)
kg


2.)A brass ring of diameter 10.00 cm at 17.3°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 17.3°C. Assume the average coefficients of linear expansion are constant.

(a) To what temperature must the combination be cooled to separate the two metals?


(b) What if the aluminum rod were 10.06 cm in diameter?

Answers

Answer:

1) m = 0.3137 kg

2a)T_f = -181.7°C

2b) T_f = -1176.97°C

Explanation:

1) We are given;

Length; l = 4.30 cm = 0.043 m

Width; w = 14.0 cm = 0.014 m

height; h = 27.0 cm = 0.027 m

density of gold; ρ = 19.3 × 10³ kg/m³

Formula for the density is known as;

ρ = mass/volume

Thus;

m =ρV

m = 19.3 × 10³ × (lwh)

m = 19.3 × 10³ × (0.043 × 0.014 × 0.027)

m = 0.3137 kg

2a) We are given;

Diameter of brass; L_br = 10 cm

Diameter of aluminum; L_al = 10.01 cm

Now, to some for change in temperature we will use the formula;

L_f = L_i + αL_i(Δt)

Where α is coefficient of expansion.

Now, for the ring to be removed from the rod, the final diameter of the brass has to be same as the aluminium.

Thus;

L_f(brass) = L_f(aluminium)

From table attached, α_brass ≈ 19 × 10^(-6) /°C

Also, α_aluminium ≈ 24 × 10^(-6) /°C

Thus;

L_f(brass) = 10 + (19 × 10^(-6) × 10 × (Δt))

Similarly,

L_f(aluminium) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))

Since L_f(brass) = L_f(aluminium), then;

10 + (19 × 10^(-6) × 10 × (Δt)) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))

Rearranging, we have;

10.01 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))

0.01 = Δt(-50.24 × 10^(-6))

Δt = 0.01/(-50.24 × 10^(-6))

Δt ≈ -199°C

Thus, temperature at which the combination must be cooled to separate the two metals is;

T_f = T_i + Δt

T_f = 17.3 + (-199)

T_f = -181.7°C

2b) Diameter of aluminum is now;

L_al = 10.06 cm

Thus;

10.06 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))

0.06 = Δt(-50.24 × 10^(-6))

Δt = 0.06/(-50.24 × 10^(-6))

Δt = -1194.27°C

T_f = 17.3 + (-1194.27)

T_f = -1176.97°C

A coil of resistance 100ohm is placed in a magnetic field of 1mWb the coil has 100 turns and a galvanometer of 400ohm resistance is connected in series with it. find the average emf and the current if the coil is moved in one tenth of a second from the given field to a field of 2.0mWb​

Answers

Answer:

C

Explanation:

C

physics grade9 teacher guide​

Answers

Answer:

huh

Explanation:

Select the correct answer.
Each square dance begins with what?
A. Handshake
B. Dosado
C. Bow or curtsy
D. Promenade

Answers

Answer:

C

Explanation:

The men bow to the women and the women curtsy to the men.

(took on test and got it right)

HURRY IM TIMED

How can you make people feel inspired?

By leading them on an emotional journey through various states to inspiration
By talking about something that interests you
By proving yourself to be a trustworthy speaker
By making them laugh and feel comfortable

Answers

Answer:

By talking about something that interesto you’

sorry if wrong

Explanation:

The classical theory of electromagnetism predicted that the energy of the electrons ejected should have been proportional to the intensity of the light.

a. True
b. False

Answers

Answer:

False

Explanation:

No, it is not true that energy of the electrons ejected should have been proportional to the intensity of the light. Perhaps the kinetic energy of the ejected electrons is independent of the intensity of the incident radiation. This is the very fact that classical theory of electromagnetism fails to explain in photoelectric effect. The kinetic energy of the electrons remains constant even if the amplitude of the incident light is increased.

A toy car rolls down a ramp. Which force causes the car to move

Answers

Answer:

Gravity

Explanation:

Gravity pulls things down to earth and it is a force

A candle is placed 50 cm from a diverging lens with a focal length of 28. What is the image distance in cm.

Answers

Answer:

v = -17.94 cm

Explanation:

Given that,

The candle is placed at a distance of 50 cm, u = -50 cm

The focal length of the diverging lens, f = -28 cm (negative in case of a concave lens)

We need to find the image distance. We know that the lens formula is as follows:

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(-28)}+\dfrac{1}{(-50)}\\\\v=-17.94\ cm[/tex]

So, the image distance is equal to 17.94 cm.

Identical satellites X and Y of mass m are in circular orbits around a planet of mass M. The radius of the planet is R. Satellite X has an orbital radius of 3R, and satellite Y has an orbital radius of 4R. The kinetic energy of satellite X is Kx . Satellite X is moved to the same orbit as satellite Y by a force doing work on the satellite. In terms of Kx , the work done on satellite X by the force is

Answers

Answer:

The work down on satellite X by the force in terms of Kx is [tex]\dfrac{-K_x}{4}[/tex].

Explanation:

The work done is given as in terms of

[tex]W=\Delta TE[/tex]

Where ΔTE is the change in total energy.

This is given as

[tex]W=\Delta TE\\W=TE_y-TE_x\\W=\dfrac{-GMm}{2(4R)}-\dfrac{-GMm}{2(3R)}\\W=\dfrac{-GMm}{8R}+\dfrac{GMm}{6R}\\W=\dfrac{-6GMm+8GMm}{48R}\\W=\dfrac{2GMm}{48R}\\W=\dfrac{GMm}{24R}[/tex]

Rearranging it in  terms of K_x gives

[tex]W=\dfrac{GMm}{24R}\\W=\dfrac{GMm}{-4\times -6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{2(3R)}\\W=\dfrac{1}{-4}K_x\\W=\dfrac{-K_x}{4}[/tex]

Describing a Wave
What does a wave carry?

Answers

Answer:

Waves carry energy from one place to another.

Explanation:

Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals.

It carries energy from one area to another

How many times a minute does a boat bob up and down on ocean waves that have a wavelength of 35.6 m and a propagation speed of 4.68 m/s? (the answer may not be a whole number)

Answers

Answer:

It will bob 7.887640449 times a minute

Explanation:

I hope this is correct!!

Think of a hydropower dam . How is electrical energy produced from potential and kinetic energy ?

Answers

hydroelectric dam converts the potential energy stored in a water reservoir behind a dam to mechanical energy—mechanical energy is also known as kinetic energy. ... The generator converts the turbine's mechanical energy into electricity.

Hope this helps!

Answer:

Potential energy and kinetic energy are constituents of mechanical energy.

When a turbine is switched on, it rotates with mechanical energy.

Since a motor runs the turbine, it converts this mechanical energy to electrical energy.

9. Mr. Smith went skiing in Maine last weekend. He traveled 523 kilometers to Sugarloaf from
Leominster. His average speed was 109 km/hr. How long did it take Mr. Smith to hit the slopes?

Answers

Answer:

Time taken by Mr. smith = 4.80 hour (Approx.)

Explanation:

Given:

Distance travel by Mr. smith = 523 kilometer

Average speed of Mr. smith = 109 km/hr

Find;

Time taken by Mr. smith

Computation:

Time taken = Distance cover / Speed

Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.

smith

Time taken by Mr. smith = 523 / 109

Time taken by Mr. smith = 4.798 hr

Time taken by Mr. smith = 4.80 hour (Approx.)

Which investigation BEST measures the gravitational force on a
toy car?
A. rolling the car down a steep ramp and measuring time
B. using a spring scale and measuring the weight of the car
C. pushing the car and measuring how far it travels before it stops
D. throwing the car in the air and measuring how far it goes before coming
down

Answers

Answer:

B : using a spring scale and measuring the weight of the car

Explanation:

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