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4 years ago

The orbits of most comets

Physics

1 answer:

Ludmilka [50]4 years ago

4 0

Answer:

option B

Explanation:

The correct answer is option B

Comets are the frozen snowball that is made of different gases and which rotates around the sun.

Comets are generally icy but when comets come near to the sum they get warm and release gasses which can be seen from the earth.

Comet produces a visible atmosphere or coma.

examples of comets are Halley's Comet, Comet Encke.

The orbit of comets is entirely beyond the orbit of Neptune.

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If you are in a spaceship that is sitting on the surface of a planet, you feel your weight. How does this compare to the weight

nordsb [41]

Answer:

You will feel more weight if it is accelerating out of the planet.

You will feel less weight if it is accelerating towards the planet.

Explanation:

The weight that you are observing or feeling is basically due to the change in acceleration of your fall or rising up in the spaceship. When the acceleration is stationary on the surface, you experience your normal weight due to the gravitational acceleration of that planet.

When the spaceship accelerates above or out of the planet you experience acceleration more than the acceleration of gravity hence more weight.

When the spaceship accelerates towards the planet you experience acceleration less than the acceleration of gravity hence less weight.

If the spaceship is free falling at the gravitational acceleration you experience a zero weight

8 0

4 years ago

Given a force of 56N and an acceleration of 7m/s2, what is the mass?

bearhunter [10]

F = ma
You need to rearrange this equation to find mass so you would have the equation :
M = f/a
Input the numbers into the equation :
M = 56/7
M = 8

8 0

3 years ago

Read 2 more answers

2.2 kg box is being dragged to right with 31.06 N force and accelerating at 3.2 m/s^2 how strong is force of friction acting on

insens350 [35]

24. 02 N force of friction acting on the box.

<h3 /><h3>What is force?</h3>

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Force = mass. acceleration

Force = (2.2).(3.2)

Force = 7.04 N

Total force applied is 31.06 N

Frictional force = 31.06 - 7.04

Frictional force = 24. 02 N

24. 02 N force of friction acting on the box.

<h3 />

To learn more about force refer to the link:

brainly.com/question/13191643

#SPJ1

4 0

1 year ago

An open cooking pot containing 0.5 liter of water at 20°C, 1 bar sits on a stove burner. Once the burner is turned on, the water

sesenic [268]

Answer:

a. the time required for the onset of evaporation is: 196.1 seconds and b. the time required for all of the water to evaporate is: 1328.3 seconds.

Explanation:

We need to stablish that there is 3 states at this problem. At the firts one, water is compressed liquid and the conditions for this state are: P1=100KPa,T1=20°C,V1=0.5m^3. From the compressed liquid chart and using extrapolation, we can get: v1=vf1=0.0010018 (m^3/Kg) and u1=uf1=83.95(KJ/kg). Now we can find the mass of water at the state 1 as: $m=\frac{V_{1} }{v_{1} } =\frac{0.5*10^{-3} }{0.0010018}=0.5(Kg)$ Then the liquid water is heated at a rate of 0.85KW, and its volume increase, while work is done by the system at the boundary, we can assume that the pressure remains constant throughout the entire process. At the second state the water is saturated liquid and the conditions are: P2=100KPa, T2=Tsat=99.63°C, v2=vf2=0.001043(m^3/Kg) and u2=uf2=417.36(KJ/Kg). Now we can find the work as: $W=mP(v_{2} -v_{1} )=0.5*100*(0.001043-0.0010018)=0.00207(KJ)$ . (a) After that we need to do an energy balance for process 1-2 and get: U=Q-W or $m(u_{2} -u_{1} )= Q*t-W$ , solving for t we get the time required for the onset of evaporation: $t=\frac{0.5*(417.36-83.95)+0.00207}{0.85}=196.1(s)$ .(b) Then continue heat transfer to the cooking pot and results in phase change getting vapor at 99.63°C. At the final state or third state the mass is zero because all liquid was evaporated and the initial mass at this state is the same for the second state: 0.5 (Kg) and doing an energy balances results in: $(m_{3} u_{3} -m_{2} u_{2})=Q*t-W+( m_{3}-m_{2})h_{e}$ , but m3=0, now solving for t we can get the time required for all of the water to evaporate as: $t=\frac{m_{2}(h_{e}-u_{2})+W}{Q}$ . We can get from the saturated liquid chart the enthalpy he=hge=2675.5(KJ/Kg) @P=100KPa. Now we need to calculate the work related with the volume decreases as vapor exits the control volume or process 2-3 work boundary as: $W=\int\limits^3_2 {p} \, dV= p*(V_{3} -V_{2} )=-m_{2} P_{2} v_{2} =-(0.5)*100*0.001043=-0.0522(KJ)$ . Now replacing every value in the time equation we get: $t=\frac{0.5(2675.5-417.36)+(-0.0522)}{0.85}=1328.3(s)$