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adell [148]
3 years ago
12

Which of the following is created when the ice of a comet starts to melt into gases?

Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
6 0
The correct answer would be the last option. It is a plasma tail that is created when the ice of a comet start to melt into gases. As the comet is near to the sun, it melts containing dust, rock and ice that are in the gas phase. They are being ionized by the photons from the sun.
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Allen and Jason are chucking a speaker around. On one particular throw, Allen throws the speaker, which is playing a pure tone o
Sholpan [36]

Answer:

Explanation:

We shall apply Doppler's effect of sound .

speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at  6 m/s .

apparent frequency = f_o\times\frac{V+v_o}{ V-v_s}

V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .

Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f

apparent frequency =  f\times \frac{340+6}{340-10}

= f\times \frac{346}{330}

So m = 346 , n = 330 .

8 0
3 years ago
What does the zigzag line in the circuit diagram represent? A. a source of electrical energy B. an electrical conductor C. an el
lora16 [44]
C. an electrical resistor 
8 0
3 years ago
Read 2 more answers
A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during
Katarina [22]

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:

  • From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.
  • From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.

We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:

  • The speed increases from 0 m/s to 1341 m/s.
  • The time elpased is 2.0 min.
  • 1 min = 60 s.

The acceleration of the ship during the first 2.0 minutes is:

a = \frac{\Delta v }{t} ) \frac{(1341m/s-0m/s)}{2.0min} \times \frac{1min}{60s}  = 11 m/s^{2}

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

Learn more: brainly.com/question/16274121

3 0
2 years ago
A motorboat traveling on a straight course slows
never [62]

Answer:

2.572 m/s²

Explanation:

Convert the given initial velocity and final velocity rates to m/s:

  • 65 km/h → 18.0556 m/s
  • 35 km/h → 9.72222 m/s

The motorboat's displacement is 45 m during this time.

We are trying to find the acceleration of the boat.

We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.

  • v² = v₀² + 2aΔx

Substitute the known values into the equation.

  • (9.72222)² = (18.0556)² + 2a(45)
  • 94.52156173 = 326.0046914 + 90a
  • -231.4831296 = 90a
  • a = -2.572

The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².

3 0
3 years ago
A trooper is moving due south along the freeway at a speed of 30.1 m/s when a red car passes the trooper. The red car moves with
romanna [79]

Answer:

The correct answer to the following question will be "41.87 m".

Explanation:

The given values are:

The speed of trooper = 30.1 \ m/s

The velocity of red car = 45.4 \ m/s

Now,

A red car goes as far as possible until the speed or velocity of the troops is the same as that of of the red car at

t=\frac{45-30}{2.7}                                                              (∵ time=\frac{distance}{time})

t=5.56 \ sec

then,

The distance covered by trooper,

t1=30\times 5.56+\frac{1}{2}\times 2.7\times (5.56)^2

   =208.33 \ m

The distance covered by red car,

= 45\times 5.56

= 250.2 \ m

Maximum distance = 250.2-208.33

                                = 41.87 \ m                                                        

6 0
3 years ago
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