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podryga [215]
2 years ago
5

An aircraft travels a distance of 50km in a straight line

Physics
1 answer:
Juliette [100K]2 years ago
6 0

Answer:

200 m/s

Explanation:

v = distance / time = 50km/250s = 50000m/250s = 200 m/s

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If the action force is the swimmer pushing water in the leftward direction, what is the reaction force?
Sphinxa [80]
  • According to Newton's Third Law of Motion, to every action, there is an equal and opposite reaction; action and reaction act on different bodies.
  • Here, the action force is in the leftward direction, so the reaction will be in the opposite direction.
  • If the action force is the swimmer pushing water in the leftward direction, then the reaction force is in the rightward direction.
  • And the reaction force will be given by the water on the swimmer.

<u>Answer</u><u>:</u>

<u>The </u><u>reaction </u><u>force </u><u>is </u><u>the </u><u>water </u><u>pushing </u><u>the </u><u>swimmer </u><u>in </u><u>the </u><u>rightward </u><u>direction</u><u>.</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

3 0
2 years ago
A car has a mass of 1600 kg. It is stuck in the snow and is being pulled out by a cable that applies a force of 7560 N due north
sergiy2304 [10]

Answer:

Acceleration of the car will be a=0.1375m/sec^2

Explanation:

We have given mass of the ball m = 1600 kg

Force in north direction F= 7560 N

Resistance force which opposes the movement of car F_R=7340N

So net force on the car F_{net}=F-F_R=7560-7340=220N

According to second law of motion we know that F=ma

So 220=1600\times a

a=0.1375m/sec^2

7 0
3 years ago
Imagine a system where a block rests on an inclined plane. The block is then given an initial push so that it starts sliding dow
Helen [10]

Answer:

statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.

Explanation:

The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.

The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction

here, the downward direction signifies the downward motion parallel to the inclined plane.

Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.

Hence, for the block to stop sliding the the above statement should be true.

6 0
3 years ago
Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of
EleoNora [17]

Answer:

tex]\lambda_{Be}[/tex] = 22.78 nm

Explanation:

Bohr's model for the hydrogen atom has been used by other atoms with a single electric charge by changing the number of charges by the charge of the new atom (atomic number)

      E_{n}= k e² / 2a₀ (1 /n²)

      ao = h'² / k m e²               h' = h/2πi

For another atom with a single electron in the last layer

      a₀ ’= h’² / k m (Ze)²  

      a₀ ’= a₀ / Z²

Therefore, when replacing in the equation

      E_{n} = - Z²  Eo/n²

     E₀ = 13,606 eV

The transition occurs when the electron stops from one level to another

         E_{n} -  E_{m} = Z² E₀ (1 / n² - 1 / m²) = Z² ΔE

Let's relate this expression to the wavelength

       c = λ f

      E = h f

      E = h c /λ

      h c / λ = Z² ΔE

     λ = 1 / Z² (hc / ΔE)

     λ = 1 / Z² λ_hydrogen

Let's apply this last equation to our case

Lithium Z = 3

     E_{n} = - 9 Eo / n²

     

      40.5 10-9 = 1/9 λ_hydrogen

Beryllium Z = 4

      λ = 1/16 λ_hydrogen

Let's write our two equations is and solve

     40.5 10-9 = 1/9 λ_hydrogen

    tex]\lambda_{Be}[/tex] = 1/ 16 λ_hydrogen

      40.5 10⁻⁹ = 1/9 (16 \lambda_{Be} )

    tex]\lambda_{Be}[/tex] = 40.5 9/16

  tex]\lambda_{Be}[/tex] = 22.78 nm

6 0
3 years ago
Ron fills a beaker with glycerin (n = 1.473) to a depth of 5.0 cm. if he looks straight down through the glycerin surface, he wi
Tju [1.3M]

By law of refraction we know that image position and object positions are related to each other by following relation

\frac{\mu_1}{h_o} = \frac{\mu_2}{h_i}

here we know that

\mu_1 = 1.473

h_o = 5 cm

\mu_2 = 1

now by above formula

\frac{1.473}{5} = \frac{1}{h_i}

h_i = 3.39 cm

so apparent depth of the bottom is seen by the observer as h = 3.39 cm

7 0
3 years ago
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