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2 years ago

On an aircraft carrier, a jet can be catapulted from 0 to 250 km/h in 2.05 s

Physics

2 answers:

Komok [63]2 years ago

7 0

Answer:

mass = 27750 kg

Explanation:

acceleration: (change in velocity)÷(time)

acceleration = (final velocity - initial velocity) ÷ time taken

Given that:

<u><em>Final velocity is 250 km/h and initial is 0 km/h and time is 2.05s</em></u>

change 2.05s to hrs. that is 0.0005694444 hrs

acceleration = [(250 - 0) ÷ 0.0005694444]

acceleration = 439024.4 km/h²

conversion to m/s² :

acceleration = 439024.4 m/s²

--------------------------------------------------------------------

Force = mass(kg) * acceleration(m/s²)

we know that force is 9.40×10^5 N

9.40×10^5 = mass(kg) * 33.875 (m/s²)

mass = 9.40×10^5 ÷ 33.875

mass = 27750 kg

sdas [7]2 years ago

6 0

Answer: The mass of the jet is 26,225

Explanation: Convert 250km to m =250,000

1 hour = 3600

Velocity = 69.4 m/s

a= change in v/change in t

69.4/2 = 34.7

9.10×105 / 34.7 = 26,225 KG

Yw and pls mark me as brainiest

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Imagine that while you and a passenger are in a deep-diving submersible in the North Pacific near Alaska’s Aleutian Islands, you

mamaluj [8]

Answer:

I would say its a deep ocean trench

Explanation:

This is because deep ocean trenches are found at the deepest part of the ocean and also at Pacific ocean margins or Rim where subduction usually occurs and Aleutian islands are part of the Pacific Rim

6 0

3 years ago

A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$

5 0

3 years ago

true or false? destructive interference occurs when a trough meets up with another trough given location along the medium

disa [49]

Answer:Poopy-di scoop

Scoop-diddy-whoop

Whoop-di-scoop-di-poop

Poop-di-scoopty

Scoopty-whoop

Whoopity-scoop, whoop-poop

Poop-diddy, whoop-scoop

Poop, poop

Scoop-diddy-whoop

Whoop-diddy-scoop

Whoop-diddy-scoop, poop

Explanation:

4 0

3 years ago

A mass on a horizontal surface is connected to the spring and pulled to the right along the surface stretching the spring by 25

solniwko [45]

Answer:

320 N/m

Explanation:

From Hooke's law, we deduce that

F=kx where F is applied force, k is spring constant and x is extension or compression of spring

Making k the subject of formula then

$k=\frac {F}{x}$

Conversion

1m equals to 100cm

Xm equals 25 cm

25/100=0.25 m

Substituting 80 N for F and 0.25m for x then

$k=\frac {80}{0.25}=320N/m$

Therefore, the spring constant is equal to 320 N/m

3 0

3 years ago

3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of

Alex Ar [27]

Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

$n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555$

Again using Snell's law for blue light as :

$n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283$

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

7 0

3 years ago