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Alex777 [14]
2 years ago
8

A 7000-kg plane is launched from an aircraft carrier in 2.0 seconds

Physics
1 answer:
ExtremeBDS [4]2 years ago
5 0

The acceleration and velocity of the plane is 78.57 m/s² and 157.14 m/s respectively

To calculate the acceleration of the plane, we use the formula below.

<h3>Formula:</h3>
  • a = F/m..................... Equation 1

Where:

  • a = Acceleration of the plane
  • F = Force applied to the plane
  • m = mass of the plane.

From the question,

Given:

  • F = 550000 N
  • m = 7000 kg

Substitute these values into equation 1

  • a = 550000/7000
  • a = 78.57 m/s²

To calculate the velocity, we use the formula below.

  • v = u+at............. Equation 2

Where:

  • v = Final velocity
  • u = initial velocity
  • a = acceleration
  • t = time.

From the question,

Given:

  • u = 0 m/s
  • a = 78.57 m/s
  • t = 2.0 seconds

Substitute these values into equation 2

  • v = 0+2(78.57)
  • v = 157.14 m/s

Hence, The acceleration and velocity of the plane is 78.57 m/s² and 157.14 m/s respectively.
Learn more about acceleration here: brainly.com/question/460763

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Noble Gases do not readily form compounds because they ___ chemically stable with 8 valence electrons
azamat

Answer : Noble Gases do not readily form compounds because they are chemically stable with 8 valence electrons.

Explanation :

Noble gases are the chemical elements that are present in group 18 in the periodic table.

The elements are helium, neon, argon, krypton, xenon and radon.

They are chemically most stable except helium due to having the maximum number of 8 valence electrons can hold their outermost shell that means they have a complete octet.

They are rarely reacts with other elements to form compounds by gaining or losing electrons since they are already chemically stable.

Hence, the noble Gases do not readily form compounds because they are chemically stable with 8 valence electrons.

8 0
3 years ago
What is the passenger's apparent weight at t=1.0s?
Fittoniya [83]

Answer:

For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.

Explanation:

The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.

From the slope of graph it is clear that acceleration at t = 1 sec is given as:

Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2

Now, there are two cases:

1- Elevator moving up

2- Elevator moving down

For upward motion:

Apparent Weight =  m(g + a)

Apparent Weight = (75 kg)(9.8 + 4)m/s^2

<u>Apparent Weight = 1035 N</u>

For downward motion:

Apparent Weight =  m(g - a)

Apparent Weight = (75 kg)(9.8 - 4)m/s^2

<u>Apparent Weight = 435 N</u>

4 0
3 years ago
Read 2 more answers
A source of sound is detected in the middle of a room. Which can account for a decrease in sound intensity for an observer?
dedylja [7]

Answer:

D

Explanation:

8 0
3 years ago
Read 2 more answers
Is pressure water a acid , basic or natural substance
Yakvenalex [24]

Anyway, water acts as a base and as a acid depending on what it is reacting with. Molecules that do this are called amphiprotic.

7 0
4 years ago
A diverging lens has a focal length of magnitude 22.8 cm. (a) Locate the images for each of the following object distances. 45.6
Nostrana [21]

(a) -15.2 cm

We can solve the problem by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f = -22.8 cm is the focal length of the lens (negative because it is a diverging lens)

p = 45.6 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving the equation for q, we find the position of the image:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{45.6 cm}=-0.066 cm^{-1}

q=\frac{1}{-0.066 cm^{-1}}=-15.2 cm

and the negative sign means that the image is virtual.

(b) -11.4 cm

In this case, the distance of the object from the lens is

p = 22.8 cm

Substituting into the lens equation, we can find the new image distance, q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{22.8 cm}=-\frac{2}{22.8 cm}

q=\frac{-22.8 cm}{2}=-11.4 cm

and the negative sign means that the image is virtual.

(c) -7.6 cm

In this case, the distance of the object from the lens is

p = 11.4 cm

Substituting into the lens equation, we can find the new image distance, q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{11.4 cm}=-0.132 cm^{-1}

q=\frac{1}{0.132 cm^{-1}}=-7.6 cm

and again, the negative sign means that the image is virtual.

3 0
3 years ago
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