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salantis [7]
3 years ago
14

A jet with mass m = 5 × 104 kg jet accelerates down the runway for takeoff at 1.9 m/s2. What is the net vertical force on the ai

rplane as it levels off?
Physics
1 answer:
OLEGan [10]3 years ago
3 0
The first sentence got me all psyched up to answer the question "What
horizontal force do the engines generate in order to accelerate it ?". 

But the actual question, in the second sentence, turned out to be
a completely different one.

When the plane levels off and continues on at a constant altitude, it's
not accelerating up or down, so the net vertical force on it is zero. 
The lift generated by the wings is exactly balancing the downward
force of gravity on the airplane.
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The amount of light that enters the pupil is controlled by the:<br> retina.<br> lens.<br> inis.
nignag [31]

Answer: The amount of light that enters the pupil is controlled by the Iris

Explanation:

5 0
3 years ago
Two toy cars (m1 = 0.200 kg and m2 = 0.250 kg) are held together rear to rear with a compressed spring between them. When they a
timurjin [86]

Answer:

Acceleration of the second particle at that moment is given as

a_2 = 2.12 m/s^2

Explanation:

As we know that both cars are connected by same spring

So on this system of two cars there is no external force

So we will have

F = 0 = m_1a_1 + m_2a_2

now we have

m_1 = 0.200 kg

m_2 = 0.250 kg

a_1 = 2.65 m/s^2

now we have

0.200(2.65) + 0.250a_2 = 0

so we have

a_2 = 2.12 m/s^2

8 0
3 years ago
Help me find the acceleration
ANEK [815]

a = 3.09 m/s²

<h3>Explanation</h3>

This question doesn't tell anything about how long it took for the car to go through 105 meters. As a result, the <em>timeless </em>suvat equation is likely what you need for this question.

In the <em>timeless</em> suvat equation,

a = \dfrac{v^2 - u^2}{2\; x}

where

  • a is the acceleration of the car;
  • v is the <em>final</em> velocity of the car;
  • u is the <em>initial</em> velocity of the car; and
  • x is the displacement of the car.

Note that <em>v</em> and <em>u</em> are velocities. Make sure that you include their signs in the calculation.

In this question,

  • a is the unknown;
  • v = -10.9 \; \text{m} \cdot \text{s}^{-2};
  • u = -27.7 \; \text{m} \cdot \text{s}^{-2}; and
  • x = - 105 \; \text{m}.

Apply the <em>timeless</em> suvat equation:

a = \dfrac{v^{2} - u^{2}}{2\; x}\\\phantom{a} = \dfrac{(-10.9)^{2} - (-27.7)^{2}}{2 \times (-105)}\\\phantom{a} = 3.09 \; \text{m} \cdot \text{s}^{-2}.

The value of a is greater than zero, which is reasonable. Velocity of the car is negative, meaning that the car is moving backward. The car now moves to the back at a slower speed. Effectively it accelerates to the front. Its acceleration shall thus be positive.

7 0
3 years ago
If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
Bas_tet [7]

Answer:

  W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

              F = G m₁ M / r²

              W = ∫ F. dr

              W = G m₁ M ∫ dr / r²

we integrate

             W = G m₁ M (-1 / r)

                 

We evaluate between the limits, lower r = R_ Moon and r = ∞

           W = -G m₁ M (1 /∞ - 1 / R_moon)

            W = G m1 M / r_moon

Body weight is

             W = mg

             m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

            m = 1000/32 = 165 / g_moon

            g_moom = 165 32/1000

            .g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system  

           W_capsule = 1000 pound (1 kg / 2.20 pounds)

           W_capsule = 454 N

           W = m_capsule g

           m_capsule = W / g

           m = 454 /9.8

           m_capsule = 46,327 kg

Let's calculate

          W = 6.67 10⁻¹¹ 46,327   7.36 10²² / 1.74 10⁶

          W = 1,307 10⁶ J

7 0
3 years ago
Which of the following statements does NOT accurately describe the Berlin Airlift?
Karo-lina-s [1.5K]

The Berlin Airlift is best described as the aircraft used to delivered needed food and supplies to the city of Berlin through the air because all other routes were blocked by the Soviet Union.

<h3>What is Berlin Airlift?</h3>

The Berlin airlift was a 1940s military operation that supplied West Berlin with food and other vital goods by air after the Soviet Union blockaded the city.

Thus, the Berlin Airlift is best described as the aircraft used to delivered needed food and supplies to the city of Berlin through the air because all other routes were blocked by the Soviet Union.

Learn more about Berlin Airlift here: brainly.com/question/1104371

#SPJ1

8 0
2 years ago
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