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lara31 [8.8K]
3 years ago
13

What is the net force acting on the airplane?

Physics
1 answer:
Whitepunk [10]3 years ago
3 0

\huge{ \mathrm{  \underline{ Answer }\:  \:  ✓ }}

Total force acting on right side = 800 N

Total force acting on left side :

  • 60 N + 40 N

  • 100 N

Now, equivalent force acting on the plane is :

  • greater force - minor force

  • 800 N - 100 N
  • 700 Newtons

And the direction of equivalent force will be the direction of greater force, that is right direction.

Hence, Correct option is :

  • 700 N right -->

_____________________________

\mathrm{ \:TeeNForeveR\:}

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Jake drops a book out of a window from a height of 10 meters. At what velocity does the book hit the ground?
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<span>Solving for velocity : </span>

velocity^{2}<span> = 2gh </span>
<span>v = </span>2gh^{ \frac{1}{2} }
<span>v = (2 x 9.81 x 10)^1/2 </span>
<span>v = 196.2 m/sec (answer)</span>
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A table-tennis ball is thrown at a stationary bowling ball. The table-tennis ball makes a one-dimensional elastic collision and
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<u>Option b. </u>A smaller magnitude of momentum and more kinetic energy.

<h3>What is a momentum?</h3>
  • In Newtonian physics, an object's linear momentum, translational momentum, or simply momentum is defined as the product of its mass and velocity.
  • It has both a magnitude and a direction, making it a vector quantity. The object's momentum, p, is defined as: p=mv if m is the object's mass and v is its velocity (also a vector quantity).
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7 0
2 years ago
If an object is an irregularly shaped solid and it is dropped into a graduated cylinder and it displaces 25 mL of water and has
zavuch27 [327]

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1250

Explanation:

7 0
3 years ago
You are performing a knee extension exercise. You hold a 20kg weight at full knee extension. The weight is 0.4m from your knee j
dmitriy555 [2]

Answer:

The moment is -78.4 N-m (clockwise).

Explanation:

Given:

Mass of the object (m) = 20 kg

Distance of the object from the knee joint (d) = 0.4 m

Weight of leg is not considered.

Acceleration due to gravity (g) = 9.8 m/s²

Now, weight of the object is equal to the product of its mass and acceleration due to gravity. So,

Weight = Mass × Acceleration due to gravity

            = mg=20\times 9.8 =196\ N

We know that, moment of a force about a point is defined as the product of force applied and the perpendicular distance between the point and the line of application of force.

Moment of the given weight about the knee joint is given as:

Moment about knee joint = Weight × Distance from knee joint to weight

Moment about knee joint = 196 × 0.4 = 78.4 Nm

Now, from the diagram below, we can observe that, the weight acts vertically down and thus the sense of rotation about the knee joint at point O is clockwise. So, moment is negative.

Therefore, the moment is -78.4 N-m (clockwise).

7 0
3 years ago
To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit
DiKsa [7]

To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

R = 350ft

a_t = 1.1ft/s^2

PART A )

a_c = \frac{V^2}{R}

a_c = \frac{V^2}{350}

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 5.25ft/s^2

a = \sqrt{a_t^2+a_r^2}

5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

27.5625 = 1.21 + \frac{v^4}{122500}

v=42.3877ft/s

Now calculate the angular velocity of the motorcycle

v = r\omega

42.3877 = 350\omega

\omega = 0.1211rad/s

Calculate the angular acceleration of the motorcycle

a_t = r\alpha

1.1 = 350\alpha

\alpha = 3.1428*10^{-3}rad/s^2

Calculate the time needed by the motorcycle to reach an acceleration of

5.25ft/s^2

\omega = \alpha t

0.1211 = 3.1428*10^{-3}t

t = 38.53s

PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a_r = \frac{v^2}{R}

a_r = \frac{21.5^2}{350}

a_r =1.3207ft/s^2

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a = \sqrt{a_t^2+a_r^2}

a = \sqrt{(1.1)^2+(1.3207)^2}

a = 1.7187ft/s^2

PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

3 0
3 years ago
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