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atroni [7]
3 years ago
14

The net force acting on a Cessna 172 airplane has a magnitude of 1900 N and points in the positive x direction. If the plane has

a mass of 860 kg, what is its acceleration?
Physics
1 answer:
Kay [80]3 years ago
4 0

Answer: 10.05m/s²

Explanation:

Force acting on the plane has a magnitude of 1900N.

Mass of plane = 860kg

We then calculate the acceleration which will be:

= 1900 / 860

= 2.21 m/s²

Since acceleration due to gravity is equal to 9.8m/s², then the net acceleration will be:

= ✓(2.21² + 9.8²)

= ✓(4.88 + 96.04)

=✓100.92

= 10.05m/s²

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3 0
3 years ago
A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting
rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is \sqrt{725} ft/ sec

Explanation:

Given:

h(t) =  25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Lets find the derivative of distance with respect to time

\frac{dD}{dt} (t)  = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

Substituting the values of h(t) and  x(t) and simplifying we get,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

\frac{dh}{dt} = 25ft/sec

\frac{dx}{dt} = 10 ft/sec

\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}  = \frac{725}{\sqrt{725}}  = \sqrt{725} ft / sec

5 0
3 years ago
A student pushes against a wall with a force of 30N. The wall does not move. What amount of force does the wall exert on the stu
True [87]

Answer:

C

Explanation:

they both have to be the same for both to not move

8 0
2 years ago
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A thin 1.5 mm coating of glycerine has been placed between two microscope slides of width 0.8 cm and length 3.9 cm . Find the fo
Radda [10]

The  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

<h3>Force required to pull one end at a constant speed</h3>

The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is determined by applying Newton's second law of motion as shown below;

F = ma

where;

  • m is mass
  • a is acceleration

At a constant speed, the acceleration of the object will be zero.

F = m x 0

F = 0

Thus, the  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

Learn more about constant speed here: brainly.com/question/2681210

3 0
2 years ago
an object weighs 98 n on earth. How much does it weigh on planet x where the acceleration due to gravity in 6 m/s^2
Degger [83]
60 N because 98N=mg (here g= 9.8 on earth) thus mass can be calculated which is 98/9.8 = 10kg Now,new weight with g = 6m/s^2 =m×g' (here g' is new acceleration of the new planet) = 10×6=60N
7 0
3 years ago
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