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lesya [120]
3 years ago
6

An airplane takes off a runway at a constant speed of 49m/s at constant angle 30 to the horizontal

Physics
1 answer:
joja [24]3 years ago
6 0

Complete Question

An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters )  is the airplane above the ground 13 seconds after takeoff?

Answer:

The height  is H  =  318.5 \ m

Explanation:

From the question we are told that

   The speed at which the plane takes off is  u = 49 \ m/s

      The angle at which it takes off is  \theta  =  30 ^o

        The time taken is t =  13  s

The vertical distance traveled is  mathematically represented as

          H  =  u sin  \theta  t

Substituting values  

         H  =  (49) * sin (30) *13

        H  =  318.5 \ m

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3 years ago
A person walks the path shown below. The total trip consists of four straight-line paths.
dmitriy555 [2]

At the end of the walk, the person's resultant displacement is 495.1 m at 63⁰ south of west.

<h3>What is resultant displacement?</h3>

The resultant displacement of an object is the change in position of the object. It can be described as the shortest distance connecting the final position of the object to the initial position of the object.

<h3>Net horizontal displacement </h3>

Path 1 = 40 m

Path 2 = 0 m

Path 3 = 110 m x cos(30) = 95.26 m

Path 4 = 180 m x cos(60) = 90 m

Total horizontal displacement, X = 40 m + 0 m + 95.26 m + 90 m = 225.26 m

<h3>Net vertical displacement </h3>

Path 1 = 0 m

Path 2 = 230 m

Path 3 = 110 m x sin(30) = 55 m

Path 4 = 180 m x sin(60) = 155.885 m

Total horizontal displacement, Y = 0 m + 230 m + 55 m + 155.885 m = 440.885 m

<h3>Resultant displacement</h3>

R = √(X² + Y²)

R = √(225.26² + 440.885²)

R = 495.1 m

<h3>Direction of the displacement</h3>

θ = arc tan (Y/X)

θ = arc tan (440.885/225.26)

θ =  63⁰

Thus, at the end of the walk, the person's resultant displacement is 495.1 m at 63⁰ south of west.

Learn more about resultant displacement here: brainly.com/question/13309193

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3 0
1 year ago
M (b) Calculate the speed of an electron that is accelerated through the same electric potential difference.
MaRussiya [10]

The speed of a electron that is accelerated from rest through an electric potential difference of 120 V  is 6.49\times10^6m/s

<h3>How to calculate the speed of the electron?</h3>

We know, that the  energy of the system is always conserved.

Using the Law of Conservation of energy,

\triangle K+\triangle U=0

Here, K is the kinetic energy and U is the potential energy.

Now, substituting the formula of U and K, we get:

(\frac{1}{2} mv^2)+(-q\triangle V) =0------(1)

Here,

m is the mass of the electron

v is the speed of the electron

q is the charge on the electron

V is the potential difference

Let v_f and v_i represent the final and initial speed.

Here, v_i =0

Solving for v_f, we get:

v_f=\sqrt{\frac{-2q\triangle V}{m} }

=\sqrt{\frac{2\times1.602\times10^{-19}\times 120}{9.109\times10^{-31}} }

=6.49\times10^6m/s

To learn more about the conservation of energy, refer to:

brainly.com/question/2137260

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4 0
1 year ago
Match the speed to the section that describes.
inn [45]

Answer:

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3 years ago
If the bonds in the reactants of Figure 7-3 contained 432 kJ of chemical energy and the bonds in the
Yuri [45]

Answer:

  • <u>The energy change would be 46kJ</u>
  • <u>The energy would be absorbed</u>

Explanation:

The <em>energy change </em>during a chemical reation, i.e. the reaction energy, is equal to the chemical energy stored in the<em> bonds of the products </em>less the chemical energy stored in the <em>bonds of the reactants</em>.

Hence:

  • <em>Energy change</em> = 478 kJ - 432kJ = 46kJ

The change is positive, this is, the chemical energy of the products is greater than the chemical energy of the reactants.

That corresponds to the second graph, where the level of the energy of the products in the graph is higher than the level of the energy of the reactants. Therefore, the conclusion is that the reaction <em>absorbed energy</em> and it is endothermic.

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