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yuradex [85]
1 year ago
12

an airplane flying due north at 90. km/h is being blown due west at 50. km/h. what is the resultant velocity of the plane?

Physics
1 answer:
mina [271]1 year ago
6 0

Given:

v_{1} = 90 \; km/h

v_{2} = 50 \; km/h

We know that the resultant velocity is,

v = \sqrt{v_{1} ^{2}+v_{2} ^{2}  }

v = \sqrt{90^{2}+50^{2}}

v=\sqrt{8100+2500}

v=\sqrt{10600}

So, the resultant velocity is,

v=102.96\;km/h

<h3>Explain Velocity and its types?</h3>

Velocity is a vector physical term that measures how quickly distance changes with respect to time.

Constant velocity: This refers to the speed at which an object moves through space and time equally.

The speed at which an object cuts various displacements at various moments is known as variable velocity.

Instantaneous velocity: This term refers to either the speed at a specific instant or the constant speed throughout a very brief period of time.

Average velocity is the sum of all distances traveled in a given period of time.

Transverse velocity is the linear speed of an item moving on a circular path. Transverse velocity is the amount of distance covered in a unit of time.

To learn more about Velocity, visit:

brainly.com/question/18084516

#SPJ4

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Which of the following statements is a consequence of the equation, E = mc2
denis23 [38]
This is D.
A - technically the matter is converted but yes E is released
B - Mass and energy are different forms of each other
C - They can now combine to form the conservation of mass AND energy

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How can the rate of evaporation of a liquid be increased?​
Nadusha1986 [10]

When air is blown above the surface of liquid, it will take away the liquid carrying air particles from the air above the liquid, resulting in decrease in humidity and increase in rate of evaporation.

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In this problem, you will apply kinematic equations to a jumping flea. Take the magnitude of free-fall acceleration to be 9.80m/
polet [3.4K]
V o - initial velocity
v = velocity at the maximum height,
v² = v o² - 2 g h
v = 0
0 = v o² - 2 g h
v o² = 2 g h = 2 · 9.80 · 0.460
v o² = 9.052
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8 0
3 years ago
A man wears convex lens glasses of focal length 30cm in order to correct his eyes defect. Instead of the optimum 25cm, his dista
omeli [17]

Answer:

14 cm

Explanation:

F = (frac{uv}{u – v})

F = +ve

v = -ve

30 = (frac {25 {times} (-v)}{25 – (-v)})

v = (frac {25 {times} (-v)}{25+v})

v = 14cm

(Note that either negative or positive values go to show the positioning and hence, they are not a strong necessity in your final answer.)

So happy that i could help you!

Now this question could turn out to be easy for you!!

7 0
2 years ago
baseball is hit into the air at an initial speed of 37.2 m/s and an angle of 49.3 ° above the horizontal. At the same time, the
Agata [3.3K]

Answer:

The average speed of the fielder is 5.24 m/s

Explanation:

The position vector of the ball after it was hit can be calculated using the following equation:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem.

When the ball is caught, its position vector will be (see r1 in the figure):

r1 = (r1x, 0.873 m)

Then, using the equation of the position vector written above:

r1x = x0 + v0 · t · cos α

0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²

Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:

r1x = v0 · t · cos α

0.873 m = v0 · t · sin α + 1/2 · g · t²

Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:

0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²

0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²

Solving the quadratic equation:

t = 0.03 s and t = 5.72 s.

It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.

With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:

r1x = v0 · t · cos α

r1x = 37.2 m/s · 5.72 s · cos 49.3°

r1x = 1.39 × 10² m

The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.

The average velocity is calculated as the traveled distance over time, then:

average velocity = treveled distance / elapsed time

average velocity = 30.0 m / 5.72 s = 5.24 m/s

8 0
3 years ago
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