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alexandr1967 [171]
2 years ago
8

A rocket is being launched straight up. Air resistance is not negligible.

Physics
2 answers:
devlian [24]2 years ago
7 0

A rocket is launching upwards. The force acting on the rocket to move upward is F. The forces acting opposite to thrust are weight and drag force. The drag force is due to the air resistance. In order to move upwards the thrust force must be greater than weight and drag force. Figure shows the free body diagram of rocket at the time of launching.There are only three forces: weight (always present), thrust (provided from the rocket’s engines to lift it up) and air resistance (said specifically to be not negligible). There are no horizontal forces.

Yanka [14]2 years ago
3 0

Answer:

hope this will help you

have a great day

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Triangle XYZ has vertices X(0, 2), Y(4, 4), and Z(3, –1). Graph △XYZ and its image after a rotation of 180° about (2, –3).
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The image of the triangle is to be formed by rotating ΔXYZ 180 degrees about the (2, -3) as shown in the graph.

<h3>What is Geometry?</h3>

It deals with the size of geometry, region, and density of the different forms both 2D and 3D.

Triangle XYZ has vertices X(0, 2), Y(4, 4), and Z(3, –1).

If the triangle is ΔXYZ. Then the image of the triangle is to be formed by rotating ΔXYZ 180 degrees about the (2, -3) as shown in the graph.

More about the geometry link is given below.

brainly.com/question/7558603

#SPJ1

6 0
2 years ago
During takeoff, an airplane climbs with a speed of 195 m/s at an angle of 15° above the horizontal. The speed and angle constitu
matrenka [14]

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The horizontal component of the velocity is 188 m/s

The vertical component of the velocity is 50 m/s.

Explanation:

Hi there!

Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:

We can find vx using the following trigonometric rule of a right triangle:

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195 m/s · cos 15° = vx

vx = 188 m/s

The horizontal component of the velocity is 188 m/s

To calculate the y-component we will use the following trigonometric rule:

sin α = opposite / hypotenuse

sin 15° = vy / 195 m/s

195 m/s · sin 15° = vy

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The vertical component of the velocity is 50 m/s.

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Hope I get a brainliest answer.

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