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MariettaO [177]
3 years ago
9

Which diagram represents deposition?

Physics
1 answer:
Vera_Pavlovna [14]3 years ago
3 0

Your answer will be gas

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675 km equals how much cm ?
irga5000 [103]
KHDMDCM.
Now go from Kilometer to Centimeter: 5.
Move the decimal 5 places to the right: 67,500,000 centimeters.
Hope this helps :)
7 0
3 years ago
The x component of vector is -27.3 m and the y component is +43.6 m. (a) What is the magnitude of ? (b) What is the angle betwee
lara [203]

Answer:51.44 units

Explanation:

Given

x component of vector is -27.3\hat{i}

y component of vector is 43.6\hat{j}

so position vector is

r=-27.3\hat{i}+43.6\hat{j}

Magnitude of vector is

|r|=\sqrt{27.3^2+43.6^2}

|r|=\sqrt{2646.25}

|r|=51.44 units

Direction

tan\theta =\frac{43.6}{-27.3}=-1.597

vector is in 2nd quadrant thus

180-\theta =57.94

\theta =122.06^{\circ}

4 0
3 years ago
An astronaut holds a rock 100m above the surface of Planet X. The rock is then thrown upward with a speed of 15m/s, as shown in
Harlamova29_29 [7]

The acceleration due to gravity of the planet X is 1 m/s².

The given parameters;

  • height above the ground, h = 100 m
  • initial velocity of the rock, u = 15 m/s
  • time of motion of the rock, t = 10 s

The acceleration due to gravity is calculated as follows;

h = ut - \frac{1}{2} gt^2\\\\100 = 15(10) - (0.5\times 10^2)g\\\\100 = 150 - 50g\\\\50g = 150-100\\\\50g = 50\\\\g = 1 \ m/s^2

Thus, the acceleration due to gravity of the planet X is 1 m/s²

Learn more here: brainly.com/question/24564606

7 0
2 years ago
What is the result of a force distributed over an area
natita [175]

Answer:

The result of force distributed over an area – Pressure = Force(in Newton's – N)/area (m 2 ) Pascal (Pa) – SI unit for Pressure – Named after.

I hope it help you,

<h3>Follow me for more answer.</h3>
8 0
3 years ago
A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 375.0 m
Gekata [30.6K]

Answer:

819.78 m

Explanation:

<u>Given:</u>

  • OA = range of initial position of the airplane from the point of observation = 375 m
  • OB = range of the final position of the airplane from the point of observation = 797 m
  • \theta = angle of the initial position vector from the observation point = 43^\circ
  • \alpha = angle of the final position vector from the observation point = 123^\circ
  • \vec{AB} = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} =  (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} =  (-708.34\ \hat{i}+412.67\ \hat{j})\ m

The magnitude of the displacement vector = \sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

8 0
3 years ago
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