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Citrus2011 [14]
3 years ago
12

Which biome's yearly rainfall mainly evaporates? A. taiga B. desert C. tropical rainforest D. temperate grassland

Physics
2 answers:
victus00 [196]3 years ago
8 0
Desert is the correct answer.
soldi70 [24.7K]3 years ago
7 0
The correct answer is desert.
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A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its potential energy (PE) when it is 2.00 m above the grou
IceJOKER [234]

The potential energy is 4.10 J

Explanation:

The potential energy of a body is given by

U=mgh

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object relative to the ground

For the lemming in this problem we have

m = 0.078 kg is its mass

h = 5.36 m is the height above the ground

g=9.8 m/s^2 is the acceleration of gravity

Solving the equation,

U=(0.078)(9.8)(5.36)=4.10 J

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7 0
4 years ago
Determine what forces are present in this situation: An elevator is rising at a constant speed; the elevator is the object in th
Alecsey [184]
Tension pointing up at the top Weight pointing down at the bottom If there is someone standing inside the elevator, there will be a normal reaction force pointing down at the bottom.
7 0
3 years ago
Read 2 more answers
What is the x-component of a vector with a magnitude of 115 km at an angle of 22°?
ruslelena [56]

The x-component of a vector are < 106.6, 43.07 >

Depending on the angle we are provided, the x-component of a vector can either be cos or sin. Cos always corresponds to the right triangle's side that contacts the specified angle.

If a vector v with magnitude ||v|| makes an angle θ with the positive x-axis then,

v = ||v|| cos θi + ||v|| sin θj

 =  < ||v|| cos θ , ||v|| sin θ >

Magnitude p = 115 km

Angle = 22°

p = ||p|| < cos θ, sin θ >

p = 115 < cos 22°, sin 22° >

p = 115 < 0.927, 0.3746 >

p = < 106.6, 43.07 >

Therefore,  the x-component of a vector are < 106.6, 43.07 >

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4 0
2 years ago
1. A 14-cm tall object is placed 26 cm from a converging lens that has a focal length of 13 cm.
AURORKA [14]

Answer:

a) Please find attached the required drawing of light passing through the lens

By the use of similar triangles;

The image distance from the lens = 26 cm

The height of the image = 14 cm

c) The image distance from the lens = 26 cm

The height of the image = 14 cm

Explanation:

Question;

a) Determine the image distance and the height of the image

b) Calculate the image position and height

The given parameters are;

The height of the object, h = 14 cm

The distance of the object from the mirror, u = 26 cm

The focal length of the mirror, f = 13 cm

The location of the object = 2 × The focal length

Therefore, given that the center of curvature ≈ 2 × The focal length, we have;

The location of the object ≈ The center of curvature of the lens

The diagram of the object, lens and image created with MS Visio is attached

From the diagram, it can be observed, using similar triangles, that the image distance from the lens = The object distance from the lens = 26 m

The height of the image = The height of the object - 14 cm

b) The lens equation is used for finding the image distance from the lens as follows;

\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}

Where;

v = The image distance from the lens

We get;

v = \dfrac{u \times f}{u - f}

Therefore;

v = \dfrac{26 \times 13}{26 - 13} = \dfrac{26 \times 13}{13} = 26

The distance of the image from the lens, v = 26 cm

The magnification, M =v/u

∴ M = 26/26 = 1, therefore, the object and the image are the same size

Therefore;

The height of the image = The height of the object = 14 cm.

5 0
3 years ago
a force of 5.5 n is applied to an object. the moment arm for the force is 0.84 m. what is the torque produced by the force?
Kipish [7]

A force of 5.5N is applied to an object. The moment arm for the force is 0.84 m, the torque produced by the force is <u>4.62N-m</u>.

A force applied perpendicularly to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. We can find the torque from a force by taking the perpendicular component of that force and multiplying by the magnitude of the R vector where this R vector is the vector that points from the axis to the point where the force is applied.

To calculate torque produced by the force, we have-

Torque= forcexmoment arm

T = fxd

T =5.5Nx0.84m

T=4.62N-m.

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6 0
2 years ago
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