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3 years ago

Helphpphphphpphphphpphphpph

Physics

1 answer:

adell [148]3 years ago

7 0

Airplane with nose up: The plane's speed through the air is the square root of (80 m/s squared) plus (120 m/s squared. The whole picture is a right triangle, and the plane's speed is the hypotenuse. The angle is the angle whose tangent is (80/120). You can get it from a calculator, a book, a slide rule, or online from the site that rhymes with floogle. The man pulling the load is also a right triangle. The horizontal component is (hypotenuse) times (cosine of the angle). The vertical component is (hypotenuse) times (sine of the same angle). Fill in what you know, look up the sin and cos of 25 degrees and write those in too, and then you can solve for what you have to find.

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Which one of the following statements is not a characteristic of a plane mirror?

OleMash [197]

Answer:

C) false. The image is formed by the prolongation of the rays, so it is VIRTUAL

Explanation:

Let's review each of the statements

A) True. The image is the same size as the object in a flat mirror, m = 1

B) True. The rays are not inverted, so the right images

C) false. The image is formed by the prolongation of the rays, so it is VIRTUAL

D) True. Flat mirrors reverse left and right

E) True. When using trigonometry the angles are equal, therefore two triangles formed have the same leg, and the distance to the object and the image are equal

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3 years ago

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3 years ago

Read 2 more answers

An astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity c

KiRa [710]

Acceleration can be defined as the change of speed in an instant of time, that is

$a = \frac{v_f-v_i}{\Delta t}$

Here,

$v_f =$ Final velocity

$v_i =$ Initial velocity

$\Delta t =$ Change in time

From this expression we will calculate the requested values replacing the variables in each of the given terms

PART A) The values under this condition are:

$v_f = 5.3m/s$

$v_i = 15m/s$

$\Delta t = 10.4s$

Replacing,

$a = \frac{5.3m/s-(15m/s)}{10.4s}$

$a = -0.93m/s^2$

PART B ) The values under this condition are:

$v_f = -15m/s$

$v_i = -5.3m/s$

$\Delta t = 10.4$

Replacing,

$a = \frac{-(15m/s)-(-5.3m/s)}{10.4s}$

$a = -0.93m/s^2$

Therefore the acceleration in the second time interval is $-0.93m/s^2$

PART C) The values under this condition are:

$v_f = -15m/s$

$v_i = 15m/s$

$\Delta t = 10.4$

Replacing,

$a = \frac{-15m/s-(15m/s)}{10.4s}$

$a = -2.9m/s^2$

6 0

3 years ago

The speeds of 22 particles are as follows: two at 5.30 cm/s, four at 1.40 cm/s, six at 7.14 cm/s, eight at 1.52 cm/s, two at 7.6

Leona [35]

Answer:

Average velocity is 3.93 cm/s

Root mean square velocity is 4.79 cm/s

Velocity peak to peak is 6.28 cm/s

Explanation:

Speed of 2 particles = 5.3 cm/s

Speed of 4 particles = 1.4 cm/s

Speed of 6 particles = 7.14 cm/s

Speed of 8 particles = 1.52 cm/s

Speed of 2 particles = 7.68 cm/s

$v_{avg}=\frac{2\times 5.3+4\times 1.4+6\times 7.14+8\times 1.52+2\times 7.68}{22}\\\Rightarrow v_{avg}=\frac{86.56}{22}\\\Rightarrow v_{avg}=3.93\ cm/s$

Average velocity is 3.93 cm/s

$v_{rms}=\sqrt{\frac{2\times 5.3^2+4\times 1.4^2+6\times 7.14^2+8\times 1.52^2+2\times 7.68^2}{22}}\\\Rightarrow v_{rms}=\sqrt{\frac{507.34}{22}}\\\Rightarrow v_{rms}=4.79\ cm/s$