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2 years ago

What is the length of AC?

Physics

1 answer:

Grace [21]2 years ago

7 0

D: 16 AB is half of AC and since you’re given AB=8 to get the full length you just multiply it by 2

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When is a secondary source more helpful than a primary source?

UNO [17]

Answer:

I think the answer is C.

Explanation:

A primary source is a first hand account of an event while a secondary source is a retelling or second hand account meaning as many details will be prevalent.

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2 years ago

What is refraction of light?

prohojiy [21]

Answer:

bending of light

Explanation:

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2 years ago

Read 2 more answers

If the work done on a object is 100 J and it comes to a rest on a surface with a mass of 10

SVETLANKA909090 [29]

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3 years ago

Plane polarized light with intensity I0 is incident on a polarizer. What angle should the principle axis make with respenct to t

Nataliya [291]

Answer:

Q = 47.06 degrees

Explanation:

Given:

- The transmitted intensity I = 0.464 I_o

- Incident Intensity I = I_o

Find:

What angle should the principle axis make with respect to the incident polarization

Solution:

- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:

I = I_o * cos^2 (Q)

- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:

Q = cos ^-1 (sqrt (I / I_o))

- Plug the values in:

Q = cos^-1 ( sqrt (0.464))

Q = cos^-1 (0.6811754546)

Q = 47.06 degrees

8 0

3 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

7 0

3 years ago