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3 years ago

Which statements apply to transverse waves? Check all that apply.

Physics

2 answers:

Levart [38]3 years ago

8 0

Correct Answers:

- The waves have a trough

- the waves have a crest

- the energy that is transferred move in a perpendicular direction

- particles move in an up and down motion

Gennadij [26K]3 years ago

6 0

Answer:

A, c, d, e

Explanation:

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That they sometimes explode?

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A plane mirror of circular shape with radius r=20cm is fixed to the ceiling. A bulb is to be placed on the axis of the mirror. A

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Answer:

0.75 m

Explanation:

Let's call the distance between the bulb and the mirror x.

The bulb and the length of the mirror form a triangle. The mirror and the illuminated area on the floor form a trapezoid. If we extend the lines from the mirror edge to the reflected image of the bulb, we turn that trapezoid into a large triangle. This triangle and the small triangle are similar. So we can say:

x / 0.4 = (3 + x) / 2

Solving for x:

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3 years ago

Which of the following is the equation for calculating daily caloric needs? A. BMR calories + Digestion calories B. BMR calories

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Explanation:

To determine your total daily calorie needs, multiply your BMR by the appropriate activity factor, as follows: If you are sedentary (little or no exercise) : Calorie-Calculation = BMR x 1.2. If you are lightly active (light exercise/sports 1-3 days/week) : Calorie-Calculation = BMR x 1.375.

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3 years ago

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Given the vector A with components Ax = 2.00, Ay = 6.00, the vector B with components Bx = 2.00, By = 22.00, and the vector D =

nekit [7.7K]

Answer:

The magnitude of vector d is 16 and the angle with the x-axis is 270°

Explanation:

Operations With Vectors

Given two vectors in rectangular components:

$\vec a=(ax,ay)\ ,\ \vec b=(bx,by)$

The sum of the vectors is:

$\vec a+\vec b=(ax+bx,ay+by)$

The difference between the vectors is:

$\vec a-\vec b=(ax-bx,ay-by)$

The magnitude of $\vec a$ is:

$|\vec a|=\sqrt{ax^2+ay^2}$

The angle $\vec a$ makes with the horizontal positive direction is:

$\displaystyle \tan\theta=\frac{ay}{ax}\\$

The question provides the vectors:

$\vec a=(2,6)$

$\vec b=(2,22)$

$\vec d=\vec a-\vec b$

Calculate:

$\vec d=(2,6)-(2,22)=(0,-16)$

The magnitude of $\vec d$ is:

$|\vec d|=\sqrt{0^2+(-16)^2}=\sqrt{0+256}=16$

The angle is calculated by:

$\displaystyle \tan\theta=\frac{-16}{0}$

The division cannot be calculated because the denominator is zero. We need to estimate the correct angle by looking at the components of the vector. Since the x-coordinate is zero and the y-coordinate is negative, the vector points downwards (south), thus the angle must be -90° or 270° if the range goes from 0° to 360°.

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