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Y_Kistochka [10]
2 years ago
10

The speed of a wave in a medium is effected by____,____, and____

Physics
1 answer:
lisov135 [29]2 years ago
7 0

\huge\mathfrak\green{answer}

As per as my knowledge

The speed of a wave in a medium is affected by <u>d</u><u>e</u><u>n</u><u>s</u><u>i</u><u>t</u><u>y</u>,<u> </u><u>w</u><u>a</u><u>v</u><u>e</u><u>l</u><u>e</u><u>n</u><u>g</u><u>t</u><u>h</u> and <u>t</u><u>e</u><u>m</u><u>p</u><u>e</u><u>r</u><u>a</u><u>t</u><u>u</u><u>r</u><u>e</u><u> </u>:)

(Good luck on your test and mark me brainliest if this helps)

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Exercise will not help maintain the health of your endocrine system.
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A. True

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2 years ago
Falls resulting in hip fractures are a major cause of injury and even death to the elderly. Typically, the hip’s speed at impact
Westkost [7]

Answer:

-5.8868501529 m/s² or -5.8868501529g

0.118909090909 s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{1.3^2-2^2}{2\times 0.02}\\\Rightarrow a=-5.8868501529\ m/s^2

Dividing by g

\dfrac{a}{g}=\dfrac{-57.75}{9.81}\\\Rightarrow a=-5.8868501529g

The acceleration is -5.8868501529 m/s² or -5.8868501529g

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{1.3-2}{-5.8868501529}\\\Rightarrow t=0.118909090909\ s

The time taken is 0.118909090909 s

7 0
3 years ago
if the emission lines in the spectrum of one object are more strongly blueshifted than those from a second object, then the firs
Lelu [443]

Answer:

Toward us faster than the second object

Explanation:

6 0
2 years ago
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p
Irina-Kira [14]

Answer:

The magnitude of each force is 2.45 x 10⁻¹⁶ N

Explanation:

The charge of proton, +q = 1.603 x 10⁻¹⁹ C

The charge of electron, -q = 1.603 x 10⁻¹⁹ C

Distance between the two charges, r = 971 nm = 971 x 10⁻⁹ m

Apply Coulomb's law;

F = \frac{kq_1q_2}{r^2}

where;

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

q₁ and q₂ are the charges of proton and electron respectively

F is the magnitude of force between them

Substitute in the given values and solve for F

F = \frac{(8.99*10^9)(1.603*10^{-19})^2}{(971*10^{-9})^2} \\\\F = 2.45*10^{-16} \ N

Therefore, the magnitude of each force is 2.45 x 10⁻¹⁶ N

4 0
3 years ago
What is bought from an electric company
Luden [163]

Answer: electricity

Explanation:

8 0
3 years ago
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