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Bond [772]
3 years ago
12

How does the sound produced by a vibrating object in a medium reach your ear? ​

Physics
1 answer:
BaLLatris [955]3 years ago
5 0

Answer:The vibrating object produces a series of compressions and rarefactions, one after the other in the medium. These pulses travel one behind the other as the sound waves move forward. When this sound wave reaches our ear, it forces the tympanic membrane to vibrate and thus causes the sensation of hearing

Explanation:

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a 3000 kg and a 7000 kg Mass attract each other with a force of 0.0015 N. What distance separates the two objects (Radius) (Plea
frutty [35]

Answer:

<h3> 3.057m</h3>

Explanation:

According to law of gravitation;

F = GMm/d²

G is the universal gravitation

M and m are the masses

d is the distance between the masses

d² = GMm/F

d² = 6.67408 × 10-11 *3000*7000/0.0015

d² = 140.15568*10^-5/0.0015

d² = 1.4016*10^-3/0.0015

d² = 1.4016*10^-3/1.5*10^-3

d²  = 0.9344*10

d² = 9.344

d = √9.344

d = 3.057m

Hence the distance between the two objects is  3.057m

3 0
3 years ago
Two identical charges q placed 2.0 mapart exert forces of magnitude 4.0 N on each other What is the value of the charge q? a) q
katen-ka-za [31]

Answer:

c) 4.2*10^{-5}C

Explanation:

Coulomb's law says that the force exerted between two charges is inversely proportional to the square of distance between them, and is given by the expression:

F=\frac{kq_{1}q_{2}}{r^{2}}

where k is a proportionality constant with the value k=9*10^{9}\frac{Nm^{2}}{C^{2}}

In this case q_{1}=q_{2}=q, so we have:

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

Solving the equation for q, we have:

kq^{2}=Fr^{2}

q^{2}=\frac{Fr^{2}}{k}

q=\sqrt{\frac{Fr^{2}}{k}}

Replacing the given values:

q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}

q=4.2*10^{-5}C

3 0
3 years ago
An 7.5 × binocular has 3.7-cm-focal-length eyepieces. What is the focal length of the objective lenses? Express your answer to t
elixir [45]

To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

\mu = \frac{f_0}{f_e}

Here,

\mu = Magnification

f_e = Focal length eyepieces

f_0 = Focal length of the Objective

Rearranging to find the focal length of the objective

f_0 = \mu f_e

Replacing with our values

f_0 = 7.5* 3.7cm

f_0 = 27.75cm

Therefore the focal length of th eobjective lenses is 27.75cm

5 0
3 years ago
What is the term for hardening or curing the sealant or composite material by exposing the material to a curing light?
natka813 [3]

Answer:

Photopolymerization is the term for hardening or curing the sealant or composite material by exposing the material to a curing light

Explanation:

It the process through which a photopolymer reacts to the radiations in presence of the Ultra-violet light. Photopolymers refers to those materials and liquid or the plastic resins that are hardened in presence of light source such as the lasers or the lamps. When these materials are irradiated, they undergo chemical reactions and they are hardened and become solid, and hence the process called photopolymerization. Photopolymerization is majorly used in the coating and printing industries and in also in dentistry to fill the top of teeth and prevent the cavities

5 0
3 years ago
Since all stars begin their lives with the same basic composition, what characteristic most determines how they will differ?
Dafna1 [17]
4. Mass they are formed with
7 0
3 years ago
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