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Lerok [7]
3 years ago
12

Mechanical waves propagate or move through a medium because

Physics
1 answer:
lys-0071 [83]3 years ago
8 0
The answer to that would be that 

they require so its mandatory for mechanical waves to travel through a medium
You might be interested in
Is it possible to have a charge of 5 x 10-20 C? Why?
ruslelena [56]

1) No

2) Yes

3) No

4) Equal and opposite

5) 32400 N

6) Repulsive

7) The electric force is 2.3\cdot 10^{39} times bigger than the gravitational force

Explanation:

1)

In nature, the minimum possible charge that an object can have is the charge of the electron, which is called fundamental charge:

e=1.6\cdot 10^{-19}C

Electrons are indivisible particles (they cannot be separated), this means that an object can have at least the charge equal to the charge of one electron (in fact, it cannot have a charge less than e, because it would meant that the object has a "fractional number" of electrons).

In this problem, the object has a charge of

Q=5\cdot 10^{-20}C

If we compare this value to e, we notice that Q, so no object can have a charge of Q.

2)

As we said in part 1), an object should have an integer number of electrons in order to be charged.

This means that the charge of an object must be an integer multiple of the fundamental charge, so we can write it as:

Q=ne

where

Q is the charge of the object

n is an integer multiple

e is the fundamental charge

Here we have

Q=2.4\cdot 10^{-18}C

Substituting the value of e, we find n:

n=\frac{Q}{e}=\frac{2.4\cdot 10^{-18}}{1.6\cdot 10^{-19}}=15

n is integer, so this value of the charge is possible.

3)

We now do the same procedure for the new object in this part, which has a charge of

Q=2.0\cdot 10^{-19}C

Again, the charge on this object can be written as

Q=ne

where

n is the number of electrons in the object

Using the value of the fundamental charge,

e=1.6\cdot 10^{-19}C

We find:

n=\frac{Q}{e}=\frac{2.0\cdot 10^{-19}}{1.6\cdot 10^{-19}}=1.25

n is not integer, so this value of charge is not possible, since an object cannot have a fractional number of electrons.

4)

To solve this part, we use Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (Action force), then object B exerts an equal and opposite force on object A (reaction force)".

In this problem, we have two objects:

- A charge Q

- A charge 5Q

Charge Q exerts an electric force on charge 5Q, and we can call this action force. At the same time, charge 5Q exerts an electric force on charge Q (reaction force), and according to Newton's 3rd law, the two forces are equal and opposite.

5)

The magnitude of the electric force between two single-point charges is

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the two charges

In this problem we have:

q_1=+4.5\cdot 10^{-6}C is charge 1

q_2=+7.2\cdot 10^{-6}C is charge 2

r = 0.30 cm = 0.003 m is the separation

So, the electric force  between the two charges is

F=(9\cdot 10^9)\frac{(4.5\cdot 10^{-6})(7.2\cdot 10^{-6})}{(0.003)^2}=32400 N

6)

The electric force between two charged objects has direction as follows:

- If the two objects have charges of opposite signs (+ and -), the force between them is attractive

- If the two objects have charges of same sign (++ or --), the force between them is repulsive

In this problem, the two charges are:

q_1=+4.5\cdot 10^{-6}C is charge 1

q_2=+7.2\cdot 10^{-6}C is charge 2

We see that the two charges have same sign: therefore, the force between them is repulsive.

7)

The electric force between the proton and the electron in the atom can be written as

F_E=k\frac{q_1 q_2}{r^2}

where

q_1 = q_2 = e = 1.6\cdot 10^{-19}C is the magnitude of the charge of the proton and of the electron

r=5.3\cdot 10^{-11} m is the separation between them

So the force can be rewritten as

F_E=\frac{ke^2}{r^2}

The gravitational force between the proton and the electron can be written as

F_G=G\frac{m_p m_e}{r^2}

where

G is the gravitational constant

m_p = 1.67\cdot 10^{-27}kg is the proton mass

m_e=9.11\cdot 10^{-27}kg is the electron mass

Comparing the 2 forces,

\frac{F_E}{F_G}=\frac{ke^2}{Gm_p m_e}=\frac{(9\cdot 10^9)(1.6\cdot 10^{-19})^2}{(6.67\cdot 10^{-11})(1.67\cdot 10^{-27})(9.11\cdot 10^{-31})}=2.3\cdot 10^{39}

8 0
3 years ago
Kali left school and traveled toward her friend's house at an average speed of 40 km/h. Matt left one hour later and traveled in
zlopas [31]

Answer:

t = 5 hr

Explanation:

Let kali moves toward east with velocity= V₁= 40 km/ h

Mat moves toward west with velocity = V₂= 50 km/hr

As Klai left one hour earlier = t₁= 1 hr

distance traveled in 1st hour = s₁ = v * t = 40 * 1 = 40 km

Remaining distance = 400 - 40 = 360 km

As they move in the opposite directions:

Relative speed= 40 + 50 = 90 km/ h

s = v * t

⇒ t = s / v

⇒ t₂ = 360 / 90

⇒ t₂ = 4 hr

Total time = t = t₁ + t₂

t = 1 hr + 4 hr

t = 5 hr

5 0
3 years ago
Read 2 more answers
Magnesium Oxide Reaction
Verizon [17]

Answer:

When magnesium reacts with oxygen, it produces light bright enough to blind you temporarily. Magnesium burns so bright because the reaction releases a lot of heat. As a result of this exothermic reaction, magnesium gives two electrons to oxygen, forming powdery magnesium oxide (MgO).

6 0
2 years ago
Read 2 more answers
With detailed explaniation
belka [17]
  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

#C

\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

Now

\\ \rm\Rrightarrow v=u-gt

\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

4 0
2 years ago
The voltage across the terminals of a generator is 5.7 v when it supplies a current of 0.3 A. It becomes 5.1 V when I=0.9A. Find
snow_tiger [21]

Answer:

  • The emf of the generator is 6V
  • The internal resistance of the generator is 1 Ω

Explanation:

Given;

terminal voltage, V = 5.7 V, when the current, I = 0.3 A

terminal voltage, V = 5.1 V, when the current, I = 0.9 A

The emf of the generator is calculated as;

E = V + Ir

where;

E is the emf of the generator

r is the internal resistance

First case:

E = 5.7   + 0.3r -------- (1)

Second case:

E = 5.1 + 0.9r -------- (2)

Since the emf E, is constant in both equations, we will have the following;

5.1 + 0.9r = 5.7   + 0.3r  

collect similar terms together;

0.9r - 0.3r = 5.7 - 5.1

0.6r = 0.6

r = 0.6/0.6

r = 1 Ω

Now, determine the emf of the generator;

E = V + Ir

E = 5.1 + 0.9x1

E = 5.1 + 0.9

E = 6 V

6 0
3 years ago
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