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3 years ago

Can you give me an example of how sound passes through a solid?please help

2 answers:

5 0

When a layer of cold air close to the ground is covered by a layer of warmer air, sound waves traveling upward may be bent, or refracted, by the difference in temperature and redirected toward the ground.

alexira [117]3 years ago

3 0

Answer:

sound travels through steel about 15 times faster than through air.

Explanation:

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If someone is driving 100 miles in 60 minutes then drives 150 miles in 100 minutes west, what is his acceleration rate.

Sliva [168]

Answer:

his acceleration rate is -0.00186 m/s²

Explanation:

Given;

initial position of the car, x₀ = 100 miles = 160, 900 m ( 1 mile = 1609 m)

time of motion, t₀ = 60 minutes = 60 mins x 60 s = 3,600 s

final position of the car, x₁ = 150 miles = 241,350 m

time of motion, t₁ = 100 minutes = 100 mins x 60 s = 6,000 s

The initial velocity is calculated as;

u = 160, 900 m / 3,600 s

u = 44.694 m/s

The final velocity is calculated as;

v = 241,350 m / 6,000 s

v = 40.225 m/s

The acceleration is calculated as;

$a = \frac{\Delta V}{\Delta t} = \frac{v- u}{t_1 - t_ 0} = \frac{40.225 - 44.694}{6000-3600} = -0.00186 \ m/s^2\\\\$

Therefore, his acceleration rate is -0.00186 m/s²

6 0

3 years ago

2. What does the term reflection mean?

Vesna [10]

Answer:

egrfeirugherhgourehgabgwehgoehborghrewuhgelkg

Explanation:

4 0

3 years ago

Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge

Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃= 2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

$\frac{k*q_{1}*q_3 }{(d_{13})^{2} } = \frac{k*q_{2}*q_3 }{(d_{23})^{2} }$

We divide by (k * q3) on both sides of the equation

$\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }$

$q_{1} = \frac{q_{2}*(d_{13})^{2} }{(d_{23} )^{2} }$

$q_{1} = \frac{5*(2)^{2} }{(4 )^{2} }$

q₁ = + 1.25 nC

3 0

3 years ago

The frequency of a wave is 200 Hz. The wavelength is 0.1 m. What is the period of the wave?

Aleks04 [339]

The formula for the period of wave is: wave period is equals to 1 over the frequency. $waveperiod=\frac{1}{frequency}$
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.

wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:

wave period= 1/ 200/s

get the reciprocal form of 200/s which is s/200

then you can start the actual computation:

wave period= 1 x s divided by 200

this would give us an answer of 0.005 s.

6 0

3 years ago

he atomic radii of Li and O2- ions are 0.068 and 0.140 nm, respectively. (a) Calculate the force of attraction in newtons betwee

Crazy boy [7]

Answer:

$F = -1.07 *10^{-8} \ N$

$F_r = 1.07 *10^{-8} \ N$

Explanation:

Generally the force of attraction between this two irons is mathematically represented as

$F = \frac{k * [Q_{Li} ] * [Q_{O} ] }{ r^2}$

Here k is known as the proportionality constant with value $k = 2.31 * 10^ {-28} J \cdot m$

substituting -2 for $Q_{O}$ i.e the charge on oxygen , +1 for $Q_{Li}$ i.e the charge on Lithium and $[0.140 + 0.068 ] nm= 0.208 nm = 0.208*10^{-9}$ for r

$F = \frac{ 2.31 * 10^ {-28}* +1 * -2 }{ ( 0.208*10^{-9} )^2 }$

$F = -1.07 *10^{-8} \ N$

Generally the force of repulsion will be the magnitude but different direction to the force o attraction

So Force of repulsionn is

$F_r = 1.07 *10^{-8} \ N$

6 0

4 years ago