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Delvig [45]
3 years ago
8

What is the frequency of a mechanical wave that has a

Physics
1 answer:
kakasveta [241]3 years ago
5 0

Explanation:

v= (f) x ( lambda)

1.7 ms^-1/12.05 m = f =o.14 hz

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. Two identical vehicles traveling at the same speed are made to collide with barriers in an insurance company collision test. T
Serggg [28]

Answer:

F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

Explanation:

For this exercise let's use the relationship between momentum and momentum.

         I = F t = Δp

in this case the final velocity is zero

        F t = 0 -m v₀

        F = m v₀ / t

in order to answer the question we must assume that the two vehicles have the same mass and speed

concrete barrier

        F₁ = -p₀ / 0.1

        F₁ = - 10 p₀

barrier collapses

         F₂ = -p₀ / 1

let's look for the relationship of the forces

        F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

5 0
3 years ago
What is the speaker’s power output if the sound intensity level is 102 dBdB at a distance of 25 mm ? Express your answer to two
lesya [120]

Answer:

Power  = 124.50 W

Explanation:

Given that:

The Sound intensity of a speaker output is 102 dB

and the distance r = 25 m

For the intensity of sound,

\beta (dB)= 10 \  log_{10 } (\dfrac{I}{I_o})

where;

the threshold of hearing   I_o = 10^{-12} (W/m^2)

\dfrac{102 }{10}= log_{10}( \dfrac{I}{10^{-12}})

10^{10.2} =  \dfrac{I}{10^{-12}}

I = 10^{10.2} \times 10^{-12}

I = 0.01585 W/m²

If we recall, we know remember that ;

Power = Intensity × A rea

Power = 0.01585 W/m² × 4 × 3.142 × (25 m)²

Power  = 124.50 W

7 0
3 years ago
According to the intuitive guess you just made, is the overall momentum of the two-cart system after the collision greater than,
Aloiza [94]

Answer:

It is equal to the overall momentum before collision, so far no external object is involved.

Explanation:

Momentum is always conserved during collision as a rule. This is equal to the product of the mass and velocity. Thank you.

5 0
3 years ago
A ball is dropped from rest at the top of a 6.10 m
natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

\frac{v_1}{v} = e ( coefficient of restitution ) = \frac{1}{\sqrt{10} }

and

\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

e = \sqrt{\frac{h_1}{6.1} }

e = \sqrt{\frac{h_2}{h_1} }

So on

e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }

(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}= .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0
3 years ago
Question 6 (5 points)
Ahat [919]

Answer:

its how life meets water and earth meets air.

Explanation:

4 0
3 years ago
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