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Naddik [55]
3 years ago
11

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t

he impact: one travels in the air and the other in the concrete, and they are 0.90s apart. The speed of sound in air is343 m/s, and in concrete is 3000 m/s.How far away did the impact occur?
Physics
1 answer:
Jet001 [13]3 years ago
4 0
<h2>The impact is occurred at a distance 348.55 m far from person.</h2>

Explanation:

Let s be the distance to the impact position and t be the time when he hears the sound though concrete after impact.

Time when he hears impact through air = t + 0.90

The distance traveled by sound wave in both concrete and air is s.

Speed of sound in air = 343 m/s

Speed of sound in concrete = 3000 m/s

Distance traveled in air = Distance traveled in concrete

343 x Time when he hears impact through air = 3000 x Time when he hears impact through concrete

        343 x (t + 0.90) = 3000 x t

                  3000t - 343 t = 343 x 0.90

                  2657t = 308.7

                      t = 0.116 s

Distance traveled in concrete = 3000 x t = 3000 x 0.116 = 348.55m

So the impact is occurred at a distance 348.55 m far from person.

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4 years ago
initially, a particle is moving at 5.33 m/s at an angle of 37.9° above the horizontal. Two seconds later, its velocity is 6.11 m
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Explanation:

Average acceleration is the change in velocity over the change in time:

a = (v − v₀) / t

In the x direction:

aₓ = (6.11 cos (-54.2°) − 5.33 cos (37.9°)) / 2.00

aₓ = -0.316 m/s²

In the y direction:

aᵧ = (6.11 sin (-54.2°) − 5.33 sin (37.9°)) / 2.00

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7 0
3 years ago
What is the pressure of a 300 lb. object on a 100 sq. in. area?
denis23 [38]

Answer:

D. 3 psi

Explanation:

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3 0
3 years ago
What is the cell emf for the concentrations given? Express your answer using two significant figures.
alisha [4.7K]

Complete Question

A  voltaic cell is  constructed with two Zn^{2+}- Zn electrodes. The  two cell compartment have  [Zn^{2+}] =  1.6 \ M and  [Zn^{2+}] =  2.00*10^{-2} \  M respectively.

What is the cell emf for the concentrations given? Express your answer using two significant figures

Answer:

The value is   E =  0.06 V

Explanation:

Generally from the question we are told that

   The  concentration of [Zn^{2+}] at the cathode is  [Zn^{2+}]_a =  1.6 \ M

    The  concentration of [Zn^{2+}] at the anode is [Zn^{2+}]_c =  2.00*10^{-2} \  M

Generally the the cell emf for the concentration is mathematically represented as

     E =  E^o - \frac{0.0591}{2} log\frac{[Zn^{2+}]a}{ [Zn^{2+}]c}

Generally the E^ois the standard emf of a cell, the value is  0 V

So

      E =  0  -  \frac{0.0591}{2}  * log[\frac{ 2.00*10^{-2}}{1.6} ]

=>      E =  0.06 V

4 0
3 years ago
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