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VLD [36.1K]
3 years ago
11

If you shout into the grand canyon, your voice travels at the speed of sound (340 m/s) to the bottom of the canyon and back, and

you hear an echo. how deep is the grand canyon at a spot where you can hear your echo 5.2 seconds after you shout?
Physics
1 answer:
Nostrana [21]3 years ago
4 0
Given:
The speed of sound is  340 m/s
Time for the echo is 5.2 s.

Let h = the depth of the canyon.

Because the sound of your voice travels to the bottom of the canyon and back to your ear, the total distance traveled is 2h.

By definition,
distance = velocity * time.
Therefore
2h = (340 m/s)*(5.2 s) = 1768 m
h = 884 m

Answer: The depth is 884 m.

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If 28.0 J of work is done by an external force to move a charge from a potential of 8.0 V 8.0 V to a potential of 4.0 V , 4.0 V,
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Answer:

Change in electric potential energy is -28.0 J

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Electric potential energy is also equal to the change in the configuration of the charge particles.

Thus,

Change in electric potential energy = - Work Done

According to the problem, Work Done is equal to 28 J. Thus,

Change in electric potential energy = -28 J  

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Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y
WITCHER [35]

Question:

A point charge of -2.14uC  is located in the center of a spherical cavity of radius 6.55cm  inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm  from the center of the cavity.  

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 3.65\times 10^5N/C

Explanation:

A point charge ,q = -2.14\times 10^{-6} C is located in the center of a spherical cavity of radius , r =6.55\times 10^{-2}  m inside an insulating spherical charged solid.  

The charge density in the solid , d = 7.35 \times 10^{-4}C/m^3.

Distance from the center of the cavity,R =9.5\times 10^{-2 }m

Volume of shell of charge= V  =(\frac{4\pi}{3})[ R^3 - r^3 ]

Charge on the shell ,Q = V \times d'

Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d

Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}

Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}

Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}

Q =1.7745 \times 10^{-6 }C

Electric field at 9.5\times 10^{-2}m due to shellE1  = \frac{k Q}{R^2}

E1 =  \frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}

E1 =1.769\times 10^6 N/C

Electric field at  9.5\times 10^{-2} due to 'q' at center E2 = \frac{kq}{R^2}

E2 =\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}

E2 =2.134\times 10^6 N/C

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

=[  2.134  - 1.769 ]\times 10^6

= 3.65\times 10^5 N/C

8 0
3 years ago
A wheel that is rotating at 33.3 rad/s is given an angular acceleration of 2.15 rad/s 2. Through what angle has the wheel turned
Neporo4naja [7]

Answer:

The angle through which the wheel turned is 947.7 rad.

Explanation:

initial angular velocity, \omega _i = 33.3 rad/s

angular acceleration, α = 2.15 rad/s²

final angular velocity, \omega_f = 72 rad/s

angle the wheel turned, θ = ?

The angle through which the wheel turned can be calculated by applying the following kinematic equation;

\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\\theta = \frac{\omega_f^2\   -\  \omega_i^2}{2\alpha } \\\\\theta = \frac{(72)^2\   -\  (33.3)^2}{2(2.15)}\\\\\theta = 947.7 \ rad

Therefore, the angle through which the wheel turned is 947.7 rad.

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