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r-ruslan [8.4K]
3 years ago
10

The volume of sound wave corresponds to which part of the wave? A. Velocity B. Frequency C. Amplitude D. Wavelength

Physics
1 answer:
Alborosie3 years ago
5 0
The answer is C i hope its right
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The relationship between power, force, and velocity is ______?
Kipish [7]

Answer: P = FV becuase all of them are equal in strength.

Explanation: In the straight forward cases where a constant force moves an object at constant velocity, the power is just P = Fv. In a more general case where the velocity is not in the same direction as the force, then the scalar product offorce and velocity must be used. P = Power, F = Force, V = Velocity.

Hoped This Helped!

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1 year ago
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A 1400 kg car is moving at 33.8 m/s when a force is applied the opposite direction of the car's motion. The car slows down to 21
zubka84 [21]

Solve for acceleration:

<em>a</em> = (21.4 m/s - 33.8 m/s) / (4.7 s)

<em>a</em> ≈ -2.6 m/s²

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<em>F</em> = (1400 kg) <em>a</em> ≈ -3700 N

The minus sign tells you the force points in the opposite direction of the car's motion. Its magnitude is always positive, so <em>F</em> = 3700 N.

3 0
3 years ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

Generally the charge is  mathematically represented as

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area which is mathematically represented as

       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 So

       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
3 years ago
Why is weathering slow in cold, dry places?
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Rate of weathering depends on temperature and moisture. Cold dry places have less water to weather things


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Explanation:

I took the test and that was the answer

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