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VLD [36.1K]
3 years ago
6

Cross-Text Discussion​

Physics
1 answer:
marysya [2.9K]3 years ago
8 0

Answer:

idk

Explanation:

AND WAZZZZZZZZZZZZZUP

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A sound wave travels at 379 m/sec and has a wavelength of 8 meters. Calculate its frequency and period.
olga nikolaevna [1]

\qquad\qquad\huge\underline{{\sf Answer}}

Wavelength, Wavespeed and frequency depends on each other in the following way :

\qquad \sf  \dashrightarrow \: frequency =  \dfrac{wave \:  \: velocity}{wavelength}

\qquad \sf  \dashrightarrow \: f =  \dfrac{389}{8}

\qquad \sf  \dashrightarrow \: f \approx 48.62 \:  \: hertz

And we know the Reciprocal relationship between frequency and period ~

So, let's find it's period •

\sf{\qquad \sf  \dashrightarrow \: t = \dfrac{8}{389}}

\sf{\qquad \sf  \dashrightarrow \: t \approx 0.02 sec }

6 0
2 years ago
HELPP WITH THESE QUESTIONS PLEASEEE!!
lubasha [3.4K]
Pe=1/2Kx^2
Half times spring constant times distance squared over time
5 0
3 years ago
If the only force exerted on a star far from the center of the Galaxy (r = 7.40 ✕ 1020 m) is the gravitational force exerted by
lina2011 [118]

Answer:

The value is  v = 1.309*10^{5}\ m/s

Explanation:

The radius is r = 7.40 *10^{20} \  m

The mass of the ordinary matter is M_{rod} =  1.90 *10^{41}\  kg

Generally the speed of the star is mathematically represented as

         v = \sqrt{\frac{G * M}{r} }

Here G is the gravitational constant with a value

        G = 6.67384 * 10^{-11}

So

      v = \sqrt{\frac{6.67384 * 10^{-11} * 1.90 *10^{41}}{7.40 *10^{20}} }

=>    v = 1.309*10^{5}\ m/s

8 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
4 years ago
Please help I don't know how to answer these questions!
Yuki888 [10]

1) The potential energy is the most at the highest position and the least at the equilibrium position

2) The kinetic energy is the most at the equilibrium position and  the least at the highest position

Explanation:

1)

The potential energy of an object is the energy possessed by the object due to its position in a gravitational field; mathematically, it is given by

PE=mgh

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object relative to the ground

For the pendulum in this problem, m is the mass of the bob, and h is the height of the above relative to the ground. We see from the formula that the potential energy is directly proportional to the height:

PE\propto h

This means that:

  • The potential energy is the most when the bob is at the highest position
  • The potential energy is the least when the bob is at the equilibrium position,  which is the lowest position

2)

We can solve this part by applying the law of conservation of energy: in fact, the total mechanical energy of the pendulum (sum of potential and kinetic energy) is constant at any time during the motion,

E=KE+PE=const.

where KE is the kinetic energy.

From the equation above, we observe that:

  • When PE is maximum, KE must be at minimum
  • When PE is minimum, KE must be maximum

Therefore, this implies that:

  • The kinetic energy is the most when the potential energy is the least, i.e. at the equilibrium position
  • The kinetic energy is the least when the potential energy is the most, i.e. at the highest position

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

6 0
3 years ago
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