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Marat540 [252]
2 years ago
10

Also confused on this can someone please help??!

Physics
1 answer:
irga5000 [103]2 years ago
7 0
The answer is 167 pounds.
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What are the size and direction of the force that is the sum of a force of 3 n acting south and a force of 6 n acting north?
V125BC [204]
The resultant force is 3 N north.
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3 years ago
A wave passes along the surface of the water in a ripple tank. Describe the motion of a molecule on the surface as the water pas
laila [671]

Answer:

Explanation:

Water waves are generally a transverse wave which do not cause permanent displacement of molecules of the medium. Transverse waves are waves in which the direction of propagation of the wave is perpendicular to the direction of vibration of the particles of the medium.

As the wave propagates from one point to another on the surface of water transferring energy, a molecule of water on its surface vibrates upwards and downwards. Its motion is perpendicular to the direction of propagation of the wave. After the vibration, it comes back to its initial position.

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2 years ago
a car traveling at a velocity of 2 m/s undergoes an acceleration of 4.5 m/s^2 over a distance of 340 m. How fast will it be goin
ra1l [238]
Vi = 2m/s
a= 4.5 m/s 
d= 340 m
vf= ?

use this equation ...  vf^2=vi<span>^2+2ad

you should get vf = 55.3
hope this helps </span>
3 0
3 years ago
A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos
lukranit [14]

Answer:

The inductor contains N = 523962.32 loops  

Explanation:

From the question we are told that

     The capacitance of the capacitor is  C =  286nF = 286 * 10^{-9} \  F

      The resonance frequency is  f = 18.0 kHz =  18*10^{3} Hz

       The diameter is  d =  1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m

       The  of the air-core inductor is l = 12 \ m

        The permeability of free space is  \mu_o = 4 \pi *10^{-7} \ T \cdot m/A

 

Generally the inductance of this air-core inductor is mathematically represented as

              L =  \frac{\mu_o * N^2 \pi d^2}{4 l}

This inductance can also be mathematically represented as

               L = \frac{1}{w^2}

Where w is the angular speed mathematically given as

             w = 2 \pi f

So

            L =  \frac{1}{4 \pi ^2 f^2}

Now equating the both formulas for inductance

         \frac{\mu_o * N^2 \pi d^2}{4 l}  =  \frac{1}{4 \pi ^2 f^2}

making N the subject of  the formula

              N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C}  }

              N =  \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }

             

 Substituting value

            N =  \frac{1}{ 3.142  * 18*10^{3} * 0.00011 }  \sqrt{\frac{12}{ 3.142  * 4 \pi *10^{-7}* 286 *10^{-9}} }

              N = 523962.32 loops  

4 0
3 years ago
A student connects four AA batteries (1.5 V each) in series to light up a light bulb. The circuit has a resistance of 35 2. How
allsm [11]

Answer:

D

Explanation:

6/35=0.17

8 0
3 years ago
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