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Neko [114]
2 years ago
11

Which of the following has the greatest inertia ?

Physics
1 answer:
Elden [556K]2 years ago
8 0

Answer:

b

Explanation:

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Generated for what there no waves shown
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The displacement of a wave traveling in the positive x-direction is y(x, t)|= (3.5 cm)cos(2.7x − 92t), where x is in m and t is
zubka84 [21]

Answer

Given,

y(x, t) = (3.5 cm) cos(2.7 x − 92 t)

comparing the given equation with general equation

y(x,t) = A cos(k x - ω t)

 A = 3.5 cm  , k = 2.7 rad/m    , ω = 92 rad/s

we know,

a) ω =2πf

   f = 92/ 2π

   f = 14.64 Hz

b) Wavelength of the wave

 we now, k = 2π/λ

      2π/λ = 2.7

      λ = 2 π/2.7

      λ = 2.33 m

c) Speed of wave

     v = ν λ

     v = 14.64 x 2.33

     v = 34.11 m/s

5 0
3 years ago
Please help! you are amazing! brainliest too!
german
I think it’s B hope it helps:)
6 0
2 years ago
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A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
3 years ago
a skydiver jumps out of a plane. describe how gravitational potential energy changes as the skydiver falls. describe how the sky
Oliga [24]

gravitational potential is directly proportional to the height of the object relative to a reference line and is given as

PE = mgh                        

where m = mass of object , g = acceleration due to                    

gravity   and  h = height of the object above the reference line .


as the skydiver falls , its height above the ground decrease and hence the gravitational potential energy of the skydiver decrease.

as per conservation of energy , total energy of the skydiver must remain constant all the time . hence the decrease in potential energy appears as increase in kinetic energy by same amount to keep the total energy constant

KE + PE = Total energy

so as the skydiver falls , it gains speed and hence the kinetic energy of skydiver increase since kinetic energy is directly proportional to the square of the speed.


when the parachute opens, the skydiver experience force in upward which tries to balance the weight of the skydiver. hence the speed of the skydiver decrease until upward force becomes equal to the downward force. hence the kinetic energy decrease just after the parachute opens

4 0
3 years ago
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