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Rus_ich [418]
4 years ago
10

In a stormy sea, 2 waves pass a fixed point every second, and the waves are 10 meters apart. What is the speed of the wave?

Physics
1 answer:
Kitty [74]4 years ago
7 0

             Speed of a wave  =  (frequency) x (wavelength)

                                         =  (2 per second) x (10 m)

                                         =      20 m/s .

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An electron and a proton are both released from rest, midway between the plates of a charged parallel-plate capacitor. The only
topjm [15]

Answer:

Explanation:

Let the potential difference between the middle point and one of the plate be ΔV .

electric potential energy will be lost and it will be converted into kinetic energy .

Electrical potential energy lost = Vq , where q is charge on charge particle .

For proton

ΔV× q = 1/2 M V² ( kinetic energy of proton )

where M is mass and V be final velocity of proton .

For electron

ΔV× q = 1/2 m v² ( kinetic energy of electron  )

where m is mass and v be final velocity of electron . Charges on proton and electron are same in magnitude .

As LHS of both the equation are same , RHS will also be same . That means the kinetic energy of both proton and electron will be same

1/2 M V² =  1/2 m v²

(V / v )² = ( m / M )

(V / v ) = √ ( m / M )

In other words , their velocities  are  inversely proportional to square root of their masses .

4 0
3 years ago
8. A car initially has 100 J of total energy. After driving down the road, the car's
sveta [45]

The work done on the car is -20 J.

Work done on the car is negative, meaning that the car actually does work on the external system.

<h3>Energy and law of conservation of energy</h3>
  • Energy is the ability to do work
  • the law of conservation of energy states that the total energy in a system is conserved

From the law of conservation of energy, the initial energy of the car before it moves down the road remains constant or unchanged.

  • Initial energy = 100 J
  • Initial energy = Final energy - work done on car
  • Final Energy = Work done on car + initial energy

80J = Work done on car + 100 J

Work done on car = 80 - 100J

Work done on car = -20 J

Hence, the work done on the car is -20 J

Work done on car is negative.

Since work done on the car is negative, it means that the car actually does work on the external system. Hence, the decrease in the energy of the car.

Learn more about energy and work at: brainly.com/question/13387946

8 0
2 years ago
QC In ideal flow, a liquid of density 850 kg / m³ moves from a horizontal tube of radius 1.00cm into a second horizontal tube of
kvv77 [185]

The  volume flow rates for ∆P is 6.81m³/s .

<h3>What is pressure?</h3>

The amount of force applied on perpendicular to the surface of an object per unit area. The unit of it is pascal.

According to bernaulli's theorem theorem

P+1/2pV²+pgy = constant

where p fluid density

g is acceleration due to gravity, pressure at elevation,v is Velocity at elevation ,y is height of elevation.

As there are two tubes then the height of tube 1 is equal to height of tube two .

P1-P2=1/2p(Vd²-Vl²)

The flow rate of liquid is  A1V1=A2V2 .

rest is attached in image

to learn more about Pressure click here brainly.com/question/12971272

#SPJ4

4 0
2 years ago
A technique in which the muscles are stretched by an outside force is called _____.
geniusboy [140]
A technique in which the muscles are stretched by an outside force is called Passive Stretching 
6 0
3 years ago
A particle with charge 6 mC moving in a region where only electric forces act on it has a kinetic energy of 1.9000000000000001 J
Vesna [10]

Answer:

16.9000000000000001 J

Explanation:

From the given information:

Let the initial kinetic energy from point A be K_A = 1.9000000000000001 J

and the final kinetic energy from point B be K_B = ???

The charge particle Q = 6 mC = 6 × 10⁻³ C

The change in the electric potential from point B to A;

i.e. V_B - V_A = -2.5 × 10³ V

According to the work-energy theorem:

-Q × ΔV = ΔK

-Q \times ( V_B - V_A) = (K_B - K_A)

-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)

15 = (K_B - 1.9000000000000001 \ J)

K_B = 15+ 1.9000000000000001 \ J

\mathbf{K_B =1 6.9000000000000001 \ J}

3 0
3 years ago
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