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IgorLugansk [536]
3 years ago
6

A commuter train travels from the airport for half an hour at 110 km/hr until it reaches the

Physics
1 answer:
nikklg [1K]3 years ago
7 0

Answer:

it would be 10km

Explanation:

because that reaches max height before it falls stupid

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The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s is
Arada [10]
The formula for kinetic energy is: KE = 1/2 x mass x velocity squared. We can plug in the information for the ball into this equation to determine the kinetic energy. The mass is 0.5 kg and the velocity is 10 meters per second. So: KE = (1/2)(0.5)(10)^2. Thus the kinetic energy of the ball is 25 J. The unit Joule (J) is the same as kgm^2/s^2 and is the SI unit for kinetic energy.
3 0
4 years ago
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How much work (MJ) must a car produce to drive 125 miles if an average force of 306 N must be maintained to overcome friction?
baherus [9]

Answer:

The right solution is "61.557 MJ". A further explanation is given below.

Explanation:

The given values are:

Force,

F = 306 N

Drive,

D = 125 miles,

i.e.,

  = 201168

meters

As we know,

The work done will be:

= F\times S

On substituting the given values, we get

= 306\times 201168

= 61557408 \ J

On converting it in "MJ", we get

= 61.557\times 10^6 \ J

= 61.557 \ MJ

6 0
3 years ago
Two material objects cannot occupy the same space at the same time, but two different disturbances of matter can. True or False?
Alex777 [14]

Answer:

false

Explanation:

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4 years ago
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How to find the energy of a wave
andreyandreev [35.5K]
Ask my friend TheBrain.
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3 years ago
A transmission line consisting of two concentric circular cylinders of metal, with radii a, and b, with b > a, is filled with
AURORKA [14]

Answer:

P = √( μ / ε ) × πa^2 |Jo|^2 ln {b/a}.

Explanation:

So, we will be making use of the data or parameters given in the question above;

=>" radii a, and b, with b > a, is filled with a uniform dielectric material with permittivity ε and permeability μ0."

=> " A TEM mode is propagated along the line and the peak value of magnetic field when rho = a is B0."

So, we will be making use of the two equations below;

Ë = ( λ/ 2πEP) × P'. --------------------------(1)..

B' = √ μE × ( λ/ 2πEP) × P' --------------(2).

Where equation (1) and (2) represent Gauss' law and magnetic field equation respectively.

Jo = 1/√ μE × λ/2 × π × a.

When we solve for charge per unit length, we have;

λ = 2 × π × Jo × a × √ μE.

The energy flux,s = E' × J'= √μE× |Jo(Z) |^2 × a^2/b^2 { cos^2 kz - wt + Avg Jo} Z'.

Hence, the time. Average power flux = 1/2× √ μE× |Jo(Z) |^2 × a^2/b^2 × Z'.

Therefore, P = ∫z' . <s> da

P = ∫ ∫ 1/2× √ μE× |Jo(Z) |^2 × a^2/b^2 × Z' pd pd p Θ.

(Take limit on the first at second integration as : 2π,0 and b,a).

P = √( μ / ε ) × πa^2 |Jo|^2 ln {b/a}.

5 0
3 years ago
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