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3 years ago

a 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?

1 answer:

iogann1982 [59]3 years ago

4 0

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)² = 40,218.75 joules

Final KE = (1/2) (1430 kg) (11.0 m/s)² = 86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec = 4,978.1 watts .
_______________________________

One interesting thing to notice while you're (I'm) solving this problem
is the fact that although the car's speed only increased by about 47%,
its kinetic energy more than doubled ... increased by about 115%.
That's because kinetic energy is proportional to the SQUARE of the
speed.

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A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu

Montano1993 [528]

Answer:

The angle of separation is $\Delta \theta = 0.93 ^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta _ i = 56^o$

The refractive index of violet light in diamond is $n_v = 2.46$

The refractive index of red light in diamond is $n_r = 2.41$

The wavelength of violet light is $\lambda _v = 400nm = 400*10^{-9}m$

The wavelength of red light is $\lambda _r = 700nm = 700*10^{-9}m$

Snell's Law can be represented mathematically as

$\frac{sin \theta_i}{sin \theta_r} = n$

Where $\theta_r$ is the angle of refraction

=> $sin \theta_r = \frac{sin \theta_i}{n}$

Now considering violet light

$sin \theta_r__{v}} = \frac{sin \theta_i}{n_v}$

substituting values

$sin \theta_r__{v}} = \frac{sin (56)}{2.46}$

$sin \theta_r__{v}} = 0.337$

$\theta_r__{v}} = sin ^{-1} (0.337)$

$\theta_r__{v}} = 19.69^o$

Now considering red light

$sin \theta_r__{R}} = \frac{sin \theta_i}{n_r}$

substituting values

$sin \theta_r__{R}} = \frac{sin (56)}{2.41}$

$sin \theta_r__{R}} = 0.344$

$\theta_r__{R}} = sin ^{-1} (0.344)$

$\theta_r__{R}} = 20.12^o$

The angle of separation between the red light and the violet light is mathematically evaluated as

$\Delta \theta = \theta_r__{R}} - \theta_r__{V}}$

substituting values

$\Delta \theta =20.12 - 19.69$

$\Delta \theta = 0.93 ^o$

6 0

3 years ago

A long, straight, horizontal wire carries current toward the east. A proton moves toward the east alongside and just south of th

kondor19780726 [428]

Answer:

north direction.

Explanation:

The wire carries a current towards the east . The magnetic field will make circular path around the wire in clockwise direction so at a point just south of wire , magnetic field will be into the plane of paper containing wire. If we take east as x -axes , north as y axes then out of plane will form z axes. Hence direction of magnetic field will be - z direction .

Magnetic field can be represented as - B k

Proton is moving towards east ie in + x direction so it can be represented as follows

velocity = V i

Force F = q( V i x -B k)

= ( BqV) j or + ve j direction or along north direction

So direction of force will be along north direction.

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2 years ago

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Reactivity is the answer

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3 years ago

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What statement applies to the horizontal rows or periods in the periodic table? A. The properties are all the same. B. The eleme

Sliva [168]

The right answer for the question that is being asked and shown above is that: "C. . The properties change going across each row. " the statement that applies to the horizontal rows or periods in the periodic table is that the properties change going across each row.

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3 years ago

Charge q1 is distance r from a positive point charge q. charge q2=q1/3 is distance 2r from q. what is the ratio u1/u2 of their p

makvit [3.9K]

We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space).

The potential energy of charge q is proportional to 
∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, 

so if r₂ = 3r₁ and q₂ = q₁/4, then 
U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁) 
= 4•3 = 12.

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2 years ago