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mariarad [96]
3 years ago
5

A ball is launched horizontally at 4 m/s

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
3 0

Answer:

3.5 seconds of flight time; 13.9 m from the base of the cliff

Explanation:

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An atom emits a light wave with a wavelength of 449 nm. what type or color of light does this represent?
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The visible spectrum ranges from 390 nm to 700 nm. the visible spectrum consist of the red ( 620 - 750 nm ) , orange ( 590 - 620 nm ) , yellow ( 570 - 590 nm ) , green ( 495 - 570 nm ) , blue ( 450 - 495 nm ) and violet ( 380 - 450 nm ) so the wave length 449 nm will produce a violet color
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You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an
a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, \mu=6\times 10^{-2}

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

KE=PE+W

\dfrac{1}{2}mv^2=mgh+W

W is the work done by the friction.

W=f\times d

f=\mu mg\ cos\theta

W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}

W=\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}

\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}

v = 8.19 m/s

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8 0
2 years ago
A velocity selector has a magnetic field of magnitude 0.22 T perpendicular to an electric field of magnitude 0.51 MV/m.
ohaa [14]

Answer:

speed of a particle  = 2.31 × 10^{6} m/s

energy of proton required = 27.77 KeV

energy of electron required =  15.171 eV

Explanation:

given data

magnetic field of magnitude = 0.22 T

electric field of magnitude = 0.51 MV/m

to find out

speed of a particle and energy must protons have to pass through undeflected and  energy must electrons have to pass through undeflected

solution

we know that force due to magnetic and electric field is express as

force due to magnetic field B = qvB    ..............1

and force due to electric field E = qE   .....................2

so without deflection force due to magnetic field  = force due to electric field  

so here qvB = qE

and V = \frac{E}{B}    ...................3

put here value

V =  \frac{0.51*10^6}{0.22}

speed of a particle  = 2.31 × 10^{6} m/s

and

now energy of proton required will be here as

energy of proton required = mass of proton × \frac{V^2}{2}

put here value

energy of proton required = 1.67 × 10^{-27} × \frac{(2.31*10^6)^2}{2}

energy of proton required = 4.45 × 10^{-15} J

energy of proton required = 4.45 × 10^{-15} J ÷ (1.602 × 10^{-19}

energy of proton required = 27777.777 eV

energy of proton required = 27.77 KeV

and

now we get here energy of electron required that is

energy of electron required = mass of electron × \frac{V^2}{2}

put here value

energy of electron required = 9.11 × 10^{-31} × \frac{(2.31*10^6)^2}{2}

energy of electron required =  24.305× 10^{-19} J

energy of electron required =  24.305 × 10^{-19} J ÷ (1.602 × 10^{-19}  

energy of electron required =  15.171 eV

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