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3 years ago

What is the deceleration of a bicycle that goes from 40m/s to 20m/s in 5 seconds

Physics

1 answer:

Alex Ar [27]3 years ago

5 0

Answer:

$-4 m/s^2$

Explanation:

The acceleration of the bycicle is given by:

$a=\frac{v-u}{t}$

where

v = 20 m/s is the final velocity of the bike

u = 40 m/s is the initial velocity

t = 5 s is the time interval

Substituting numbers, we find

$a=\frac{20 m/s-40 m/s}{5s}=-4 m/s^2$

and the acceleration is negative, which means that the bycicle is slowing down.

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A circuit consists of an ideal ac generator, a capacitor, and an ideal inductor, all connected in series. The charge on the capa

Serjik [45]

Answer:

$-16\omega sin(\omega t + \pi/4)$

Explanation:

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3 years ago

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In a pith ball experiment, the two pith balls are at rest. The magnitude of the tension in each string is |T|=0.55N, and the ang

Len [333]

Answer:

Explanation: Two pith ball will repel each other . they will remain balaced due to tension in the spring whose one component balances the weight and the other balances the repulsive force on each.

The gravitational force will be balanced by T cos 27.33 and the electrostatic repulsive force will be balanced by T sin27.33

Fg =T cos 27.33

= .55 X .888

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Fq = T sin27.33

=.55 x .459

= .25 N.

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3 years ago

What term refers to a universal fact sometimes based on mathematical equations?

noname [10]

Answer:

Scientific law

Explanation:

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2 years ago

A typical helium-neon laser found in supermarket checkout scanners emits 633-nm-wavelength light in a 1.0-mm-diameter beam with

Fofino [41]

Answer:

Eo = 9.796 x 10^2 N/C

Bo = 3.266 x 10^-6 T

Explanation:

Given

Wavelength λ = 633 nm

Diameter of the beam D = 1.0 mm

Power P = 1.0 mW

Solution

Radius of the beam r = D/2 = 0.5 mm = 0.0005 m

Area of cross section

$A = \pi r^{2} \\A = 3.15 \times 0.0005^{2}\\A = 7.58 \times 10^{-7} m^{2}\\$

Intensity

$I = \frac{P}{A} \\I = \frac{0.001}{7.85\times 10^{-7}} \\I = 1273.885 {W}/{m^{2} }$

Amplitude of Electric Field

$E_{o} = \sqrt{\frac{2I}{ \epsilon_{o}c } } \\E_{o} = \sqrt{\frac{2 \times 1273.88}{ 8.85 \times 10^{-12} \times 3 \times 10^{8} } }\\E_{o} = 9.796 \times 10^{2}N/C$

Amplitude of Magnetic Field

$B_{o} = \sqrt{\frac{2 \mu_{o}I}{c } } \\B_{o} = \sqrt{\frac{2 \times 4 \times \pi \times 10^{-7} \times 1273.88}{ 3 \times 10^{8} } }\\B_{o} = 3.266 \times 10^{-6} T$