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2 years ago

How much time does it take light from a flash camera to reach a subject 6.0 meters across a room in scientific notation?

1 answer:

6 0

The time it takes light from a flash camera to reach a subject 6.0 meters across a room in scientific notation is 2.0 *10^-8 s.

Explanation:

Given

t=?

d=6m

v=3*10^8 m/s

we have, v=d/t

here t=d/v

t=6m/3*10^8 m/s

v=2*10^-8 m/s

The time it takes light from a flash camera to reach a subject 6.0 meters across a room in scientific notation is 2.0 *10^-8 s.

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Which process binds together sediment to form new rock?

vekshin1

Answer:

its cementation i got it correct

Explanation:

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3 years ago

If an object falling freely were somehow equipped with an odometer to measure the distance it travels, then the amount of distan

ioda

Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

$h=ut+\frac{1}{2}gt^2$

here u=0 and t=time taken

$h=\frac{1}{2}gt^2$

for $t=1 s$

$h_1=\frac{1}{2}g$

for $t=2 s$

$h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g$

distance traveled in 2 nd sec $=2g-\frac{1}{2}g=\frac{3}{2}g$

for $t=3 s$

$h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g$

distance traveled in 3 rd sec $=\frac{9}{2}g-2g=\frac{5}{2}g$

so we can see that distance traveled in each successive second is increasing

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2 years ago

A transformer consists of 290 primary windings and 824 secondary windings. Part A: If the potential difference across the primar

jasenka [17]

Answer:

Part 1) Voltage in secondary windings is 61.08 Volts

Part 2) Current in secondary windings is 0.53 Amperes

Explanation:

The potential developed in the primary and secondary winding of a transformer are related as

$\frac{N_{p}}{N_{s}}=\frac{V_{p}}{V_{s}}$

where

Np no of turns in primary coil

Ns no of turns in secondary coil

Vp Voltage of turns in primary coil

Vs Voltage of turns in secondary coil

Applying values in the formula we get

$\frac{290}{824}=\frac{21.5}{V_{s}}\\\\\therefore V_{s}=21.5\times \frac{824}{290}=61.08V$

Part 2)

Using Ohm's law the current is given by

$I=\frac{V_{s}}{R}\\\\I=\frac{61.089}{115}=0.53A$

5 0

2 years ago

A particle moving in simple harmonic motion with a period T = 1.5 s passes through the equilibrium point at time t0 = 0 with a v

ch4aika [34]

Answer

given,

Time period= T = 1.5 s

If it's moving through equilibrium point at t₀= 0 with v = 1.0 m/s

v_max=1.00 m/s

we know,

v_ max=A ω

v = A sin (ωt)

-0.50= -1.00 sin (ωt)

sin (ωt) = 0.5

$\omega t = sin^{-1}(0.5)$

$\dfrac{2\pi}{T}\times t =0.524$

$\dfrac{2\pi}{1.5}\times t =0.524$

t = 0.125 s

we have time period T=1.5 it is the time to complete one oscillation

means from eq to right,then left,then eq,then left,then from right to eq

time taken for left = t/4 = 0.125/4 = 0.375 s

smallest value of time

=0.375 + 0.125

= 0.50 sec

7 0

2 years ago

A 15.5 kg mass vibrates in simple harmonic motion with a frequency of 9.73 Hz. It has a maximum displacement from equilibrium of

algol [13]

Answer:

12.14 cm

Explanation:

mass, m = 15.5 kg

frequency, f = 9.73 Hz

maximum amplitude, A = 14.6 cm

t = 1.25 s

The equation of the simple harmonic motion

y = A Sin ωt

y = A Sin (2 x π x f x t)

put, t = 1.25 s, A = 14.6 cm, f = 9.73 Hz

y = 14.6 Sin ( 2 x 3.14 x 9.73 x 1.25)

y = 14.6 Sin 76.38

y = 12.14 cm

Thus, the displacement of the particle from the equilibrium position is 12.14 cm.

6 0

3 years ago