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3 years ago

Convert the speed of light 3.0x10^8 m/s to km/day

Physics

2 answers:

Aleksandr [31]3 years ago

7 0

Answer: $2.592 \times 10^{8}km/day$

$1 m = 0.001 km\\ 1 s= 1.157\times10^{-5} days\\ 1 m/s = \frac {0.001}{1.157\times10^{-5}} km/day = 86.4 km/day \\ 3.0\times 10^{8} m/s = 3.0\times 10^{8} m/s \times \frac {86.4 km/day}{1m/s} =2.592 \times 10^{8}km/day$

just olya [345]3 years ago

6 0

Answer:

$2592*10^7\frac{km}{day}$ .

Explanation:

We need to know that 1km = 1000 m= 10^3m and 1 day = 86400 = 864 * 10^2s.

$3*10^8 \frac{m}{s} = 3*10^8\frac{m}{s} \frac{86400 s}{1 day}$

$= 3*10^8*864*10^2\frac{m}{day} = 2592 * 10^{10} \frac{m}{day}$

Now, $2592 * 10^{10} \frac{m}{day}\frac{1km}{10^3m}$

$= 2592 * 10^{10}*10^{-3}\frac{km}{day} = 2592*10^7\frac{km}{day}$ .

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Ph11_UnitPacket2019

frozen [14]

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2 years ago

An 88 kg worker stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale reads 900 N. Fi

Aleks [24]

Answer:

The magnitude is "3.8 m/s²", in the upward direction.

Explanation:

The given values are:

Mass,

m = 88 kg

Scale reads,

T = 900 N

As we know,

⇒ $N=mg$

On substituting the given values, we get

⇒ $=88\times 9.8$

⇒ $=862.4 \ N$

Now,

⇒ $T=mg-ma$

On substituting the given values in the above equation, we get

⇒ $900=862.4-9.8 a$

On subtracting "862.4" from both sides, we get

⇒ $900-862.4=862.4-9.8 a-862.4$

⇒ $37.6=-9.8a$

⇒ $a=-\frac{37.6}{9.8}$

⇒ $a=3.8 \ m/s^2$ (upward direction)

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2 years ago

A light bulb uses 0.5 amperes from a source of 120 volts. How much power is used by the bulb?

Sladkaya [172]

Power = voltage(V) * current(I)

= 120 * 0.5

Power = 60 watts

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2 years ago

When a pendulum is at the midpoint of its oscillation, hanging straight down, which statement is true?

Svetach [21]

When a pendulum is at the midpoint of its oscillation, hanging straight down ...

-- that's the fastest it's going to swing, so its kinetic energy is maximum;
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'c' is your choice.

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2 years ago

Read 2 more answers

Assume this 1.20-mm-radius copper wire is electrically neutral in the Earth reference frame, in which it is at rest and carrying

agasfer [191]

Answer:

The charge density in the system is $4.25*10^4C/m$

Explanation:

To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.

Our data given correspond to:

$r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C$