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3 years ago

U. O. IS

Physics

1 answer:

juin [17]3 years ago

7 0

b. 6,900 km/hr

Explanation:

Given parameters:

Mass of George = 70kg

Mass of bus = 5700kg

Speed of bus = 85km/hr

Unknown:

Speed of George to stop the bus = ?

Solution:

For George to stop the bus after collision, his momentum must be equal to that of the bus.

Momentum is the product of mass and velocity of a moving body.

Momentum = mass x velocity

Momentum of George = Momentum of bus

mass of bus x velocity of bus = mass of George x velocity of George

Velocity of George = $\frac{Mass of bus x Velocity of bus}{mass of George}$

Velocity of George = $\frac{5700 x 85}{70}$ = 6,921.4km/hr

This is approximately 6,900 km/hr

learn more:

Momentum brainly.com/question/2990238

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calculate the energy spent on spraying a drop of mercury of 1 cm radius into 10^6 droplets all of same radius. surface tension o

WARRIOR [948]

Answer:

ANS : .Energy spent on spraying = $4.3542*10^{-4}J$

Explanation:

Given:

Radius of mercury = 1cm initially ;
split into $10^{6}$ drops ;

Thus, volume is conserved.

i.e ,

$\frac{4}{3} \pi R_{o}^{3} = 10^{6}*\frac{4}{3} \pi R_{n}^{3}\\R_{n}=\frac{R_{o}}{10^{2}} = \frac{1cm}{100} = 0.01 cm$

Energy of a droplet = $T$ Δ $A$

Where ,

T is the surface tension
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Initial energy $E_{i} = T*A_{i}\\= 0.0035 * 4 *\pi *0.01^{2}\\=4.398*10^{-6}J$

Final energy $E_{f}=10^{6}*T*A_{f}\\=10^{6}*0.0035*4*\pi *(0.0001)^2\\=4.39823*10^{-4}$

∴ .Energy spent on spraying = $=E_{f}-E{i}\\=(439.823-4.39823)*10^{-6}\\=4.3542*10^{-4}J$

ANS : .Energy spent on spraying = $4.3542*10^{-4}J$

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3 years ago

Label each statement 1 2 or 3 for the law of motion it illudtrates

KiRa [710]

Answer:

Explanation:

I'm pretty sure these are right. you might want to go back and check the first and third, but the other 4 are right

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4 years ago

The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The p

Rudiy27

Answer:

a) How many parsecs are there in one astronomical unit?

$4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$3.081x10^{16}m$

$9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$63325AU$

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ( $1.496x10^{11} m$ ) is defined as 1 astronomical unit:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

Since p is small it can be represent as:

$p(rad) = \frac{1AU}{d}$ (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

$p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}$

$p('') = 2.06x10^5 p(rad)$

$p(rad) = \frac{p('')}{2.06x10^5}$ (2)

Then, equation 2 can be replace in equation 1:

$\frac{p('')}{2.06x10^5} = \frac{1AU}{d}$

$\frac{d}{1AU} = \frac{2.06x10^5}{p('')}$ (3)

From equation 3 it can be see that $1pc = 2.06x10^5 AU$

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(b) How many meters are in a parsec?

$2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}$ ⇒ $3.081x10^{16}m$

(c) How many meters in a light-year?

To determine the number of meters in a light-year it is necessary to use the next equation:

$x = c.t$

Where c is the speed of light ( $c = 3x10^{8}m/s$ ) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

$x = (3x10^{8}m/s)(31536000s)$

$x = 9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}$ ⇒ $63325AU$

(e) How many light-years in a parsec?

$2.06x10^{5}AU . \frac{1ly}{63235AU}$ ⇒ $3.26ly$

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2 years ago

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3 years ago

Read 2 more answers