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3 years ago

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To celebrate a victory, a pitcher throws her glove straight upward with an initial speed of 5.3 m/s. How long does it take for t

he glove to reach its maximum height

1 answer:

Elena-2011 [213]3 years ago

3 0

Hello!!

For the maximum height the final velocity is zero, <u>because can't up more.</u>

Then, use the formula:

V = Vi + gt

Replacing, we have:

0 m/s = 5,3 m/s + (-9,8 m/s² * t)

0 m/s - 5,3 m/s = -9,8 m/s² * t

(-5,3 m/s) / -9,8 m/s² = t

t = 0,54 s

The time it will take to reach the maximum height is <u>0,54 seconds.</u>

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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?

$3.081x10^{16}m$

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

$9.46x10^{15}m$

(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 $1pc = 2.06x10^5 AU$

<em>a) How many parsecs are there in one astronomical unit? </em>

$1AU . \frac{1pc}{2.06x10^5AU}$ ⇒ $4.85x10^{-6}pc$

<em>(b) How many meters are in a parsec? </em>

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

<em>(c) How many meters in a light-year? </em>

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

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

$x = 9.46x10^{15}m$

<em>(d) How many astronomical units in a light-year?</em>

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

<em>(e) How many light-years in a parsec?</em>

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

5 0

3 years ago

In this circuit (see picture), which resistor will draw the least power?

Basile [38]

A few different ways to do this:

Way #1:
The current in the series loop is (12 V) / (total resistance) .
(Turns out to be 2 Amperes, but the question isn't asking for that.)

In a series loop, the current is the same at every point, so it's
the same current through each resistor.

The power dissipated by a resistor is (current)² · (resistance),
and the current is the same everywhere in the circuit, so the
smallest resistance will dissipate the least power. That's R1 .

And by the way, it's not "drawing" the most power. It's dissipating it.

Way #2:
Another expression for the power dissipated by a resistance is

(voltage across the resistance)² / (resistance) .

In a series loop, the voltage across each resistor is

[ (individual resistance) / (total resistance ] x battery voltage.

So the power dissipated by each resistor is

(individual resistance)² x [(battery voltage) / (total resistance)²]

This expression is smallest for the smallest individual resistance.
(The other two quantities are the same for each individual resistor.)
So again, the least power is dissipated by the smallest individual resistance.
That's R1 .

Way #3: (Einstein's way)
If we sat back and relaxed for a minute, stared at the ceiling, let our minds
wander, puffed gently on our pipe, and just daydreamed about this question
for a minute or two, we might have easily guessed at the answer.

===> When you wire up a battery and a light bulb in series, the part
that dissipates power, and gets so hot that it radiates heat and light, is
the light bulb (some resistance), not the wire (very small resistance).

3 0

3 years ago

what is the kinetic energy of a 1 kilogram ball is thrown into the air with an initial velocity of 3 m/sec

s344n2d4d5 [400]

Data:
m (mass) = 1 Kg
s (speed) = 3 m/s
Kinetic energy = ? (Joule)

Formula (Kinetic energy)
$E_{k} = \frac{m*s^2}{2}$

Solving:
$E_{k} = \frac{m*s^2}{2}$
$E_{k} = \frac{1*3^2}{2}$
$E_{k} = \frac{1*9}{2}$
$E_{k} = \frac{9}{2}$
$\boxed{\boxed{E_{k} = 4.5\:J}}\end{array}}\qquad\quad\checkmark$

3 0

3 years ago

Sonar in pipeline inspections

butalik [34]

Makes no sense get a better question

4 0

3 years ago

A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?

garri49 [273]

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

$U_i +K_i = U_f + K_f$

where:

$U_i$ is the initial potential energy, at the bottom

$K_i$ is the initial kinetic energy, at the bottom

$U_f$ is the final potential energy, at the top

$K_f$ is the final kinetic energy, at the top

We can rewrite the equation as:

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m is the mass of the hopper

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 0$ is the initial height

u is the launch speed of the hopper

$h_f = 1.20 m$ is the maximum altitude reached by the hopper

v = 0 is the final speed (which is zero when the hopper reaches the maximum height)

Solving the equation for u, we find the launch speed of the hopper:

$u=\sqrt{2gh_g}=\sqrt{2(9.8)(1.20)}=4.8 m/s$

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0

3 years ago