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2 years ago

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Neil Armstrong, the first human being to step on the moon, left a foot print that is essentially unchanged nearly 50 years later

. How is it possible for this, and all the other footprints, tracks and markings, left on the lunar surface to remain unchanged for so long?

1 answer:

Likurg_2 [28]2 years ago

6 0

Since the boot-print was left there nearly 50 years ago, there has been very little wind and very little rain in that area, and plus, there have been very few people or other animals walking around in that spot to disturb it.

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An athlete kicks a soccer ball that starts at rest so that it leaves their foot with a speed of 10m/s from the top o f a rectang

kirza4 [7]

Answer:

$a=500m/s^2$

Explanation:

We need only to apply the definition of acceleration, which is:

$a=\frac{v_f-v_i}{t_f-t_i}$

In our case the final velocity is $v_f=10m/s$ , the initial velocity is $v_i=0m/s$ since it departs from rest, the final time is $t_f=0.02s$ and the initial time we are considering is $t_i=0s$

So for our values we have:

$a=\frac{10m/s-0m/s}{0.02s-0s}=500m/s^2$

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2 years ago

If earth stops rotating then what will be the value of 'g' at poles

Marianna [84]

The value of 'g' is not affected by rotation at any place on Earth.

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3 years ago

Most nuclear waste in the US is ____.

solmaris [256]

The most waste is Yucca mtn. in Nevada
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2 years ago

Read 2 more answers

PLSS HELP AND I WILL GIVE BRAINLY TO THE ONE WHO ANSWERS CORRECTLY

Ierofanga [76]

The potential difference,electric current ,resistance and new electric current will be 12 V,4 A,3 Ω,2 A.

<h3>What is resistance?</h3>

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.

The energy in terms of the charge and potential difference is;

E= qV

60=5 C × V

V= 12 V

The electric current is found as;

$\rm I= \frac{q}{t} \\\\ I= \frac{5}{1.25}\\\\ I=4 \ A$

From the ohm's law;

V=IR

12=4 ×R

R=3Ω

If the voltage is constant and the resistance is doubled, then the new electric current is half of the previous condition;

$\rm I'=\frac{I}{2} \\\\ I'=\frac{4}{2} \\\\\ I'=2 \ A$

Hence, the potential difference,electric current ,resistance and new electric current will be 12 V,4 A,3 Ω,2 A.

To learn more about the resistance, refer to the link;

brainly.com/question/20708652

#SPJ1

7 0

2 years ago

The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

irina [24]

Part a)

At t = 0 the position of the object is given as

$x = 0$

At t = 2

$x = 2 sin(\pi/2) = 2cm$

so displacement of the object is given as

$d = 2 - 0 = 2cm$

so average speed is given as

$v_{avg} = \frac{2}{2} = 1 cm/s$

Part b)

instantaneous speed is given by

$v = \frac{dy}{dt}$

$v = 2cos(\pi t/4 ) * \frac{\pi}{4}$

now at t= 0

$v = \frac{\pi}{2} cm/s$

at t = 1

$v = 2 cos(\pi/4) * \frac{\pi}{4}$

$v = \frac{\pi}{2\sqrt2}$

at t = 2

$v = 0$

Part c)

Average acceleration is given as

$a_{avg} = \frac{v_f - v_i}{t}$

$a_{avg} = \frac{0 - \frac{\pi}{2}}{2}$

$a = -\frac{\pi}{4} cm/s^2$

Part d)

Now for instantaneous acceleration

As we know that

$a =- \omega^2 y$

at t = 0

$a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2$

at t = 1

$y = \sqrt2 cm$

now we have

$a = -\frac{\pi^2}{16}*\sqrt2$

At t = 2 we have

$y = 2 cm$

$a = -\frac{\pi^2}{16}*2$

$a = -\frac{\pi^2}{8}$

<em>so above is the instantaneous accelerations</em>

7 0

3 years ago