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3 years ago

9

An atom that has 117 protons in its nucleus has not yet been made. Once this atom is made, to which group will element 117 belon

g ? Explain ur answer

1 answer:

Pavel [41]3 years ago

3 0

In nomine patris, et filii, et spiritus sancti.

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Consider the system shown in fig. 6-26. the rope and pulley have negligible mass, and the pulley is frictionless. the coefficien

Anestetic [448]

Need and answer choice if you have one

4 0

3 years ago

The Palo Verde nuclear power generator of Arizona has three reactors that have a combined generating capacity of 3.937×109 W . H

den301095 [7]

Answer:

t = 2.68 x 10¹⁴ years

Explanation:

First we need to find the amount of energy that Sun produce in one day.

Energy = Power * Time

Energy of Sun in 1 day = (3.839 x 10²⁶ W)(1 day)(24 hr/1 day)(3600 s/ 1 hr)

Energy of Sun in 1 day = 3.32 x 10³¹ J

Now, the time required by the nuclear power generator, in years, will be:

Energy of power generator = Energy Sun in 1 day = 3.32 x 10³¹ J

3.32 x 10³¹ J = Power * Time

3.32 x 10³¹ J = (3.937 x 10⁹ W)(t years)(365 days/1 year)(24 hr/1 day)(3600 s/ 1 hr)

t = 3.32 x 10³¹ /1.24 x 10¹⁷

<u>t = 2.68 x 10¹⁴ years</u>

8 0

3 years ago

4. A net force of 15 N is exerted on a book to cause it to accelerate at a rate of 5 m/s2. Determine the mass of the

kenny6666 [7]

Answer:

<h2>3 kg </h2>

Explanation:

The mass of an object given it's force and acceleration can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{15}{5} = 3 \\$

We have the final answer as

<h3>3 kg</h3>

Hope this helps you

4 0

2 years ago

PLEASE HELP MEEE!!! QUESTION 3

siniylev [52]

2 is the answer so mark it

7 0

3 years ago

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac

Anarel [89]

Answer:

$R = 24.3 m$

Explanation:

As we know that the orbital speed is given as

$v = \sqrt{\frac{GM}{R + h}}$

here we know that

v = 5500 m/s

$R = 4.48 \times 10^6 m$

$h = 630 km$

now we have

$5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}$

$M = 2.34 \times 10^24 kg$

now acceleration due to gravity of planet is given as

$a = \frac{GM}{R^2}$

$a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}$

$a = 7.7 m/s^2$

now range of the projectile on the surface of planet is given as

$R = \frac{v^2 sin2\theta}{g}$

$R = \frac{14.6^2 sin(2\times 30.8)}{7.7}$

$R = 24.3 m$

3 0

3 years ago