What is the gravitational force on a 35.0 kg object standing on the Earth's surface?

1. A carbon atom contains 6 protons and 6 electrons. In order for the carbon atom to become

jek_recluse [69]

It must gain an electron because if the proton number was to change it would no longer be the same element.

5 0

2 years ago

When the electron in a hydrogen atom moves from n = 6 to n = 1, light with a wavelength of ________ nm is emitted?

Oduvanchick [21]

First we find the energy level with the following formula, where a is the energy level, n1 is the final energy level, n2 is the starting energy level and r is Rydberg's constant in Joules
$a = r \times ( \frac{1}{n1} - \frac{1}{n2} )$
We insert the values

$a = 2.18 \times {10}^{ - 18} \times ( \frac{1}{ {1}^{2} } - \frac{1}{ {6}^{2} } )$
$= 2.12 \times {10}^{ - 18}$
The wavelength is found with this formula, where h is Planck's constant and c is the speed of light
$wavelength = \frac{h \times c}{a}$
Finally we insert the values
$\frac{6.626 \times {10}^{ - 34} \times 3 \times {10}^{8} }{2.12 \times {10}^{ - 18} } = 9.376 \times {10}^{ - 8}$
Which is the same as 93.8 nm

3 0

2 years ago

We know that every object exerts an attraction on every other object and the heavier the object ___ the attraction.

hichkok12 [17]

I don't know what the exact word is, but I do know that the bigger an objects mass is the more it will attract other objects toward it, mainly smaller objects with less mass. it might be gravity or something around those lines....is it a multiple choice question?

6 0

2 years ago

A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a c

Leya [2.2K]

Answer:

$T=83.37N$

Explanation:

Since the object is under a circular motion, according to Newton's second law, when the object is at the top of the circle we have:

$\sum F_y: T-mg=F_c$

Where $F_c$ is the centripetal force and is given by:

$F_c=ma_c=m\frac{v^2}{r}$

Replacing and solving for T:

$T=m\frac{v^2}{r}+mg\\T=0.2kg\frac{(6.38\frac{m}{s})^2}{0.1m}+0.2kg(9.8\frac{m}{s^2})\\T=83.37N$

8 0

3 years ago

Mercury has a radial acceleration of 3.96 × 10−2 m/s2 and its orbital period is T = 88 days. What is the radius of Mercury’s orb

Maslowich

Answer: 58,045,522,878.8 meters

Explanation:

Ok, the data we have is

Period = T = 88 days

Radial acceleration = ar = 3.96x10^-2 m/s^2

And we know that the equation for the radial acceleration is:

ar = v^2/r = r*w^2

Where v is the velocity. r is the radius and w is the angular velocity.

And we know that:

w = 2*pi*f

where f is the frequency, and:

T = 1/f.

Then we can write:

w = 2*pi/T

and our equation becomes:

ar = r*(2*pi/T)^2

Now we solve this for r.

First we need to use the same units in both equations, so we want to write T in seconds.

T = 88 days,

A day has 24 hours, and one hour has 3600 seconds:

T = 88*24*3600 s =7,603,200s

Then:

3.96x10^-2 m/s^2 = r*(2*3.14/7,603,200s)^2

r = (3.96x10^-2 m/s^2) /(2*3.14/7,603,200s)^2 = 58,045,522,878.8 meters

5 0

3 years ago