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artcher [175]
3 years ago
8

Balancing Chemical Equations?

Physics
1 answer:
klio [65]3 years ago
8 0

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=2ahUKEwiZxdDzwILfAhUq8IMKHdeMD1oQwqsBMAB6BAgEEAQ&url=https%3A%2F%2Fwww.khanacademy.org%2Fscience%2Fchemistry%2Fchemical-reactions-stoichiome%2Fbalancing-chemical-equations%2Fv%2Fbalancing-chemical-equations-introduction&usg=AOvVaw3kUY-9qu0UPe7-KfbRVBbd        

  go here this is a really good video to help with balacing chemical equations

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The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t
dem82 [27]

Answer:

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

T = 2\pi \sqrt{\frac{r^3}{GM}}

now for the time period of moon around the earth we can say

T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}

here we know that

T_1 = 0.08 year

r_1 = 0.0027 AU

M_e = mass of earth

Now if the same formula is used for revolution of Earth around the sun

T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}

here we know that

r_2 = 1 AU

T_2 = 1 year

M_s = mass of Sun

now we have

\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}

\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}

12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

4 0
3 years ago
A pen rolls off a 0.55–meter high table with an initial horizontal velocity of 1.2 meters/second. At what horizontal distance fr
NARA [144]
To find the horizontal distance multiple the horizontal velocity by the time. Since there is no given time it must be calculated using kinematic equation.

Y=Yo+Voyt+1/2at^2
0=.55+0+1/2(-9.8)t^2
-.55=-4.9t^2
sqrt(.55/4.9)=t
t=0.335 seconds

Horizontal distance
=0.335s*1.2m/s
=0.402 meters
8 0
3 years ago
A 30° incline permanently sits on a 1.1 meter high table. Starting from rest a ball rolls off the incline with a velocity of 2m/
exis [7]
It would be a good game for you but if I get a pic I don’t want you can you come to my crib I just
3 0
3 years ago
In your own words explain coefficient of friction?​
fomenos

Answer:

A coefficient friction is a value that shows the relationship between two objects and the normal reaction between the objects that are involved.

Explanation:

3 0
3 years ago
A horizontal 745 N merry-go-round of radius
Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is  544J

Explanation:

Given :

Weight w = 745 N

Radius r =  1.45 m

Force =  56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62  = ?

Solution:

Step 1:  Finding the Mass of merry-go-round

m = \frac{ weight}{g}

m = \frac{745}{9.81 }

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I =0.5 \times m \times r^2

Substituting the values

Moment of Inertia of solid cylinder I  

=>0.5 \times 76.02 \times (1.45)^2

=> 0.5 \times 76.02\times 2.1025

=> 79.91 kg.m^2

Step 3: Finding the Torque applied T

Torque applied T = F \times r

Substituting the values

T = 56.3  \times 1.45

T = 81.635 N.m

 Step 4: Finding the Angular acceleration

Angular acceleration ,\alpha  = \frac{Torque}{Inertia}

Substituting the values,

\alpha  = \frac{81.635}{79.91}

\alpha = 1.021 rad/s^2

 Step 4: Finding the Final angular velocity

Final angular velocity ,\omega = \alpha \times  t

Substituting the values,

\omega = 1.021 \times  3.62

\omega = 3.69 rad/s

Now KE (100% rotational) after 3.62s is:

KE = 0.5 \times I \times \omega^2

KE =0.5 \times 79.91 \times 3.69^2

KE = 544J

6 0
3 years ago
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