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il63 [147K]
3 years ago
10

A book is sliding to a stop while moving across the classroom floor

Physics
1 answer:
Margarita [4]3 years ago
3 0

Answer:

can you please elaborate the question

Explanation:

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The relationship between body mass body weight and body density
nordsb [41]

Answer:

The formula for calculating Density is:

= Mass / Volume

From this formula, we can say that the relationship between Mass and Density is a direct one. In other words, if mass is increasing - all else being equal - then density will increase as well.

If mass however was decreasing, density would have to decrease as well.

For example, assume 3 bricks have masses of 5kg, 10kg and 15kg. Also assume that the bricks all have the same volume of 5 m³.

Density of 5kg brick = 5 / 5 = 1 kg/m³

Density of 10kg brick = 10 / 5 = 2kg / m³

Density of 15kg brick = 15 / 5 = 3 kg /m³

<em>Notice how density increases as mass increases and decreases when mass decreases. </em>

7 0
3 years ago
Plssss help Do you guys have any other more interesting, funny or creative ideas of incline problems than something sliding down
VashaNatasha [74]
U can always just do the classic roller coaster going up an incline and create some sort of story from that.
8 0
3 years ago
Kinetic energy of an object is equal to
Arte-miy333 [17]
Choice-'b' says the formula for kinetic energy in words.

     KE = (1/2) · (M) · (S²)
7 0
3 years ago
Read 2 more answers
A student dips a strip of metal into a liquid. Which is evidence that only a physical change has occurred?
sattari [20]

Answer:

PLS add the picture too.....

3 0
3 years ago
An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th
kolbaska11 [484]

Answer:

mg=200.4 N.

Explanation:

This problem can be solved using Newton's law of universal gravitation: F=G\frac{m_{1}m_{2}}{r^{2}},

where F is the gravitational force between two masses m_{1} and m_{2}, r is the distance between the masses (their center of mass), and G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2}) is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting w_{s} be the weight on the surface:

w_{s}=mg,

mg=8*10^{2},

m=(8*10^{2})/g,

since we now that g=9.8m/s^{2} we get that the mass is

m=81.6kg.

Now we can use Newton's law of universal gravitation

F=G\frac{Mm}{r^{2}},  

where m is the mass of the astronaut and M is the mass of the earth. From Newton's second law we know that

F=ma,

in this case the acceleration is the gravity so

F=mg, (<u>becarefull, gravity at this point is no longer</u> 9.8m/s^{2} <u>because we are not in the surface anymore</u>)

and this get us to

mg=G\frac{Mm}{r^{2}}, where mg is his new weight.

We need to remember that the mass of the earth is M=5.972*10^{24}kg and its radius is 6.37*10^{6}m.

The total distance between the astronaut and the earth is

r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6} meters.

Now we can compute his weigh:

mg=G\frac{Mm}{r^{2}},

mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}},

mg=200.4 N.

5 0
3 years ago
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