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kati45 [8]
3 years ago
13

Does density depend on the amount of substance that you have? Explain your answer.

Physics
1 answer:
saul85 [17]3 years ago
3 0

Answer:

No Density does not depend on the amount of substance you have because density is the ratio of mass and volume in an object

Explanation:

ROCK AND ROLL!!

~Monty

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A cat has a mass of 4.6 kg. What is its weight on Earth's surface?
andreyandreev [35.5K]
<h3>✽ - - - - - - - - - - - - - - -  ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ Earth's gravity is approximately 9.81

weight = mass x gravity

weight = 4.6 x 9.81

weight = 45.126

Answer is B. 45N

➶ Hope This Helps You!

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4 0
3 years ago
Read 2 more answers
Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet
sveta [45]

Answer:

The level of the root beer is dropping at a rate of 0.08603 cm/s.

Explanation:

The volume of the cone is :

V=\frac {1}{3}\times \pi\times r^2\times h

Where, V is the volume of the cone

r is the radius of the cone

h is the height of the cone

The ratio of the radius and the height remains constant in overall the cone.

Thus, given that, r = d / 2 = 10 / 2 cm = 5 cm

h = 13 cm

r / h = 5 / 13

r = {5 / 13} h

V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h

V=\frac {550}{3549}\times h^3

Also differentiating the expression of volume w.r.t. time as:

\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}

Given: \frac {dV}{dt} = -4 cm³/sec (negative sign to show leaving)

h = 10 cm

So,

-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}

\frac{55000}{1183}\times \frac {dh}{dt}=-4

\frac {dh}{dt}=-0.08603\ cm/s

<u>The level of the root beer is dropping at a rate of 0.08603 cm/s.</u>

3 0
3 years ago
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Answer:

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4 0
3 years ago
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Which state of matter has atoms that are spread out and bouncy?
NikAS [45]
The stage where atoms are spread out and bouncy is the gas stage.

7 0
3 years ago
The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh
Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

v^2=v^2_o-2ax

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}

Next, consider that the formula for the force is:

F=ma

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

F=(820kg)(0.66\frac{m}{s^2})=542.20N

Hence, the required force to stop the car is 542.20N

4 0
1 year ago
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