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Eddi Din [679]
1 year ago
7

you are about to work some magic with matter. you have a rectangular cube of matter in its solid state. it measures 3 in x 4 in

x 5 in. you have available a rectangular container measuring 5 in x 6 in x 3 in. you also have a cylindrical container witb base radius of 2.5 in and height 8 in. illustrate how that matter "fits" in each container when it is in its solid, liquid, and gaseous state.
Physics
1 answer:
Elenna [48]1 year ago
6 0

In solid and liquid the matter can occupy the 90 in³ and 157.1 in³ volume.

The matter in gaseous state can be expanded to occupy the volumes of the container.

<h3>Volume of each of the container</h3>

The volume of each of the container is calculated as follows;

<h3>Volume of the rectangular container</h3>

V = 5 in x 6 in x 3 in

V = 90 in³

<h3>Volume of the cylindrical container</h3>

V = πr²h

V = (π)(2.5 in)²(8 in)

V = 157.1 in³

<h3>Volume of the matter</h3>

Vm = 3 in x 4 in x 5 in

Vm = 60 in³

<h3>Matter in solid and liquid state</h3>

Matter has fixed volume in solid and liquid state.

In solid and liquid the matter can occupy the 90 in³ and 157.1 in³ volume.

<h3>Matter in gaseous state</h3>

Matter has no definite volume in gaseous state.

The matter in gaseous state can be expanded to occupy the volumes of the container.

Learn more about states of matter here:

#SPJ1

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An 80-kgkg quarterback jumps straight up in the air right before throwing a 0.43-kgkg football horizontally at 15 m/sm/s . How f
vladimir1956 [14]

Answer:

V = 0.0806 m/s

Explanation:

given data

mass quarterback = 80 kg

mass football = 0.43 kg

velocity = 15 m/s

solution

we consider here momentum conservation is in horizontal direction.

so that here no initial momentum of the quarterback

so that final momentum of the system will be 0

so we can say

M(quarterback) ×  V = m(football) × v (football)   ........................1

put here value we get

80 ×  V  = 0.43  × 15

V = 0.0806 m/s

5 0
3 years ago
How much force is required to accelerate a 9.0-g object at 10000 g's?
Yuki888 [10]
Hey give us m = 9.0 g = 9.0 x 10-3 kg, and a = 10,000 "g's" = 98000 m/s/s so:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N
6 0
3 years ago
Which of the following statements is true for a cell placed in beaker containing an isotonic solution?
Korvikt [17]
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The correct answer is <u>C. w</u></span><span><u>ater molecules flow in both directions at the same rate.</u></span>
3 0
3 years ago
Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m
Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC

In words, the value of q₃ must be 5.3nC.

7 0
3 years ago
40 POINTS EASY
11111nata11111 [884]
We know, Mechanical Energy = K.E. + P.E.
As ball is at ground, P.E. would be zero. But as it is in motion, it must have some K.E. and that is:

K.E. = 1/2 mv²
K.E. = 1/2 * 1 * 2²
K.E. = 4/2
K.E. = 2 J

In short, Your Answer would be Option B

Hope this helps!
7 0
3 years ago
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