If the valence electrons were removed, what would be the ion charge of the element?

Three boxes in contact rest side-by-side on a smooth, horizontal floor. Their masses are 5.0-kg, 3.0-kg, and 2.0-kg, with the 3.

Ivahew [28]

Answer:

(a)Look at the attached graphic

(b)

(b)-1 Equation 1 : m1= 5kg

50-F1= 5 *a

(b)-2 Equation 2 : m2= 3kg

F1-F2= 3 *a

(b)-3 Equation 3 : m3= 2kg

F2 = 2*a

(c) F1 =25 N

(d) F2 =10 N

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a) Draw the free-body diagrams for each of the boxes

Look at the attached graphic

(b) Write Newton’s equation for each mass along the horizontal direction.

Data: m1= 5.0-kg ,m2= 3.0-kg , ,m3= 2.0-kg

Look m1 free-body diagram:

∑Fx = m1*a

50-F1= 5 *a Equation 1

Look m2 free-body diagram:

∑Fx = m2*a

F1-F2= 3 *a Equation 2

Look m3 free-body diagram:

∑Fx = m3*a

F2 = 2*a Equation 3

(c) What magnitude force does the 3.0-kg box exert on the 5.0- kg box?

Look Free body diagram of the mass set

∑Fx = m*a m= m1+m2+m3= 5+3+2 = 10 kg

50 = 10*a

a= 50/10 = 5 m/s²

We replace a = 5 m/s² in the equation 1:

50-F1= 5 *5

50-25= F1

F1 = 25 N

(d) What magnitude force does the 3.0-kg box exert on the 2.0kg box?

We replace a= 5 m/s² in the equation 3

F2 = 2*5 = 10 N

4 0

3 years ago

One who voluntarily donates tissues/ organs striped

Artist 52 [7]

A tissue donor is what you're probably looking for. :)

8 0

3 years ago

How do nerves tie the other parts of the nervous system together?

Svetradugi [14.3K]

They do it by followinng the centeral nervous system

5 0

3 years ago

0.16 mol of argon gas is admitted to an evacuated 70 cm^3 container at 30°C. The gas then undergoes an isothermal expansion to a

Semmy [17]

Answer:

The final pressure of the gas is 9.94 atm.

Explanation:

Given that,

Weight of argon = 0.16 mol

Initial volume = 70 cm³

Angle = 30°C

Final volume = 400 cm³

We need to calculate the initial pressure of gas

Using equation of ideal gas

$PV=nRT$

$P_{i}=\dfrac{nRT}{V}$

Where, P = pressure

R = gas constant

T = temperature

Put the value in the equation

$P_{i}=\dfrac{0.16\times8.314\times(30+273)}{70\times10^{-6}}$

$P_{i}=5.75\times10^{6}\ Pa$

$P_{i}=56.827\ atm$

We need to calculate the final temperature

Using relation pressure and volume

$P_{2}=\dfrac{P_{1}V_{1}}{V_{2}}$

$P_{2}=\dfrac{56.827\times70}{400}$

$P_{2}=9.94\ atm$

Hence, The final pressure of the gas is 9.94 atm.

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The picture shows a loop of wire rotating in a magnetic field in a generator. Determine the direction of the current through the

BaLLatris [955]

I believe the answer is b

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