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d1i1m1o1n [39]
3 years ago
8

20 POINTS WILL BE MARK BRAINLIEST ATTACHED BELOW GAS LAW WORKSHEET

Physics
1 answer:
german3 years ago
7 0

Explanation:

I will do two of each as examples.

Boyle's law says that at constant temperature, the product of the initial pressure and volume equals the product of the final pressure and volume.

1. P₁ V₁ = P₂ V₂

(1.5 atm) (10.0 L) = (0.75 atm) V

V = 20.0 L

2. P₁ V₁ = P₂ V₂

(100.0 kPa) (500.0 mL) = P (1,000.0 mL)

P = 50.0 kPa

Charles' law says that at constant pressure, the quotient of the initial volume and temperature equals the quotient of the final volume and temperature.

6. V₁ / T₁ = V₂ / T₂

(10.0 L) / (1500 K) = V / (750 K)

V = 5.0 L

7. V₁ / T₁ = V₂ / T₂

(500.0 mL) / (100 K) = (1000.0 mL) / T

T = 200 K

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Answer:

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Explanation:

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3 0
3 years ago
A positive charge +q1 is located to the left of a negative charge -q2. On a line passing through the two charges, there are two
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Answer:

please the answer below

Explanation:

(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is

V_1=k\frac{q_1}{r-1.0}-k\frac{q_2}{1.0}=0\\\\V_2=k\frac{q_1}{r+5.2}-k\frac{q_2}{5.2}=0\\\\

k=8.89*10^9

For both cases we have

k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}

(b) by replacing this values of r in the expression for V we obtain

k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}

hope this helps!!

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3 years ago
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A block with mass m = 0.450 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth
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Answer:

k = 26.25 N/m

Explanation:

given,

mass of the block= 0.450

distance of the block = + 0.240

acceleration = a_x = -14.0 m/s²

velocity = v_x = + 4 m/s

spring force constant (k) = ?

we know,

x = A cos (ωt - ∅).....(1)

v = - ω A cos (ωt - ∅)....(2)

a = ω²A cos (ωt - ∅).........(3)

\omega = \sqrt{\dfrac{k}{m}}

now from equation (3)

a_x = \dfrac{k}{m}x

k = \dfrac{m a_x}{x}

k = \dfrac{0.45 \times (-14)}{0.24}

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hence, spring force constant is equal to k = 26.25 N/m

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