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zhannawk [14.2K]
2 years ago
8

A projectile is shot on level ground with a

Physics
1 answer:
erastova [34]2 years ago
5 0

Answer:

time rising = 34 / 9.8 = 3.47 sec

total time in air = 2 * 3.47 sec = 6.94 sec

(time rising must equal time falling)

R = 17 m/s * 6.94 s = 118 m

Can also use range formula

R = v^2 sin (2 theta) / g

tan theta = 34 / 17 = 2

theta = 63.4 deg

2 theta = 126.9 deg

sin 126.9 = .8

v^2 = 17^2 + 34^2 = 1445 m^2/s^2

R = 1445 * .8 / 9.8 = 118 m    agreeing with answer found above

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The temperature of the oxygen gas is  243.75 K.

Using ideal gas law to explain the answer, the absolute temperature of the gas will decrease if the number of moles of the gas increases and it will increase if the volume and/or pressure of the gas increases.

The reaction of the given elements;

H_2 \ + \ \frac{1}{2} O_2 \ --->\ \ H_2O

volume of the collected oxygen gas, V = 10 L

pressure of the gas, P = 1 atm

number of moles of the gas, n = 0.5

Using ideal law the temperature of the oxygen gas is calculated as follows;

PV = nRT\\\\T = \frac{PV}{nR} \\\\where;\\R \ is \ the \ ideal \ gas \ constant = 0.08205 \ L.atm/K.mol\\\\T = \frac{1 \times 10 }{0.5 \times 0.08205} \\\\T = 243.75 \ K

Thus, the temperature of the gas is 243.75 K.

Using ideal gas law to explain the answer. The absolute temperature of the oxygen gas is directly proportional to the product of its pressure and volume and inversely proportional to its number of moles. That is the absolute temperature of the gas will decrease if the number of moles of the gas increases and it will increase if the volume and/or pressure increases.

Learn more here: brainly.com/question/16617695

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3 years ago
A ball whose mass is 0.3 kg hits the floor with a speed of 7 m/s and rebounds upward with a speed of 4 m/s. if the ball was in c
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Newton's second law is stated as:
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A Carnot engine operates between two heat reservoirs at temperatures TH and TC. An inventor proposes to increase the efficiency
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Answer:

A) Efficiency = 1 - (TC/TH)

B) Comparing this with the efficiency of the original carnot engine, the efficiency is the same

Explanation:

The formula for efficiency of an original carnot engine is;

e = 1 - T(C) /T(H) ——— eq 1

In like manner, for a composite engine, the efficiency is;

e(12) = (W1 + W2)/Q(H1)

Where W1 is work done by 1st engine; W2 is work done by second engine and Q(H1) is the heat input to the first engine.

Now the total work done is;

W = Q(H) + Q(C)

Where Q(H) is the heat input and Q(C) is the heat released.

Thus,

e(12) = [Q(H1) + Q(C1) + Q(H2) + Q(C2)] / Q(H1)

Now, from the earlier e(12) equation compared to this, QH2 = -QC1

Thus;

e(12) = [Q(H1) + Q(C1) - Q(C1) + Q(C2)] / Q(H1)

So e(12) = [Q(H1) + Q(C2)] / Q(H1)

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So, simplifying that,

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-Q(H1) (T'/TH)

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Q(C2) = -Q(H1) (T'/TH)(TC/T')

So T' will cancel out to remain;

Q(C2) = -Q(H1)(TC/TH)

Replacing this with Q(C2) in eq 2 to obtain;

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e(12) = 1 - TC/TH

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