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Tresset [83]
2 years ago
5

A projectile is fired from the ground at a velocity of 30.0 m/s, 35.0 º from the horizontal. What is the maximum height the proj

ectile reaches in the air? How far away from the point it was fired does it hit the ground? What is the time the projectile is in the air?
Physics
1 answer:
Norma-Jean [14]2 years ago
8 0

Answer:

Vy = V sin theta = 30 * ,574 = 17.2 m/s

t1 = 17.2 / 9.8 = 1.76 sec to reach max height

Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m

H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m

Time to fall from zero speed to ground = rise time = 1.76 sec

Vx = V cos 35 = 24.6 m / sec     horizontal speed

Time in air = 1.76 * 2 = 3.52 sec before returning to ground

S = 24.6 * 3.52 = 86.6 m

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A crane lifts a 1,750 kg mass using a steel cable whose mass per unit length is 0.88 kg/m. What is the speed of transverse waves
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139.6m/s

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This formula equation is unbalanced.
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6 0
2 years ago
Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy
Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

P_{1}V_{1}=nRT

V_{1}=\dfrac{nRT}{P_{1}}

Put the value into the formula

V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}

V_{1}=2.62\times10^{-3}\ m^3

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}

Put the value into the formula

T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}

T_{2}=233.7\ K

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

P_{1}V_{1}=P_{2}V_{2}

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

Put the value into the formula

V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}

V_{2}=0.00786\ m^3

V_{2}=7.86\times10^{-3}\ m^3

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}

V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}

V_{2}=0.0057\ m^3

V_{2}=5.7\times10^{-3}\ m^3

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

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