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Pachacha [2.7K]
3 years ago
12

A gas expands from an initial volume of 30.0 L to a final volume of 65.0 L at a constant pressure of 110kPa. How much work is do

ne by the gas?
Physics
1 answer:
Brilliant_brown [7]3 years ago
8 0

Answer:

3850 J

Explanation:

V_{i} = initial volume of the gas = 30 L = 0.03 m³

V_{f} = final volume of the gas = 65 L = 0.065 m³

P = Constant pressure of the gas = 110 kPa = 110000 Pa

W = Work done by the gas

Since the pressure is constant, Work done by the gas is given as

W = P (V_{f} - V_{i})\\W = (110000) (0.065 - 0.03)\\W = 3850 J

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Answer:

just search up a ven-diagram and then try to draw it or trace it then use it for ur question

Explanation:

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3 years ago
3 An un calibrated mercury in glass thermometer immersed in melting ice. The length of the mercury thread is 25 mm when the ther
sammy [17]

Answer:

25 mm = 0 deg C

200 mm = 100 deg C

200 - 25 = 175 = change in thread per 100 deg C

95 - 25 = 70 mm - change in thread from 0 deg C

70 / 175 * 100 = 40 deg C    final temperature at 95 mm

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Projectile Motion—A tennis ball is thrown out a window 28 m above the ground at an initial velocity of 15.0 m/s and 20.0° below
NNADVOKAT [17]

Answer:

The distance will be x = 41.7 [m]

Explanation:

We must first find the components in the x & y axes of the initial velocity.

(v_{o})_{x} = 15*cos(20)= 14.09[m/s]\\(v_{o})_{y} = 15*sin(20)= 5.13[m/s]

The acceleration is the gravity acceleration therefore.

g = 9.81 [m/s^2]

Now we can calculate how long it takes to fall.

y=(v_{o})_{y}*t-0.5*g*t^2\\-28 = 5.13*t-0.5*9.81*t^2\\-28=-4.905*t^2+5.13*t\\4.905*t^2-5.13*t=28\\t = 2.96[s]

With this time we can find the horizontal distance that runs the projectile.

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5 0
2 years ago
What is the average velocity of the object?
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2 years ago
5/137 Under the action of its stern and starboard bow thrusters, the cruise ship has the velocity vB = 1 m/s of its mass center
quester [9]

The image is missing, so i have attached it;

Answer:

A) V_rel = [-(2.711)i - (0.2588)j] m/s

B) a_rel = (0.8637i + 0.0642j) m/s²

Explanation:

We are given;

the cruise ship velocity; V_b = 1 m/s

Angular velocity; ω = 1 deg/s = 1° × π/180 rad = 0.01745 rad/s

Angular acceleration;α = -0.5 deg/s² = 0.5 x π/180 rad = -0.008727 rad/s²

Now, let's write Velocity (V_a) at A in terms of the velocity at B(V_b) with r_ba being the position vector from B to A and relative velocity (V_rel)

Thus,

V_a = V_b + (ω•r_ba) + V_rel

Now, V_a = 0. Thus;

0 = V_b + (ω•r_ba) + V_rel

V_rel = -V_b - (ω•r_ba)

From the image and plugging in relevant values, we have;

V_rel = -1[(cos15)i + (sin15)j] - (0.01745k * -100j)

V_rel = - (cos15)i - (sin15)j - 1.745i

Note that; k x j = - i

V_rel = [-(2.711)i - (0.2588)j] m/s

B) Let's write the acceleration at A with respect to B in terms of a_b.

Thus,

a_a = a_b + (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel

a_a and a_b = 0.

Thus;

0 = (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel

a_rel = - (α*r_ba) - (ω(ω•r_ba)) - (2ω*v_rel)

Plugging in the relevant values with their respective position vectors, we have;

a_rel = - (-0.008727k * -100j) - (0.01745k(0.01745k * -100j)) - (2*0.01745k * [-(2.711)i - (0.2588)j])

a_rel = 0.8727i - (0.01745² x 100)j + 0.0946j - 0.009i

Note that; k x j = - i and k x i = j

Thus,simplifying further ;

a_rel = 0.8637i + 0.0642j m/s²

4 0
3 years ago
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