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ivolga24 [154]
2 years ago
12

4. A cylindrical tube has a length of 14.4cm and a radius of 1.5cm and is filled with a colorless gas. If the density of the gas

is known to be 0.00123g/cm', what is the mass of the gas in the tube in mg?​
Physics
1 answer:
professor190 [17]2 years ago
5 0

Answer:

Mass, m of gas is 0.2504 grams.

Explanation:

First, we need to solve for the volume of the cylindrical tube.

Volume of cylinder is given by the formula;

V = 2\Pi r^{2}h

Where, V represents volume.

π represents pie

r represents radius.

h represents height or length.

Given the following data;

Radius, r = 1.5cm

Length, h = 14.4cm

Density, d = 0.00123g/cm³

Substituting into the equation;

V = 2 * 3.142 * (1.5)^{2}14.4

V = 2 * 3.142 * 2.25 * 14.4

V = 203. 6016

Therefore, the volume of the cylindrical tube is 203. 6016cm³

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the formula;

Density = \frac{mass}{volume}

Mass = density  *  volume

Substituting into the equation, we have;

Mass = 0.00123 * 203. 6016

Mass = 0.2504g.

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Answer:

(A) = 3.57 m

Explanation:

from the question we are given the following:

diameter (d) = 3.2 m

mass (m) == 42 kg

angular speed (ω) = 4.27 rad/s

from the conservation of energy

mgh = 0.5 mv^{2} + 0.5Iω^{2} ...equation 1

where

Inertia (I) = 0.5mr^{2}

ω = \frac{v}{r}

equation 1 now becomes

mgh = 0.5 mv^{2} + 0.5(0.5mr^{2})(\frac{v}{r})^{2}

gh = 0.5 v^{2} + 0.5(0.5)(v)^{2}

4gh = 2v^{2} + v^{2}

h = 3v^{2} ÷ 4 g .... equation 2

  from ω = \frac{v}{r}

 v  = ωr  = 4.27 x (3.2 ÷ 2)

v = 6.8 m/s

now substituting the value of v into equation 2

h = 3v^{2} ÷ 4 g

h = 3 x (6.8)^{2} ÷ (4 x 9.8)

h = 3.57 m

8 0
2 years ago
What is the mass of an object weighing 63 N on Earth?
avanturin [10]
Weight expressed in Newtons is expressed in the equation whereby Weight= the mass of an object * the force of gravity. The force of gravity on earth is a constant 9.8 meters per second squared. Therefore if weight (w) = 63 N and the force of gravity is 63 N then the mass must equal 6.43 kg. Because the equation for weight is w=mg so 63 N (w) = m * 9.8 m/s^2. 
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2 years ago
4. In a closed system consisting of a cannon and a cannonball, the kinetic energy of a cannon is 72,000 J. If the cannonball is
FromTheMoon [43]

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D an B

Explanation:

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A physics student walks 100 meters in 80 seconds. The student stops for 30 seconds, and then walks 200 meters farther in 90 seco
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Answer:

B i think is the answer

Explanation:

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A particle moves along a straight line with a velocity in meters per second given by v = 12 - 3t + 8t, where t is in seconds. Wh
elena-14-01-66 [18.8K]

Answer:

Explanation:

Given the velocity of a particle modeled by the equation

v = 13-3t+8t² where t is in seconds

Given t = 0 and position s = +5m

A) To get the position as a function of time, we will integrate the function with respect to t ad shown;

v = 13-3t+8t²

S = ∫13-3t+8t² dt

S = 13t-13t²/2+8t³/3 + C

at t = 0 and S = +5m

5 = 13(0)-13(0)²/2+8(0)³/3+C

5 = 0-0+0+C

C = 5

Substituting c = 5 into the displacement function

S = 13t-13t²/2+8t³/3 + C

S = 13t-13t²/2+8t³/3 + 5

B) acceleration is the change in velocity with respect to time.

a = dv/dt

Given v = 13-3t+8t²

a= dv/dt = -3+16t

a = 16-3t

C) acceleration at t = 6s is derived by plugging in t = 6 into the resulting equations in (B)

a = 16-3t

a = 16-3(6)

a = 16-18

a = -2m/s²

D) net displacement from t = 0 to t = 6s

At t = 0:

S(0) = 13(0)-13(0)²/2+8(0)³/3 + 5

S(0) = 0+5

S(0) = 5m

At t = 6s

S(6) = 13(6)-13(6)²/2+8(6)³/3 + 5

S(6) = 78-234+576+5

S(6) = 425m

Net displacement from t = 0s to t = 6s is s(6)-s(0)

= 425-5

= 420m

E) Total distance travelled D = S(6)+S(0)

= 425+5

= 430m

F) Average velocity = ∆S/∆t

Average velocity = S(6)-S(0)/6-0

Average velocity = 425-5/6

Average velocity = 420/6

Average velocity = 70m/s

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