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4 years ago

6

A flywheel with a very low friction bearing takes 1.6 h to stop after the motor power is turned off. The flywheel was originally

rotating at 55 rpm. What was the initial rotation rate in radians per second?

1 answer:

nadezda [96]4 years ago

3 0

Answer: 5.76 rads/s

Explanation:

The initial rotation is 55 rpm

1 rev = 2π radians

55 revs = 55 × 2π/1 = 345.58 radians/min

345.58 rads/min = 345.58rads/60s = 5.76 rads/s

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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o

matrenka [14]

Answers:

(a) $0.0073kg$

(b) Volume gold: $3.79(10)^{-7}m^{3}$ , Volume cupper: $7.6(10)^{-8}m^{3}$

(c) $17633.554kg/m^{3}$

Explanation:

<h2>(a) Mass of gold </h2><h2 />

We are told the total mass $M$ of the coin, which is an alloy of gold and copper is:

$M=m_{gold}+m_{copper}=7.988g=0.007988kg$ (1)

Where $m_{gold}$ is the mass of gold and $m_{copper}$ is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats $K$ and mass is:

$K=24\frac{m_{gold}}{M}$ (2)

Finding ${m_{gold}$ :

$m_{gold}=\frac{22}{24}M$ (3)

$m_{gold}=\frac{22}{24}(0.007988kg)$ (4)

$m_{gold}=0.0073kg$ (5) This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density $\rho$ of an object is given by:

$\rho=\frac{mass}{volume}$

If we want to find the volume, this expression changes to: $volume=\frac{mass}{\rho}$

For gold, its volume $V_{gold}$ will be a relation between its mass $m_{gold}$ (found in (5)) and its density $\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}$ :

$V_{gold}=\frac{m_{gold}}{\rho_{gold}}$ (6)

$V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}$ (7)

$V_{gold}=3.79(10)^{-7}m^{3}$ (8) Volume of gold in the coin

For copper, its volume $V_{copper}$ will be a relation between its mass $m_{copper}$ and its density $\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}$ :

$V_{copper}=\frac{m_{copper}}{\rho_{copper}}$ (9)

The mass of copper can be found by isolating $m_{copper}$ from (1):

$M=m_{gold}+m_{copper}$

$m_{copper}=M-m_{gold}$ (10)

Knowing the mass of gold found in (5):

$m_{copper}=0.007988kg-0.0073kg=0.000688kg$ (11)

Now we can find the volume of copper:

$V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}$ (12)

$V_{copper}=7.6(10)^{-8}m^{3}$ (13) Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density $\rho_{coin$ will be a relation between its total mass $M$ and its total volume $V$ :

$\rho_{coin}=\frac{M}{V}$ (14)

Knowing the total volume of the coin is:

$V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}$ (15)

$\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}$ (16)

Finally:

$\rho_{coin}=17633.554kg/m^{3}}$ (17) This is the total density of the British sovereign coin

6 0

3 years ago

LiRa [457]

Snjajsisoaowu seisoosw w

8 0

3 years ago

If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun

zvonat [6]

Answer:

H = 1/2 g t^2 where t is time to fall a height H

H = 1/8 g T^2 where T is total time in air (2 t = T)

R = V T cos θ horizontal range

3/4 g T^2 = V T cos θ 6 H = R given in problem

cos θ = 3 g T / (4 V) (I)

Now t = V sin θ / g time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V) from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97 6 within rounding

3 0

3 years ago

A small satellite being designed requires a nitrogen storage tank to store propellant for the cold gas thruster used to maintain

ycow [4]

Answer ;

Minimum required volume = 0.635m3

Maximum internal pressure = 74.35bar

Explanation:

The detailed step by step calculation using the vanderwaal's equation of state for ideal gases is as shown in the attachment.

7 0

3 years ago

A 45.0-kg sample of ice is at 0.00°C. How much heat is needed to melt it? For water, Lf=334 kJ/kg and Lv=2257 kJ/kg

Aleonysh [2.5K]

Heat required to change the phase of ice is given by

Q = m* L

here

m = mass of ice

L = latent heat of fusion

now we have

m = 45 kg

L = 334 KJ/kg

now by using above formula

$Q = 45 * 334 * 10^3$

$Q = 1.5 * 10^7 J$

In KJ we can convert this as

$Q = 1.5 * 10^4 kJ$

so the correct answer is D option

7 0

3 years ago

Read 2 more answers