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scoundrel [369]
3 years ago
15

Given that a fluid at 260°F has a kinematic viscosity of 145 mm^2/s, determine its kinematic viscosity in SUS at 260°F.

Physics
1 answer:
Oduvanchick [21]3 years ago
5 0

Answer:

kinematic viscosity in SUS is = 671.64 SUS

Explanation:

given data

kinetic viscosity = 145 mm^2/s

we know

1 mm = 0.1 cm

so kinetic viscosity in cm is \nu =145 (0.1)^{2} =1.45 cm^{2}/s

other unit of kinetic viscosity is centistokes

1 cm^{2}/s = 100 cst

so 1.45 cm^2/s will be 145 cst

if the temperature is 260°f , then cst value should be multiplied by 4.632. therefore kinematic viscosity in SUS is = 4.362 *145 = 671.64 SUS

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A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a
svlad2 [7]

Answer:

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

the apple should be tossed after \Delta t=0.0173\ s

Explanation:

Given:

  • velocity of arrow in projectile, v=30\ m.s^{-1}
  • angle of projectile from the horizontal, \theta=20^{\circ}
  • distance of the point of tossing up of an apple, d=30\ m

<u>Now the horizontal component of velocity:</u>

v_x=v\ cos\ \theta

v_x=30\times cos\ 20^{\circ}

v_x=28.191\ m.s^{-1}

<u>The vertical component of the velocity:</u>

v_y=v.sin\ \theta

v_y=30\times sin\ 20^{\circ}

v_y=10.261\ m.s^{-1}

<u>Time taken by the projectile to travel the distance of 30 m:</u>

t=\frac{d}{v_x}

t=\frac{30}{28.191}

t=1.0642\ s

<u>Vertical position of the projectile at this time:</u>

h=v_y.t-\frac{1}{2}g.t^2

h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2

h=5.3701\ m

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

v'^2=u'^2-2g.h

0^2=u'^2-2\times 9.8\times 5.3701

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

v'=u'-g.t'

0=10.259-9.8\times t'

t'=1.0469\ s

<u>We observe that </u>t>t'<u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

\Delta t=t-t'

\Delta t=1.0642-1.0469

\Delta t=0.0173\ s

8 0
3 years ago
What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 29.5 km/h and the μs
Yakvenalex [24]

Answer:

Radius=15.773 m

Explanation:

Given data

v=29.5 km/h=8.2 m/s

μs=0.435

To find

Radius R

Solution

The acceleration is a centripetal acceleration  which is experienced by the bicycle given by

a=v^{2}/R

This acceleration is only due to static force which given as

f=ma\\f=m(v^{2}/R )

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FN is normal force equal to mass*gravity

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Therefore the minimum radius should be found by the bicycle move without  sliding

So

v^{2}/R=u_{s}g\\  R=v^{2}/u_{s}g\\R=(8.2)^{2}/(0.435*9.8)\\R=15.773m

8 0
3 years ago
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