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Rom4ik [11]
3 years ago
9

A 75-g mass sits 75 cm from the center of a rotating platform undergoing a uniform angular acceleration of 0.125 rad/s^2. The co

efficient of static friction between the mass and the platform is 0.250.
What is the speed of the mass when it slides off?
a. 0.889 m/s
b. 1.36 m/s
c. 1.44 m/s
d. 1.58 m/s
e. It will never slide off.
Physics
1 answer:
jasenka [17]3 years ago
3 0

Answer:

velocity of mass when it slides of will be 1.36 m/sec

So option (b) will be correct option.

Explanation:

We have given mass of the slits m = 75 gram = 0.075 kg

Radius r = 75 cm = 0.75 m

Coefficient of kinetic friction \mu =0.250

Acceleration due to gravity g=9.8m/sec^2

When the mass slides of

\frac{mv^2}{r}=\mu mg

v=\sqrt{\mu rg}=\sqrt{0.250\times 0.75\times 9.8}=1.36m/sec

So velocity of mass when it slides of will be 1.36 m/sec

So option (b) will be correct option.

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3 years ago
0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a
klio [65]

Answer:

The boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}

(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }

where h_{fg is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu

T is the temperature ( 15 + 460 )R

m is the mass of water ( 0.5 Ibm )

V_{fg is specific volume ( 1.5 ft³ )

we substitute

(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5}) / ( (15+460)\frac{1.5}{0.5})  

(\frac{dP}{dT} )_{sat } = 272.98 Ibf-ft²/R

Now,

(\frac{dP}{dT} )_{sat } = (\frac{P_2 - P_1}{T_2 - T_1})_{sat

where P₁ is the initial pressure ( 50 psia )

P₂ is the final pressure ( 60 psia )

T₁ is the initial temperature ( 15 + 460 )R

T₂ is the final temperature = ?

we substitute;

T_2 = ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}

T_2 = 475 + 5.2751\\

T_2 = 480.275 R

Therefore, boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

3 0
3 years ago
Pls answer I will make brainlist
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3 0
3 years ago
The radius of a planet is 2400 km, and the acceleration due to gravity at its surface is 3.6 m/s2.
kiruha [24]

Answer:

3.1\cdot10^{23}\:\mathrm{kg}

Explanation:

We can use Newton's Universal Law of Gravitation to solve this problem:

g_P=G\frac{m}{r^2}., where g_P is acceleration due to gravity at the planet's surface, G is gravitational constant 6.67\cdot 10^{-11}, m is the mass of the planet, and r is the radius of the planet.

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Now plugging in our values, we get:

3.6=6.67\cdot10^{-11}\frac{m}{(2,400,000)^2},

Solving for m:

m=\frac{2,400,000^2\cdot3.6}{6.67\cdot 10^{-11}},\\m=\fbox{$3.1\cdot10^{23}\:\mathrm{kg}$}.

6 0
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