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Svetllana [295]

3 years ago

10

An ant clings to the outside edge of the tire of an exercise bicycle. When you start pedaling, the ant's speed increases from ze

ro to 12 m/s in3.3 s . The wheel's rotational acceleration is11 rad/s2 .
1: Determine the average tangential acceleration of the tire.
2: Determine the radius of the tire.
3: Determine the angle the ant has turned during this time interval.
4: Determine the distance the ant has traveled along the arc during this time interval.

1 answer:

konstantin123 [22]3 years ago

3 0

Answer:

1- 3.64 m/s

2- 0.28 m

3- 70.8 rad

4- 19.8 m

Explanation:

see attachment

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When rocks experience high temperatures and differential stresses deep in the earth, their grains tend to ________?

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Break into small particles. The heat causes the rock to break up and form pebbles or sand. Hope that helped. Have a nice day

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Someone places a chocolate bar near a working radar set that is used to locate ships and airplanes. Which best describes what is

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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

2 years ago

What are the strength and direction of an electric field that will balance the weight of a 0.9 g plastic sphere that has been ch

Vladimir79 [104]

(a) The force exerted by the electric field on the plastic sphere is equal to
$F=qE$
where $q=-3.4 nC=-3.4 \cdot 10^{-9} C$ is the charge of the sphere and E is the strength of the electric field. This force should balance the weight of the sphere:
$F=mg =0.9 g$
where m is the mass of the sphere and g is the gravitational acceleration.

Since the two forces must be equal, we have:
$qE=mg$
and so we find the intensity of the electric field
$E= \frac{mg}{q}= \frac{0.9 \cdot 9.81 m/s^2}{3.4 \cdot 10^{-9} C} =2.6 \cdot 10^9 N/C$

(b) Now let's find the direction of the field. The electric force must balance the weight of the sphere, which is directed downward, so the electric force should be directed upward. Since the charge is negative, the force is opposite to the electric field direction, and so the direction of the electric field is downward.

4 0

3 years ago

Suppose that we use a heater to boil liquid nitrogen (N2 molecules). 4480 J of heat turns 20 g of liquid nitrogen into gas. Note

love history [14]

Answer:

The energy is $9.4\times10^{-21}\ J$

(a) is correct option

Explanation:

Given that,

Energy = 4480 j

Weight of nitrogen = 20 g

Boil temperature = 77 K

Pressure = 1 atm

We need to calculate the internal energy

Using first law of thermodynamics

$Q=\Delta U+W$

$Q=\Delta U+nRT$

Put the value into the formula

$4480=\Delta U+\dfrac{20}{28}\times8.314\times77$

$\Delta U=4480-\dfrac{20}{28}\times8.314\times77$

$\Delta U=4022.73\ J$

We need to calculate the number of molecules in 20 g N₂

Using formula of number of molecules

$N=n\times \text{Avogadro number}$

Put the value into the formula

$N=\dfrac{20}{28}\times6.02\times10^{23}$

$N=4.3\times10^{23}$

We need to calculate the energy

Using formula of energy

$E=\dfrac{\Delta U}{N}$

Put the value into the formula

$E=\dfrac{4022.73}{4.3\times10^{23}}$

$E=9.4\times10^{-21}\ J$

Hence, The energy is $9.4\times10^{-21}\ J$

4 0

2 years ago