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Savatey [412]
4 years ago
7

You ride on an elevator that is moving with constant upward acceleration while standing on a bathroom scale. the reading on the

scale is you ride on an elevator that is moving with constant upward acceleration while standing on a bathroom scale. the reading on the scale is
Physics
1 answer:
Blababa [14]4 years ago
4 0

The reading on the scale is greater than your actual weight.

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A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos
lukranit [14]

Answer:

The inductor contains N = 523962.32 loops  

Explanation:

From the question we are told that

     The capacitance of the capacitor is  C =  286nF = 286 * 10^{-9} \  F

      The resonance frequency is  f = 18.0 kHz =  18*10^{3} Hz

       The diameter is  d =  1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m

       The  of the air-core inductor is l = 12 \ m

        The permeability of free space is  \mu_o = 4 \pi *10^{-7} \ T \cdot m/A

 

Generally the inductance of this air-core inductor is mathematically represented as

              L =  \frac{\mu_o * N^2 \pi d^2}{4 l}

This inductance can also be mathematically represented as

               L = \frac{1}{w^2}

Where w is the angular speed mathematically given as

             w = 2 \pi f

So

            L =  \frac{1}{4 \pi ^2 f^2}

Now equating the both formulas for inductance

         \frac{\mu_o * N^2 \pi d^2}{4 l}  =  \frac{1}{4 \pi ^2 f^2}

making N the subject of  the formula

              N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C}  }

              N =  \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }

             

 Substituting value

            N =  \frac{1}{ 3.142  * 18*10^{3} * 0.00011 }  \sqrt{\frac{12}{ 3.142  * 4 \pi *10^{-7}* 286 *10^{-9}} }

              N = 523962.32 loops  

4 0
3 years ago
A person on a daily diet of 2,500 calories should get no_more than <br> calories from fat each day.
ivann1987 [24]

Answer:

35%

Explanation:

3 0
3 years ago
Read 2 more answers
Air in the amount of 2 lbm is contained in a well-insulated, rigid vessel equipped with a stirring paddle wheel. The initial sta
vladimir1956 [14]

Explanation:

It is known that energy balance relation is as follows.

           \Delta E_{system} = E_{in} - E_{out}

Also,   W_{in} = \Delta U

so,       W_{in} = mC_{v}(T_{2} - T_{1})  

According to the ideal gas equation,

           T_{2} = T_{1} \frac{P_{2}}{P_{1}}

Putting the values into the above equation as follows.

             T_{2} = T_{1} \frac{P_{2}}{P_{1}}

                         = (520R) \frac{40psia}{30psia}

                         = 693.3 R

Now, we will convert the temperature into degree Fahrenheit as follows.

            693.3 - 458.67

          = 234.63^{o}F

From table A-2E_{a}

 C_{p} = 0.240 Btu/lbm R  and   C_{v} = 0.171 Btu/lbm

Now, we will substitute the energy balance as follows.

            W_{in} = mC_{v}(T_{2} - T_{1})  

                         = 2 lbm \times 0.171 Btu/lbm R (693.3 - 520)

                         = 59.3 Btu

Thus, we can conclude that final temperature of air is 59.3 Btu.

6 0
4 years ago
When air is blown across the top of an open
Tanzania [10]
When air is blown across the top of an open <span>water bottle, air molecules in the bottle vibrate at a particular frequency and sound is produced in a process called "refraction". 
</span>
8 0
4 years ago
A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a
EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2} (1)

Where:

p_{S,1}, p_{S,2} - Initial and final momemtums of the small object, measured in kilogram-meters per second.

p_{B,1}, p_{B,2} - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that p_{S,1} = 7\,\frac{kg\cdot m}{s}, p_{B,1} = 0\,\frac{kg\cdot m}{s} and p_{S, 2} = 4\,\frac{kg\cdot m}{s}, then the final momentum of the big object is:

7\,\frac{kg\cdot m}{s} + 0\,\frac{kg\cdot m}{s} = 4\,\frac{kg\cdot m}{s}+p_{B,2}

p_{B,2} = 3\,\frac{kg\cdot m}{s}

The magnitude of the large object's momentum change is:

p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}-0\,\frac{kg\cdot m}{s}

p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

4 0
3 years ago
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