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slamgirl [31]
3 years ago
5

Shelley is in an elevator that is traveling downward and slowing down at a rate of

Physics
1 answer:
liraira [26]3 years ago
8 0

Answer:

N = 648.55[N]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

∑F =  Forces applied [N]

m = mass = 73.2 [kg]

a = acceleration = 0.950 [m/s²]

Let's assume the direction of the upward forces as positive, just as if the movement of the box is upward the acceleration will be positive.

By performing a summation of forces on the vertical axis we obtain all the required forces and other magnitudes to be determined.

-m*g + N = -m*a\\

where:

g = gravity acceleration = 9.81 [m/s²]

N = normal force (or weight) measured by the scale = 83.4 [N]

Now replacing:

-(73.2*9.81)+N=-73.2*0.950\\-718.092+N=-69.54\\N = -69.54+718.092\\N = 648.55[N]

The acceleration has a negative sign, this means that the elevator is descending at that very moment.

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How much heat is needed to change 1.25 kg of steak at 100°C to water at 100°C?
cricket20 [7]

The heat required to change 1.25 kg of steak is 2825 kJ /kg.

<u>Explanation</u>:

Given, mass m = 1.25 kg,     Temperature t = 100 degree celsius

To calculate the heat required,

                                 Q = m \times L

where m represents the mass in kg,

          L represents the heat of vaporization.

When a material in the liquid state is given energy, it changes its phase from liquid to vapor and the energy absorbed in this process is called heat of the vaporization. The heat of vaporization of the water is about 2260 kJ/kg.

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3 years ago
A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.
kap26 [50]

Answer:

The  frequencies are (f, f_1) =  (6615.4 \ Hz , 19846.2\ Hz)

Explanation:

From the question we are told that

  The  length of the ear canal is  l = 1.3 \ cm  =\frac{1.3}{100}  =  0.013 \ m

   The  speed of sound is assumed to be  v_s  =  344 \ m/s

Now  taking look at a typical  ear canal  we see that we assume it is  a  closed pipe

   Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

            f = \frac{v_s}{4 * l }

 substituting values  

          f = \frac{344}{4 * 0.013 }

         f = 6615.4 \ Hz

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

        f_1 =  \frac{3v_s}{4 * l}

 substituting values  

       f_1 =  \frac{3 *  344}{4 * 0.013}

       f_1 =   19846.2 \ Hz

Given that sound would be loudest in the pipe at the frequency, it implies that the child  will have an increased audible sensitivity at this  frequencies

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