Ask question

Ask question

All categories

3 years ago

13

Using your best estimates, how many times would you have to slap a 1 kg rotisserie chicken in order to cook it? You can assume t

he chicken is mostly water, starts at 0 ∘C, and that it must reach an internal temperature of 170∘ F in order to be cooked.

1 answer:

FromTheMoon [43]3 years ago

8 0

Answer:

n= 16021.03 slaps

Explanation:

Using law of Energy conservation

E_{thermal}= Kinetic energy of hand

⇒ $mc\Delta T= n\frac{1}{2}m_hv_h^2$

m_h= mass of the hand = 0.4 kg

v_h= velocity of the hand = 10 m/s

n= number of slaps

c= 4180 J/Kg °C

m= mass of chicken = 1 kg

Assuming all the energy of hand goes into chicken

Given Ti=0°C and T_f= 170 F= 76.66°C

Now putting the values in above equation to get n

$1\times4180(76.66)= n\frac{1}{2}0.4\times10^2$

n= 16021.03 slaps

You might be interested in

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the c

puteri [66]

Correct question:

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?

Answer:

the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

Explanation:

Given;

length of solenoid, L= 0.35 m

diameter of the solenoid, d = 0.04 m

current through the solenoid, I = 5.0 A

magnetic field in the center of the solenoid, 2.8 x 10⁻² T

The number of turns per meter for the solenoid is calculated as follows;

$B =\mu_o I(\frac{N}{L} )\\\\B = \mu_o I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m$

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

3 0

3 years ago

A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you

soldi70 [24.7K]

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

$F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}$ (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

$|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s$

The interaction time to avoid that the water balloon breaks is 0.029s

5 0

3 years ago

Which changes to observed time and length occur when objects approach the speed of light?

mixas84 [53]

Answer:

Time moves slower and length decreases.

Explanation:

3 0

3 years ago

One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe

Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

$\mu = \frac{m}{l}$

$\mu = \frac{8.79*10^{-3}}{70*10^{-2}}$

$\mu = 0.01255kg/m$

The expression for the wavelength of the standing wave for the second overtone is

$\lambda = \frac{2}{3} l$

Replacing we have

$\lambda = \frac{2}{3} (70*10^{-2})$

$\lambda = 0.466m$

The frequency of the sound wave is

$f_s = \frac{v}{\lambda_s}$

$f_s = \frac{344}{0.768}$

$f_s = 448Hz$

Now the velocity of the wave would be

$v = f_s \lambda$

$v = (448)(0.466)$

$v = 208.768m/s$

The expression that relates the velocity of the wave, tension on the string and linear mass density is

$v = \sqrt{\frac{T}{\mu}}$

$v^2 = \frac{T}{\mu}$

$T= \mu v^2$

$T = (0.01255kg/m)(208.768m/s)^2$

$T = 547N$

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the $n^{th}$ harmonic frequency is

$f_n = nf_1$

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

$n=3$

Then,

$f_3 = 3f_1$

Rearranging to find the fundamental frequency

$f_1 = \frac{f_3}{3}$

$f_1 = \frac{448Hz}{3}$

$f_1 = 149.9Hz$

7 0

3 years ago

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axi

Hitman42 [59]

Answer:

the balls reached a height of 4.9985 m

Explanation:

Given the data in the question;

mass one m = 3.8 kg

mass two M = 2.1 kg

Initial velocities

u = 22 m/s

U = { moving downward} = 12 m/s

Now, using the law conservation of linear moment;

mu + MU = v( m + M )

we solve for "v" which is the velocity of the ball s after collision;

v = (mu + MU) / ( m + M )

so we substitute our given values into the equation

v = ( ( 3.8 × 22 ) + ( 2.1 × -12) ) / ( 3.8 + 2.1 )

v = ( 83.6 - 25.2 ) / 5.9

v = 58.4 / 5.9

v = 9.898 m/s

Now, we determine required height using the following relation;

v"² - v² = 2gh

where v" is the velocity at the top which is 0 m/s and g = -9.8 m/s²

0 - v² = 2gh

v² = -2gh

so we substitute

( 9.898 )² = -2 × -9.8 × h

97.97 = 19.6 × h

h = 97.97 / 19.6

h = 4.9985 m

Therefore, the balls reached a height of 4.9985 m

8 0

3 years ago