Ask question

Ask question

All categories

2 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]2 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

If you are asked to determine an object’s speed, what information must you have?

o-na [289]

Speed = Distance/Time. So you are required to know A. Distance and period of time

3 0

2 years ago

A spring with k = 15.3 N/cm is initially stretched 1.81 cm from its equilibrium length. a) How much more energy is needed to fur

leonid [27]

Answer:

2.31J

Explanation:

the energy for a spring system is given by:

$E=\frac{1}{2} kx^2$

where $k$ is the spring constant: $k=15.3N/cm=1530N/m$ and $x$ is the distance stretched from the equilibrium position.

In the first case $x=1.81cm=0.0181m$

so the energy to stretch the spring 1.81cm is:

$K_{1}=\frac{1}{2} (1530N/m)(0.0181m)^2=0.25J$

and for the second case, the energy to stretch the spring 5.79cm:

$x=5.79cm=0.0579m$

$K_{1}=\frac{1}{2} (1530N/m)(0.0579m)^2=2.56J$

so to answer a) we must find the difference between these energies:

$2.56J-0.25J=2.31J$

6 0

2 years ago

An elevator exerts a force of 8000 N to lift the passengers 15 m. How much work is the elevator doing?

Rasek [7]

Answer:120,000

Explanation:

7 0

2 years ago

An infinitely long line of charge has linear charge density 6.00×10−12 C/m . A proton (mass 1.67×10−27 kg,charge +1.60×10−19 C)

Bogdan [553]

Answer:

$A) \,K.E=1.405\times 10^{-20}J$

$B)\,r_f=0.268\,m$

Explanation:

$Charge\,\,density=\lambda=6\times 10^{-12}C/n\\\\Mass\,\, of \,\,proton=m_p=1.67\times 10^{-27}kg\\\\charge\,\, of\,\, proton=q_p=1.609\times 10^{-19}C\\\\r=12\,cm=0.12\, m\\\\v=4.103\times 10^{3}m/s$

A) Initial kinetic energy of proton

$K.E=\frac{1}{2}m_pv^2\\\\K.E=1.405\times 10^{-20}J$

B) How close does the proton get to the line of charge?

Potential energy and kinetic energy are related as:

$K_i+U_i=K_f+U_f\\\\U_f-U_i=K_i-K_f\\\\q(V_f-V_i)=1.40\times 10^{-20}\\\\V_f-V_i=0.087--(1)\\$

Change in voltage is

$V_f-V_i=\frac{\lambda}{2\pi \epsilon_o}ln\frac{r_f}{r_i}\\\\ln|\frac{r_f}{r_i}|=(0.087)(\frac{2\pi \epsilon_o}{\lambda})\\\\ln|\frac{r_f}{r_i}|==0.8059\\\\\frac{r_f}{r_i}=2.24\\\\r_f=(2.24)(.12)\\\\r_f=0.268 m$

5 0

2 years ago

Io and Europa are two of Jupiter's many moons. The mean distance of Europa from Jupiter is about twice as far as that for Io and

Anastaziya [24]

The universal gravitation law and Newton's second law allow us to find that the answer for the relation of the rotation periods of the satellites is:

$\frac{T_{Eu}}{T_{Io}}$ = 2.83

The universal gravitation law states that the force between two bodies is proportional to their masses and inversely proportional to their distance squared

$F = G \frac{Mm}{r^2}$

Where G is the universal gravitational constant (G = 6.67 10⁻¹¹ $\frac{N m^2 }{kg^2}$ ), F the force, m and m the masses of the bodies and r the distance between them

Newton's second law states that force is proportional to the mass and acceleration of bodies

F = m a

Where F is the force, m the mass and the acceleration

In this case the body is the satellites of Jupiter and the planet,

$G \frac{Mm}{r^2} = m a$

Suppose the motion of the satellites is circular, then the acceleration is centripetal

a = $\frac{v^2}{r}$ r

Where v is the speed of the satellite and r the distance to the center of the planet

we substitute

$G \frac{Mm}{r^2} = m \frac{v^2}{r} \\G \frac{M}{r} = v^2$

Since the speed is constant, we can use the uniform motion ratio

v = $\frac{\Delta x}{t}$

In the case of a complete orbit, the time is called the period.

The distance traveled is the length of the orbit circle

Δx = 2π r

We substitute

$G \frac{M}{r} = (\frac{2 \pi r}{T} )^2 \\T^2 = (\frac{4 \pi ^2}{GM}) \ r^3$

Let's write this expression for each satellite

Io satellite

Let's call the distance from Jupiter is

r = $r_{Io}$

$T_{Io}^2 = (\frac{4 \pi ^2}{GM} ) \ r_{Io}^3$ TIo² = (4pi² / GM) rIo³

Europe satellite

Distance from Jupiter is

$r_{Eu} = 2 \ r_{Io}$

We calculate

$T_{Eu} = ( \frac{4\pi ^2 }{GM} (2 \ r_{Io})^3\\T_{Eu} = ( \frac{4 \pi ^2 }{GM}) r_{Io} \ 8$

$T_{Eu}^2 = 8 T_{Io}^2$

$\frac{ T_{Eu}}{T_{Io}} = \sqrt{8} = 2.83$

In conclusion, using the universal gravitation law and Newton's second law, we find that the answer for the relationship of the relation periods of the satellites is:

$\frac{T_{Eu}}{T_{Io}} = 2.83$

Learn more about universal gravitation law and Newton's second law here:

brainly.com/question/10693965

6 0

1 year ago