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LUCKY_DIMON [66]
3 years ago
9

How do liquids and gases transfer heat?

Physics
2 answers:
defon3 years ago
8 0
I believe it’s d. By convention.
adoni [48]3 years ago
4 0
I believe it’s C, by the air movement
You might be interested in
A magnetic field is applied to a freely floating uniform iron sphere with radius R = 2.00 mm. The sphere initially had no net ma
grigory [225]

Answer:

\omega=4.2704*10^-^5 rad/s

Explanation:

Angular Momentum Formula For atoms=L_{atom}=0.12Nm_{s}h

Where:

m_{s}h is the momentum for one atom (m_s is the spin quantum number)

N is the number of atoms=\frac{N_{A}*m}{M}

Where:

N_A is Avogadro Number

m is the mass of sphere

M is the molar mass of iron

Angular Momentum Formula For atoms will be=L_{atom}=0.12\frac{N_{A}m}{M} m_{s}h

Angular Momentum of Sphere=L_{sphere}=I\omega

where:

I=\frac{2mR^{2}}{5}

So,Angular Momentum of Sphere=L_{sphere}=\frac{2mR^{2}}{5}\omega

Angular Momentum of sphere=Angular Momentum of atoms

L_{sphere}=L_{atom}

\frac{2mR^{2}}{5}\omega=0.12\frac{N_{A}m}{M} m_{s}h

For iron, m_s =\frac{1}{2}. So above equation will become:

\omega=\frac{0.12*5*N_{A}h}{4*M*R^{2} }

Where R=2mm, M=0.0558Kg/mol (Molar Mass of iron),h=Planck's Constant/2π

\omega=\frac{0.12*5*(6.022*10^{23})(6.63*10^{-34}/2*\pi)}{4*0.0558*(2*10^{-3})^{2}}

\omega=4.2704*10^-^5 rad/s

5 0
3 years ago
Describe how work is done by a skater pulling in her arms during a spin. In particular, identify the force she exerts on each ar
liberstina [14]

Answer:

 ∑ τ =0,  L₀ = L_{f}

Explanation:

In a circular turning movement, when the arms are extended and then contracted in two possibilities:

- They are lowered the force of gravity is what pulls them, the tension of the muscle becomes zero to allow this movement.

    In this movement the force is vertical(gravity) and the movement of the center of mass of each arm is vertical, so that the work is the weight value of the arm by the distance traveled by the center of mass.

- Another possibility is that the arms have stuck to the body, in this case the person's muscles perform the force, this force is horizontal and the displacement is the horizontal of the center of mass of the arms from the extended position to the contracted

 

In these movements the torque of the external force is equal for each arm, but in the opposite direction, so they are canceled where a net torque of zero, this causes the angular momentum to be preserved, which changes is the moment of inertia of the system and therefore you must also change the angular velocity to keep your product constant

          ∑ τ =0

                L₀ = L_{f}

              I₀ w₀ = I w

4 0
3 years ago
You are at home in your air conditioned garage. You are planning a family road trip from New York to Florida. You are lnfating t
Alex
The tires deflated and so that means that you won’t be able to travel
6 0
3 years ago
A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

4 0
3 years ago
Two boxers are fighting. Boxer 1 throws his 5 kg fist at boxer 2 with a speed of 9 m/s.
Sladkaya [172]

Answer:

0.001 s

Explanation:

The force applied on an object is equal to the rate of change of momentum of the object:

F=\frac{\Delta p}{\Delta t}

where

F is the force applied

\Delta p is the change in momentum

\Delta t is the time interval

The change in momentum can be written as

\Delta p=m(v-u)

where

m is the mass

v is the final velocity

u is the initial velocity

So the original equation can be written as

F=\frac{m(v-u)}{\Delta t}

In this problem:

m = 5 kg is the mass of the fist

u = 9 m/s is the initial velocity

v = 0 is the final velocity

F = -45,000 N is the force applied (negative because its direction is opposite to the motion)

Therefore, we can re-arrange the equation to solve for the time:

\Delta t=\frac{m(v-u)}{F}=\frac{(5)(0-9)}{-45,000}=0.001 s

4 0
3 years ago
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