What kind of front means air masses and their boundaries stop moving?

On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a froze

sp2606 [1]

Answer:

0.78333 m/s in the opposite direction

1.566 m/s in the same direction

Explanation:

$m_1$ = Mass of penny = 0.0025 kg

$m_2$ = Mass of nickel = 0.005 kg

$u_1$ = Initial Velocity of penny = 2.35 m/s

$u_2$ = Initial Velocity of nickel = 0 m/s

$v_1$ = Final Velocity of penny

$v_2$ = Final Velocity of nickel

As momentum and Energy is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

${\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}$

From the two equations we get

$v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.0025-0.005}{0.0025+0.005}\times 2.35+\frac{2\times 0.5}{0.4005+0.5}\times 0\\\Rightarrow v_1=-0.78333\ m/s$

The final velocity of the penny is 0.78333 m/s in the opposite direction

$v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.0025}{0.0025+0.005}\times 2.35+\frac{0.005-0.0025}{0.005+0.0025}\times 0\\\Rightarrow v_2=1.566\ m/s$

The final velocity of the nickel is 1.566 m/s in the same direction

6 0

3 years ago

A light bulb emits light that travels uniformly in all directions. Detailed measurements show that at a distance of 56 m from th

vladimir1956 [14]

Complete question:

A light bulb emits light that travels uniformly in all directions. Detailed measurements show that at a distance of 56 m from the bulb, the amplitude of the electric field is 3.78 V/m. What is the average intensity of the light?

Answer:

The average intensity of the light is 0.02 W/m²

Explanation:

Given;

Amplitude of the electric field, E₀ = 3.78 V/m

The average intensity of the light is calculated as follows;

$I_{avg} = \frac{c\epsilon_0 E_0^2}{2}$

where;

$I_{avg}$ is the average intensity of the light

c is speed of light = 3 x 10⁸ m/s

$I_{avg} = \frac{(3\times 10^8)(8.85 \times 10^{-12}) (3.78)^2}{2} \\\\I_{avg} = 0.01897 \ W/m^2\\\\I_{avg} = 0.02 \ W/m^2$

Therefore, the average intensity of the light is 0.02 W/m²

4 0

3 years ago

a sealed cylinder contains a sample of ideal gas at a pressure of 2.0 atm. The rms speed of the molecules is v0. If the rms spee

fredd [130]

Answer:

P_2 = 1.62 atm

Explanation:

We know the formula for the rms speed of the ideal gas is given by

$v_{rsm}=\sqrt{\frac{3PV}{m} }$

P= pressure of the surrounding

V= volume of the vessel

m= mass of the gas

Now, From this formula rms speed (v_rms) is directly proportional to square root is pressure.

Then

$\frac{v_{rsm,1}}{v_{rsm,2}} =\sqrt{\frac{P_1}{P_2} }$

given that v_rsm,1= v0

and v_rsm,2=0.9v0

putting these values we get

$\frac{v0}{0.9v0} =\sqrt{\frac{2}{P_2} }$

P_2 = 1.62 atm

8 0

3 years ago

Read 2 more answers

A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05

7nadin3 [17]

Answer:

$W_{drag} = 4.223\,J$

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

$U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}$

The work done on the ball due to drag is:

$W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})$

$W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})$

$W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]$

$W_{drag} = 4.223\,J$

7 0

3 years ago

Read 2 more answers

1. How far does a skydiver fall

GREYUIT [131]

Free fall without air resistance:

g=9.81 m/s²

t=235 s

h=?

h=0.5*g*t²

h=0.5*9.81*235²

h=270879m =270 km

5 0

3 years ago