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3 years ago

Where is the switch located on this diagram?

Physics

1 answer:

Fed [463]3 years ago

6 0

For this case, the switch is located at point B of the diagram.
Remember that point D is the universal symbol for resistance.
In A what you have is a source of power and in C what you have is a cable.
Therefore, the answer for this case is B.

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A 310-g air track cart is traveling at 1.25 m/s and a 260-g cart traveling in the opposite direction at 1.33 m/s. What is the sp

UNO [17]

Answer:

$v_{CM}=0.0732\ m/s$

Explanation:

given,

mass of the cart 1, m₁ = 310 g

speed of car 1 , v₁ = 1.25 m/s

mass of cart 2, m₂ = 260 g

speed of cart 2, v₂ = -1.33 m/s

speed of center of mass

$v_{CM}=\dfrac{m_1v_1 + m_2 v_2}{m_1 + m_2}$

$v_{CM}=\dfrac{0.31\times 1.25 +0.26\times (-1.33)}{0.31+0.26}$

$v_{CM}=\dfrac{0.0417}{0.57}$

$v_{CM}=0.0732\ m/s$

Hence, speed of center of mass of the system is equal to 0.0732 m/s

7 0

4 years ago

Heat is extracted from a certain quantity of steam at

vodomira [7]

Answer: $v=2452.91 m/s$

Explanation:

Given

initially steam is at $100^{\circ}C$ and converted to $0^{\circ} C$ ice

Let m be the mass of steam

latent heat of fusion and vaporization for water is

$L_f=3.33\times 10^5 J/kg$

$L_v=2.26\times 10^6 J/kg$

Heat required to convert steam in to water at $100^{\circ}C$

$Q_1=m\times L_v=m\cdot 2.26\times 10^6 J$

Heat required to lower water temperature to $0^{\circ}C$

$Q_2=m\times c\times \Delta T$

$Q_2=m\times 4.184\times (100)$

$Q_2=4.184m\times 10^5 J$

Heat required to convert $0^{\circ}C$ water to ice at $0^{\circ}C$ is

$Q_3=m\times L_f$

$Q_3=m\times 3.33\times 10^5=3.33m\times 10^5 J$

$Q=Q_1+Q_2+Q_3$

$Q=(2.26+0.4184+0.33)m\times 10^6 J$

$Q=3.0084m\times 10^6 J$

So this energy is equal to kinetic energy of bullet of mass m moving with velocity v

$Q=\frac{1}{2}mv^2$

$3.0084m\times 10^6=\frac{1}{2}mv^2$

$v^2=3.0084\times 2\times 10^6$

$v=2.452\times 10^3 m/s$

$v=2452.91 m/s$

5 0

4 years ago

PLEASE ANSWER THIS WILL MARK AS BRAINLIEST PLEASE USE TRUTHFUL ANSWERS PLEASE

luda_lava [24]

Australia separated from other continents and species there evolved independently

8 0

3 years ago

Hi.

docker41 [41]

<h3>Answer :</h3>

Let the final temperature be "T".

For the piece of copper :

mass, $\sf{m_c=40\ g.}$

specific heat capacity, $\sf{c_c=0.4\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_c=200^{\circ}C.}$

Then the heat of copper :

$\sf{\dashrightarrow Q_c=m_cc_c\,\Delta\!T_c}$

$\sf{\dashrightarrow Q_c =16(T-200)\ J}$

For copper calorimeter :

mass, $\sf{m_{cc} =60\ g.}$

specific heat capacity, $\sf{c_{cc} =0.4\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_{cc} =25^{\circ}C.}$

Then the heat of copper calorimeter :

$\sf{\dashrightarrow Q_{cc} =m_{cc}c_{cc}\,\Delta\!T_{cc}}$

$\sf{\dashrightarrow Q_{cc} =24(T-25)\ J}$

For water :

mass, $\sf{m_w=50\ g. }$

specific heat capacity, $\sf{c_w= 4.2\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_w=25^{\circ}C.}$

Then heat of water :

$\sf{\dashrightarrow Q_w=m_wc_w\,\Delta\!T_w}$

$\sf{\dashrightarrow Q_w=210(T-25)\ J}$

By energy conservation, the sum of all these energies should be zero as there were no heat energy change before the process, i.e.,

$\sf{\dashrightarrow Q_c+Q_{cc}+Q_w=0}$

$\sf{\dashrightarrow16(T-200)+24(T-25)+210(T-25)=0}$

$\sf{\dashrightarrow 250T- 9050=0}$

$\sf{\dashrightarrow T=36.2^{\circ}C}$

$\large \underline{\underline{\boxed{\sf T=36.2^{\circ}C}}}$

<u>____________________________</u>

[Note: in case of considering temperature difference it's not required to convert the temperatures from $\sf{^{\circ}C}$ to K or K to $\sf{^{\circ}C}$ .]

7 0

3 years ago

What happens to the atom if one proton and one electron is removed from the atom?

solong [7]

In that case, it will change it's identity to the atom of one degree less than that!

Hope this helps!

6 0

3 years ago