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fiasKO [112]
2 years ago
8

A bond between two copper atoms would be a bond? *

Physics
2 answers:
mihalych1998 [28]2 years ago
8 0
A metallic bond would be formed
Tems11 [23]2 years ago
7 0
Metallic bond is formed
You might be interested in
Please help
inysia [295]

Explanation:

Start with what you know and list your knowns and unknowns

F = ma

F= 3N

m = 6kg

a =?

3N = 6kg x a

solve for a

3N / 6kg = a

3 0
3 years ago
A 150 g copper bowl contains 210 g of water, both at 24.0°C. A very hot 430 g copper cylinder is dropped into the water, causing
Dahasolnce [82]

Answer:

A. 15969.22 cal

B. 1052,22 cal

C. 528,87 °C

Explanation:

To solve this kind of question, a proper method is to work from the data that you have towards the data that you need. Also, it is recommended to analyze related equations as they could give us clues on how to find the missing information or the information that the problem is asking us.

Let us start with Question A. It is important to remember that energy transfers with the environment are being neglected; this means that all the energy that the cylinder lose is picked up by the water and the copper bowl. To find the amount of energy transferred to the water, we first find the amount of energy necessary to raise the water’s temperature to 100°C and then we find the amount of energy necessary to evaporate the 17.1 g of water indicated by the question. This would be:

Q = m_water * CP_water *∆T =210g *1 cal/(g K) * (100°C-24°C) = 15960 cal

Q_evap = m_wat * L = 17,1 g * 539 cal/kg* (1 kg)/(1000 g) =9.2169 cal

Therefore, the total energy that was transferred to the water is the sum of these components, that would be Q_tot = 15960 cal + 9.2159 cal = 15969.22 cal.  Let´s also remember that a temperature difference in K is equal to a temperature difference in ° C

To solve Question B, we use the same method. We must find the amount of energy necessary to raise the temperature from its initial temperature to the one stated by the problem to be the equilibrium temperature of the system (100°C):

Q= m_copper *CP_copper *∆T = 150g * 0.0923 cal/(g K) * (100°C-24°C) = 1052,22 cal

If we add the components we just found in questions A and B, we can find the amount of energy than the Copper cylinder lost, this would be: Q_tot = 15969.22 cal + 1052.22 cal = 17021.44 cal.

The question C asks us to find the initial temperature of the cylinder and Q_tot will help us to find it.

We know that Q_tot is the energy lost by the cylinder and we also know that Q_tot = m_cylinder * CP_copper * ∆T. Therefore, what we need to do  is clear the last term of the equation and find the initial temperature.

Q_tot = m_cylinder *CP_copper *∆T → T_fin-T_initial = Q_tot/(m_cylinder*CP_copper ) = (-17021.44 cal)/(430g*0.0923 cal/(g K))

→ T_initial = 100°C + (-17021.44 cal)/(430g * 0.0923 cal/(g K)) = 528,87 °C

If we convert the 100°C to K before we do the calculation, the result would be the same one, You would only need to add 273,15 to the final result to check it out.  

Hope everything was clear. If you have any further question, I'll be happy to help :D

5 0
3 years ago
A ball is tossed with a velocity of 10 m/s directed vertically upward from a window located 20 m above the ground. Determine the
marusya05 [52]

Answer:

Explanation:

Given

Initial velocity of ball u=10\ m/s

height of window h=20\ m

Using Equation of motion

y=ut+\frac{1}{2}at^2

where u=initial velocity

t=time

a=acceleration

As ball is already is at a height of 20 m so

Y=ut+\frac{1}{2}at^2+20

Y=10\times t+0.5\times (-9.8)t^2+20

Y=-4.9t^2+10t+20

(b)highest point is obtained at v=0

v^2-u^2=2as

where

v=final velocity

u=initial velocity

a=acceleration

s=displacement

(0)-10^2=2\times (-9.8)\times s

s=\frac{100}{19.6}

s=5.102\ m

Highest Point will be s+20=25.102\ m

(c)Time taken when the ball hit the ground i.e. at Y=0

-4.9t^2+10t+20=0

t=3.28\ s

impact velocity v=\sqrt{2\times 9.8\times 25.102}

v=22.181\ m/s

7 0
3 years ago
A cylinder with a piston contains 0.200 mol of oxygen at 2.00×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The
ikadub [295]

Answer:

A. W=600\ J

B. Q=2112\ J

C. \Delta U=1512\ J

D. W=0\ J

Explanation:

Given:

  • no. of moles of oxygen in the cylinder, n=0.2
  • initial pressure in the cylinder, P_i=2\times 10^5\ Pa
  • initial temperature of the gas in the cylinder, T_i=360\ K

<em>According to the question the final volume becomes twice of the initial volume.</em>

<u>Using ideal gas law:</u>

P.V=n.R.T

2\times 10^5\times V_i=0.2\times 8.314\times 360

V_i=0.003\ m^3

A.

<u>Work done by the gas during the initial isobaric expansion:</u>

W=P.dV

W=P_i\times (V_f-V_i)

W=2\times 10^5\times (0.006-0.003)

W=600\ J

C.

<u>we have the specific heat capacity of oxygen at constant pressure as:</u>

c_v=21\ J.mol^{-1}.K^{-1}

Now we apply Charles Law:

\frac{V_i}{T_i} =\frac{V_f}{T_f}

\frac{0.003}{360} =\frac{0.006}{T_f}

T_f=720\ K

<u>Now change in internal energy:</u>

\Delta U=n.c_p.(T_f-T_i)

\Delta U=0.2\times 21\times (720-360)

\Delta U=1512\ J

B.

<u>Now heat added to the system:</u>

Q=W+\Delta U

Q=600+1512

Q=2112\ J

D.

Since during final cooling the process is isochoric (i.e. the volume does not changes). So,

W=0\ J

7 0
2 years ago
If you wanted to continue the flow of electricity, which of the following objects should you use?
Delicious77 [7]

penny being its conductive

7 0
3 years ago
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