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WITCHER [35]
2 years ago
9

What is the name of the compound Cu(HCO3)2​

Physics
1 answer:
Naddik [55]2 years ago
5 0

Answer:

Copper(II) Hydrogen Carbonate Cu(HCO3)2 Molecular Weight

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4) Block A has a mass of 3kg and velocity of 13m/s, catching up with a second block B that has a mass of 3kg and is moving with
serious [3.7K]

Answer:

Option A is the correct answer.

Explanation:

Here momentum is conserved.

That is \left (m_Av_A+m_Bv_B \right )_{initial}=\left (m_Av_A+m_Bv_B \right )_{final}

Substituting values

    3\times 13+3\times 5=3v_A+3\times 8\\\\3v_A=39+15-24\\\\3v_A=30\\\\v_A=10m/s

Speed of block A after collision = 10 m/s

Option A is the correct answer.

5 0
3 years ago
A 1.50-m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 1
nydimaria [60]

Answer:

a) 1.06*10^-5

b) 0.00105 °C^-1

Explanation:

Given that

Length of the cylinder, L = 1.5 m

Radius of the cylinder, r = 0.25 cm

Voltage across the rod, V = 15 V

I• at Temperature T• = 20° C is 18.5 A

I at Temperature T = 90° C is 17.2 A

See attachment for calculations

5 0
3 years ago
Radiant energy comes from the _______.
snow_lady [41]

Answer:

C. Sun

Explanation:

7 0
3 years ago
Read 2 more answers
Moving the probe 1 cm towards the non-grounded electrode changes the value the potential from about 0.90 V to about 1.2 V. Expla
svp [43]

Answer:

-30 N/C

Explanation:

Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m

Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V

Since E = -ΔV/Δx

substituting the values of the variables into the equation, we have

E = -ΔV/Δx

E = -0.30 V/0.01 m

E = -30 V/m

Since 1 V/m = 1 N/C.

E = -30 N/C

So, the average electric field is -30 N/C

6 0
3 years ago
A rocket moves upward from rest with an acceleration of 40 m/s2 for 5 seconds. It then runs out of fuel and continues to move up
Snezhnost [94]

Answer:

Maximum height of rocket  = 2538.74 m

Explanation:

We have equation of motion s = ut + 0.5 at²

For first 5 seconds

          s = 0 x 5 + 0.5 x 40 x 5² = 500 m

Now let us find out time after 5 seconds rocket move upward.

We have the equation of motion v = u + at

After 5 seconds velocity of rocket

         v = 0 + 40 x 5 = 200 m/s

After 5 seconds the velocity reduces 9.8m/s per second due to gravity.

Time of flying after 5 seconds

          t=\frac{200}{9.81}=20.38s

Distance traveled in this 20.38 s

          s = 200 x 20.38 - 0.5 x 9.81 x 20.38² = 2038.74 m

Maximum height of rocket = 500 +2038.74 = 2538.74 m

6 0
3 years ago
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