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galben [10]
2 years ago
15

Please can anybody tell me what is this lab equipment​

Physics
1 answer:
Umnica [9.8K]2 years ago
5 0

Answer:

hii there

its a test tube holder

Explanation:

hope it helps

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Calculate the work done on a Pressure / Volume (PV) isothermal expansion where 400 Pa and 0.08 volume in m3?
Ainat [17]
PV = 400 x 0.08 = 32 J

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7 0
2 years ago
Belly-flop Bernie dives from atop a tall flagpole into a swimming pool below. His potential energy at the top is 7000 J (relativ
elena55 [62]

Answer:

KE₂ = 6000 J

Explanation:

Given that

Potential energy at top U₁= 7000 J

Potential energy at bottom U₂= 1000 J

The kinetic energy at top ,KE₁= 0 J

Lets take kinetic energy at bottom level =  KE₂

Now from energy conservation

U₁+ KE₁= U₂+ KE₂

Now by putting the values

U₁+ KE₁= U₂+ KE₂

7000+ 0 = 1000+ KE₂

KE₂ = 7000 - 1000 J

KE₂ = 6000 J

Therefore the kinetic energy at bottom is 6000 J.

5 0
2 years ago
What kind of energy involves the flow of charged particles?
Citrus2011 [14]

Answer:

Electrical energy is answer

Explanation:

hope it helps

Mark me as brainliest plz.

3 0
2 years ago
How many coulombs of positive charge are there in 4kg of plutonium, given its atomic mass is 244 and that each plutonium atom ha
Ne4ueva [31]

Answer:

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Explanation:

8 0
2 years ago
For this problem, we assume that we are on planet-i. the radius of this planet is r =4200 km, the gravitational acceleration at
Minchanka [31]
The expression commonly used for potential gravitational energy is just simplification. It is actually just the first term in Taylor expansion of the real expression. 
In general, the potential energy of gravitational field is defined as:
U=-G \frac{mM}{r}
Where G is universal gravitational constant, and r is the distance between the objects centers of mass. Negative sign represents the bound state.
Since we are not given the mass of the planet we have to calculate it.
F_g=G\frac{mM}{r_p^2}\\ mg=G\frac{mM}{r_p^2}\\ g=G\frac{M}{r_p^2}
This formula can be used for any planet. It gives you the gravitational acceleration on the planet's surface. We can use it to calculate the planet's mass:
g=G\frac{M}{r_p^2}\\ M=\frac{gr_p^2}{G}=2.41\cdot 10^{24}kg
Now we can calculate the potential energy of that cannonball when it reaches its maximum height.
U=-G \frac{mM}{r}\\ U=-G \frac{mM}{r_p+h}
When we plug in the numbers we get:
U=-4.99\cdot 10^{10} J
The potential energy has to be equal to the kinetic energy.
E_k=4.99\cdot 10^{10} J

3 0
3 years ago
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