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2 years ago

How many moles of MgCl2 are there in 302 g of the compound?

Physics

2 answers:

VashaNatasha [74]2 years ago

8 0

MgCl2 molar mass is about 94g/mol so 302/94 is about 3.2. So the answer is 3.2 moles in 302 g.

Hope this helps:)

lisov135 [29]2 years ago

3 0

Answer:

‍♀️‍♀️‍♀️‍♀️sorrryyyyy

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A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming

Monica [59]

Answer:

The deceleration is $a = - 76.27 m/s^2$

Explanation:

From the question we are told that

The height above firefighter safety net is $H = 14 \ m$

The length by which the net is stretched is $s = 1.8 \ m$

From the law of energy conservation

$KE_T + PE_T = KE_B + PE_B$

Where $KE_T$ is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

and $PE_T$ is the potential energy of the before jumping which is mathematically represented at

$PE_T = mg H$

and $KE_B$ is the kinetic energy of the person just before landing on the safety net which is mathematically represented at

$KE_B = \frac{1}{2} m v^2$

and $PE_B$ is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

So the above equation becomes

$mgH = \frac{1}{2} m v^2$

=> $v = \sqrt{2 gH }$

substituting values

$v = 16.57 m/s$

Applying the equation o motion

$v_f = v + 2 a s$

Now the final velocity is zero because the person comes to rest

$0 = 16.57 + 2 * a * 1.8$

$a = - \frac{16.57^2 }{2 * 1.8}$

$a = - 76.27 m/s^2$

4 0

2 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .