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3 years ago

What Type of Wave is Shown Below? PLZ ANSWER I'M ALREADY FAILING AT THIS SUBJECT!

Physics

1 answer:

uranmaximum [27]3 years ago

8 0

Answer:

A......................

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A cup of water is warmed from 21 °C to 85 °C. What is the difference between these two temperatures, in kelvins?

Ilya [14]

Answer:

337k

Explanation:

First, let us find the difference between the given two temperatures.

Difference = 85°C - 21°C

= 64°C

And now we have to write the temperature in kelvins.

To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.

Let us find it now.

64°C + 273 = 337k

Therefore,

64°C ⇒ 337k

8 0

1 year ago

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separa

Elanso [62]

Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

$\frac{C_2}{C_1}$ = $\frac{2.3}{1.15}$

a ) when charge remains constant

energy = $\frac{q^2}{2C}$

q is charge and C is capacity

energy stored initially E₁= $\frac{q^2}{2C_1}$

energy stored finally E₂ = $\frac{q^2}{2C_2}$

$\frac{E_1}{E_2} = \frac{C_2}{C_1}$ = $\frac{2.3}{1.15}$

$E_2$ = $\frac{1.15}{2.3 } \times E_1$

= $\frac{1.15}{2.3 } \times 8.38$

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

$\frac{E_2}{E_1} = \frac{C_2}{C_1}$

$\frac{E_2}{8.38} = \frac{2.3}{1.15}$

$E_2$ = 16.76 .

8 0

3 years ago

A car has a uniform velocity of 108km/h. How far does it travel in 30 seconds?

natita [175]

Velocity=108km/h

Convert to m/s

$\\ \tt\longmapsto 108\times \dfrac{5}{18}$

$\\ \tt\longmapsto 30m/s$

Time=30s.

$\\ \tt\longmapsto Distance=Velocity\times Time$

$\\ \tt\longmapsto Distance=30(30)$

$\\ \tt\longmapsto Distance=900m$

6 0

3 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:

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

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3 years ago

How can global climate change be studied

Marina CMI [18]

Not sure... need help with it

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3 years ago