What traits did Sir William Gilbert have that made him a good scientist?

According to Boyle’s law, PV= K, what was the volume at the time of the first measurement given the following information? Round

alexandr402 [8]

Answer:

Iam a learning

6 0

3 years ago

17 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What

Lubov Fominskaja [6]

Answer:

$\frac{D}{d} = 4.12$

Explanation:

As we know that resistance of one copper wire is given as

$r = \rho \frac{L}{a}$

here we know that

$a = \pi (\frac{d}{2})^2$

now we have

$r = \rho \frac{L}{\pi (\frac{d^2}{4})}$

$r = \rho \frac{4L}{\pi d^2}$

now we know that such 17 resistors are connected in parallel so we have

$R = \frac{r}{17}$

$R = \rho \frac{4L}{17 \pi d^2}$

Now if a single copper wire has same resistance then its diameter is D and it is given as

$R = \rho \frac{4L}{\pi D^2}$

now from above two equations we have

$\rho \frac{4L}{\pi D^2} = \rho \frac{4L}{17 \pi d^2}$

$D^2 = 17 d^2$

now we have

$\frac{D}{d} = 4.12$

3 0

3 years ago

If all this energy is added to 50 kg of water (the amount of water in a 165-lb person) at 37 ∘C, what is the final state of the

Reika [66]

Answer:

Vapors

Explanation:

We take into account that all the energy from the lightning has been transformed into steam.

$\Delta U = Q - W\\Q = mC \Delta T\\Q = mL$

We calculate the amount of energy required by water to convert into steam.

$Q_{water} = 50 \times \times 4180 \times (100-37)\\= 1.132 \times 10^7 \ J$

$Q_{change\ to\ steam} = 50 \times 2.256 \times 106 \\= 1.128 \times 10^8 \ J$

$Q_{total} = 1.132 \times 10^7 + 1.128 \times 10^8\\= 1.126 \times 10^8 \ J$

From the lightning we received $10^{10} \ J$ of energy, out of which $1.126 \times 10^8$ has been used to convert the water into steam.

Energy left = $10^{10} - 1.126 \times 10^8 = 9.88 \times 10^9 \ J$

We use this energy to convert steam into vapors.

$Q = \Delta E$

$Q = \Delta E = mc (T_{f} - T{i})\\T_{f} = \frac {\Delta E}{mc} + T_{i}\\ \\T_{f} = 100 + \frac{9.88 \times 10^{10}}{50 \times 1970}\\T_{f} = 100 + 10^8\\T_{f} = 10 ^{ \ 8} \ {^ \circ } C$

With this temperature, we can easily interpret that the vapors will be dissociated in hydrogen and oxygen particles.

5 0

3 years ago

The melting point of a solid is 90.0C. What is the heat required to change 2.5 kg of this solid at 30.0C to a liquid? The specif

Neko [114]

Hey again!

Ok..

Now... The melting Point of this solid is 90°C.

Meaning That as soon as it gets to this temp... It STARTS Melting.

So at that temp... It still has some solid parts in it.

You can say its a Solid Liquid Mixture.

Additional Heat being applied at that point is not raising the temperature;rather its used in breaking the bonds in the solid. This is the Fusion stage.

After Fusion...It'd then Be a Pure Liquid with no solids in it.

So

Q'=MC∆0----- This is the heat needed to take the solid's temp from 30°c - 90°c

Q"=ml ----- This is the heat used in breaking the bonds holding the solids in the solid-liquid phase.

So

Q= Q' + Q"

Q= mc∆0 + ml

∆0 = 90°c - 30°c = 60°c

Q= 2.5(390)(60) + (2.5)(4000)

Q=6.9 x 10⁴Joules

7 0

3 years ago

shelly starts from rest on her bicycle at the top of a hill. After 6.0s she has reached a final velocity of 14m/s. what is shell

earnstyle [38]

Divide 14 by 6 and there is your answer with the unit of m

8 0

3 years ago

Read 2 more answers