What percentage of the world's energy needs is supplied by fossil fuels?

An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its pla

Vikentia [17]

Answer:

The new energy density is same as initial energy density u0 as it is independent of plate dimensions

Explanation:

As we know that energy density is total energy per unit volume

so we know that isolated plates of capacitor is placed here

$C = \frac{\epsilon_0 A}{d}$

here we have energy density given as

$u = \frac{Q^2}{2(\epsilon_0 A/d)} Ad$

so we have

$u = \frac{Q^2d^2}{2\epsilon_0}$

so it is independent of the dimensions of the plate while total charge on the plates is constant always

so energy density will not change and remains the same

8 0

3 years ago

By measuring the amounts of parent isotopes and daughter product in the minerals contained in a rock, and by knowing the half-li

sertanlavr [38]

The asker of the second question needs a tutorial in radiometric dating. There is little likelihood that the daughter isotope has the same atomic weight as the parent isotope. To measure the mass isotopes doesn't tell us how many atoms of each exist. To get around that let's pretend — which will likely serve the purpose ineptly intended — that the values give an the particle ratio, 125:875.

The original parent isotope count was 125 + 875 = 1000. The remaining parent isotope is 125/1000 or 1/8. 1/8 = (1/2)^h, where h is the number of half-lives. 

h = log (1/8) ÷ log(1/2) = 3 

And 3 half-lives • 150,000 years/half-life = 450,000 years.

4 0

3 years ago

Which trial’s cart has the greatest momentum at the bottom of the ramp?

Licemer1 [7]

The momentum of each cart is given by:
$p=mv$
where
m is the mass of the cart
v is its velocity (at the bottom of the ramp)

To answer the problem, let's calculate the momentum of each of the 4 carts:
1) $p=(200 kg)(6.5 m/s)=1300 kg m/s$
2) $p=(220 kg)(5.0 m/s)=1100 kg m/s$
3) $p=(240 kg)(6.4 m/s)= 1536 kg m/s$
4) $p=(260 kg)(4.8 m/s)=1248 kg m/s$

Therefore, the cart with greatest momentum is cart 3, so the right answer is
- trial 3, because this trial has a large mass and a large velocity

8 0

3 years ago

A 1.80-m -long uniform bar that weighs 531 N is suspended in a horizontal position by two vertical wires that are attached to th

Readme [11.4K]

Answer:

(a) 498.4 Hz

(b) 442 Hz

Solution:

As per the question:

Length of the wire, L = 1.80 m

Weight of the bar, W = 531 N

The position of the copper wire from the left to the right hand end, x = 0.40 m

Length of each wire, l = 0.600 m

Radius of the circular cross-section, R = 0.250 mm = $.250\times 10^{- 3}\ m$

Now,

Applying the equilibrium condition at the left end for torque:

$T_{Al}.0 + T_{C}(L - x) = W\frac{L}{2}$

$T_{C}(1.80 - 0.40) = 531\times \frac{1.80}{2}$

$T_{C} = 341.357\ Nm$

The weight of the wire balances the tension in both the wires collectively:

$W = T_{Al} + T_{C}$

$531 = T_{Al} + 341.357$

$T_{Al} = 189.643\ Nm$

Now,

The fundamental frequency is given by:

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

$\mu = A\rho = \pi R^{2}\rho$

(a) For the fundamental frequency of Aluminium:

$f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\mu}}$

$f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\pi R^{2}\rho_{Al}}}$

where

$\rho_{l} = 2.70\times 10^{3}\ kg/m^{3}$

$f = \frac{1}{2\times 0.600}\sqrt{\frac{189.643}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 498.4\ Hz$

(b) For the fundamental frequency of Copper:

$f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\mu}}$

$f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\pi R^{2}\rho_{C}}}$

where

$\rho_{C} = 8.90\times 10^{3}\ kg/m^{3}$

$f = \frac{1}{2\times 0.600}\sqrt{\frac{341.357}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 442\ Hz$

7 0

4 years ago

A 31 cm tall object is placed in front of a concave mirror with a radius of 34 cm. The distance of the object to the mirror is 9

Setler [38]

Focal length is the distance between the center of a convex lens or a concave mirror and the focal point of the lens or mirror — the point where parallel rays of light meet, or converge. From the optics the focal length of the mirror can be defined as the radius of the mirror divided between two, or in other words, half the radius of the mirror.

$f = \frac{R}{2}$

$f = \frac{34}{2}$

$f = 17cm$

Therefore the focal length of the mirror is 17cm

5 0

3 years ago