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2 years ago

What is one difference between using a solar cell for electricity and using a windmill?

Physics

1 answer:

wlad13 [49]2 years ago

8 0

Answer:

Solar cell generates DC

Windmill generates AC

Explanation:

Solar cell generates DC from the panel, to use this DC for electricity it has to be passed to an inverter which convert DC to AC.

Windmill generates AC from the wind blades. This type of energy can be used directly with household appliances using AC, or passed to a rectifier to convert it to DC.

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How to I isolate t? (time value)

Vlad1618 [11]

Heya!!!

Answer to your question:

All you need to do is to compare both the equations...
On comparing, you'll get
d=6m
v=6.5m/s
a=-9.81m/s^2
we know,

6= 6.5t -0.5 (9.81)t^2
6=6.5t-4.905 t^2
4.905 t^2-6.5t +6=0
now you can use quadratic eq. to solve this.

Hope it helps ^_^

4 0

3 years ago

A horizontal spring with spring constant of 9.80 N/m is attached to a block with a mass of 1.20 kg that sits on a frictionless s

Andreas93 [3]

Answer:

Explanation:

Given

at certain point displacement and velocity is 0.345 m and 0.54 m/s

Therefore Potential and Kinetic Energy associated is

$U=\frac{1}{2}kx^2$

$U=\frac{1}{2}\times 9.8\times (0.345)^2$

$U=0.583 J$

Kinetic Energy $K=\frac{1}{2}mv^2$

$k=\frac{1}{2}\times 1.2\times (0.54)^2=0.174 J$

Total Energy $T=U+K=0.583+0.174=0.757 J$

Total Energy is conserved hence at maximum displacement all energy will be Potential energy

$T=\frac{1}{2}kA^2$

where A=maximum displacement

$0.757=\frac{1}{2}\times 9.8\times A^2$

$0.1544=A^2$

$A=0.393 m$

Maximum speed occurs at equilibrium Position where Potential Energy is zero

thus $T=\frac{1}{2}mv^2$

$0.757=\frac{1}{2}\times 1.2\times v_{max}^2$

$v_{max}=1.123 m/s$

(c)When block is at 0.2 m from Equilibrium speed then its Potential Energy is

$U=\frac{1}{2}kx^2=\frac{1}{2}\times 9.8\times (0.2)^2=0.196 J$

$T=U+K$

$K=0.757-0.196=0.561 J$

$K=\frac{1}{2}mv^2$

$0.561=\frac{1}{2}\times 1.2\times v^2$

$v=0.966 m/s$

7 0

3 years ago

Evaluate the efficiency of two identical-looking conveyor belts. Belt A can move a 10 newtons weight one meter in 3 seconds. Bel

LUCKY_DIMON [66]

The Belt A has an efficiency of 50% while the Belt B has an efficiency of 25%

Why?

To calculate the efficiency of the belts, we need to consider that both conveyor belts are identical looking and both weights are equal (10 Newtons).

We need to remember that the formula to calculate work is equal to:

$Work=Force*distance(Joule)$

Also,

$Joules=\frac{Newton}{m}$

We know the following:

For belt A,

$weight=10N\\time=3seconds\\distance=1meter\\Joule=\frac{N}{m}=\frac{10N}{1m}=10J$

For belt B,

$weight=10N\\time=3seconds\\distance=2meter\\Joule=\frac{N}{m}=\frac{10N}{2m}=5J$

Now, to calculate the efficiencty of the belts, we can use the following formula:

$Efficiency=\frac{Work(Out)}{Work(Starting)}*100$

We know that for both belts, the starting work was 20 J.

So, calculating we have:

$Efficiency_{BeltA}=\frac{10J}{20J}*100=50(percent)$

$Efficiency_{BeltB}=\frac{5J}{20J}*100=25(percent)$

Hence, we can see that The Belt A has an efficiency of 50% while The Belt B has an efficiency of 25%

Have a nice day!

3 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

3 years ago

What causes light to bend when it moves from one transparent medium to another?

lapo4ka [179]

Whenever light goes from transparent to another medium or from rarer to denser medium or vice versa, it's speed changes and hence light bends.

3 0

3 years ago

Read 2 more answers