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2 years ago

11

What is the difference between wind energy and solar energy?

1 answer:

Viktor [21]2 years ago

8 0

Answer: Wind energy is powered by wind, and solar energy is powered by the Sun.

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An alert driver can apply the brakes fully in about 0.5 seconds. How far would the car travel if it

dybincka [34]

Answer:

The car would travel after applying brakes is, d = 14.53 m

Explanation:

Given that,

The time taken to apply brakes fully is, t = 0.5 s

The velocity of the car, v = 29.06 m/s

The distance traveled by the car in 0.5 s, d = ?

The relation between the velocity, displacement, and time is given by the formula

d = v x t m

Substituting the values in the above equation,

d = 29.06 m/s x 0.5 s

= 14.53 m

Therefore, the car would travel after applying brakes is, d = 14.53 m

8 0

4 years ago

Need help with stu*pid science thing that I keep getting wrong, augh!!

Kryger [21]

Answer:

i think it is iron

Explanation:

its the only one that makes sense to me

4 0

3 years ago

Read 2 more answers

Describe the difference between potencial and kinetic energy

avanturin [10]

Potential energy is energy stored in an object. kinetic energy is energy of motion

8 0

3 years ago

A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.

Kruka [31]

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

$a=\dfrac{v-u}{t}$ .............(1)

Now, the second equation of motion is :

$s=ut+\dfrac{1}{2}at^2$

Put the value of a in above equation as :

$s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2$

$s=\dfrac{t(u+v)}{2}$

$u=\dfrac{2s}{t}-v$

$u=\dfrac{2\times 47}{8.6}-2.3$

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

8 0

4 years ago

A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3

Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

$\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}$

$\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

For calculating number of turns

$\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

$N_{S}=\frac{4407}{65}$ $N_{S}=67.8$

The number of turns in the secondary winding are 67.8 turns.

6 0

3 years ago