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3 years ago

One problem with wind energy as a major source of electricity is _____. (1 point)

Physics

1 answer:

Alborosie3 years ago

8 0

the problems that harnessing wind for energy is the expense of large tracts of land in populated areas.

hope this helps you

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Consider the uniform electric field \vec{E} =(4.00~\hat{j}+3.00~\hat{k})\times 10^3~\text{N/C} E ⃗ =(4.00 j ^ +3.00 k

Tema [17]

Answer:

Electric flux, $\phi=6.668\times 10^4\ Nm^2/C$

Explanation:

It is given that,

Electric field, $E=(4j+3k)\times 10^3\ N/C$

We need to find the electric flux through a circular area of radius 2.66 m that lies in the xy-plane. $A=Ak$

The electric flux is given by :

$\phi=E{\cdot}A$

$\phi=(4j+3k)\times 10^3{\cdot}Ak$

Since, k.k=i.i=j.j = 1

So,

$\phi=3\times 10^3\times \pi\times (2.66)^2\ k$

$\phi=6.668\times 10^4\ Nm^2/C$

So, the electric flux through a circular area is $\phi=6.668\times 10^4\ Nm^2/C$ .Hence, this is the required solution.

6 0

3 years ago

Why were republicans able to dominate politics from 1860 until 1932?

snow_lady [41]

Answer:First the Republicans were dominant from 1860 to around 1882, then not again until 1920 time 1932. The reason for the first part is the Democratic party started a civil war to keep their slaves and the Democratic party lost that war.

Explanation:

This is what i found.

5 0

3 years ago

How long does it take a wave to travel 1200 meters with the speed of 3 X 108m/sec?

Vladimir79 [104]

Around 3.70 seconds unless traveling through media

5 0

3 years ago

Which are heterogeneous mixtures?

inna [77]

C. Sugars dissolved in water

4 0

3 years ago

A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately

Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car $V_{c}$ = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

$V_{f}$ = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

$V_{f}$ = $V_{i}$ + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁ = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁ will be

$d_{c}$ = $V_{c}$ × t₁

we substitute

$d_{c}$ = 22.352 × 7

$d_{c}$ = 156.46 m

now distance between polive car and speeding car

Δd = $d_{c}$ - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( $V_{f}$ - $V_{c}$ )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = $V_{f}$ × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b) how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0

3 years ago