Solve 13.004+3.09+112.947

Over the last several decades scientists have considered the problem of nonrenewable natural resources, such as fossil fuels. Hu

sesenic [268]

The statement "Good locations for turbines are limited" describes a drawback to wind energy.

Answer: Option B

Explanation:

Wind energy is one of the most useful and efficient renewable energy sources. But nothing is ideal in this universe and the same thing applies for wind energy also. The generation of electricity from wind energy requires setting up of turbines.

And these turbines can be set up in plane areas which is free from any disturbance except wind flow. In open area and flat plane surface only the turbines can rotate freely with the effect of wind.

But regions where the wind flow is minimum due to snow formation like the northern region of earth, the turbines cannot be set up there. So the locations for setting up of turbines are limited for good outcome in wind energy. This is one of the drawback of wind energy.

8 0

3 years ago

Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal

kirza4 [7]

At t =0, the velocity of A is greater than the velocity of B.

We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;

A: vA (t) = ť^2 – 5t + 20

B: vB (t) = t^2+ 3t + 10

Given that t = 0 in both cases;

vA (0) = 0^2 – 5(0) + 20

vA = 20 m/s

For vB

vB (0) = 0^2+ 3(0) + 10

vB = 10 m/s

We can see that at t =0, the velocity of A is greater than the velocity of B.

Learn more: brainly.com/question/24857760

Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?

3 0

3 years ago

A delivery truck travels 2.8 km North, 1.0 km East, and 1.6 km South. The final displacement from the origin is ___km to the ___

34kurt

Answer:

The final displacement from the origin is 1.6 km to the NE

Explanation:

The directions in which the delivery truck travels are;

1) 2.8 km North = 2.8· $\hat j$ , in vector form

2) 1.0 km East = 1.0· $\hat i$ , in vector form

3) 1.6 km South = -1.6· $\hat j$ , in vector form

Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;

Final displacement, Δd = 2.8· $\hat j$ + 1.0· $\hat i$ +(-1.6· $\hat j$ ) = 1.2· $\hat j$ + 1.0· $\hat i$

Final displacement, = 1.0· $\hat i$ + 1.2· $\hat j$

Where;

Δx = The displacement in the x-direction = 1.0· $\hat i$

Δy = The displacement in the y-direction = 1.2· $\hat j$

The magnitude of the resultant displacement vector is given as follows

$\left | d \right |$ = √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)

The magnitude of the resultant displacement vector ≈ 1.6 km

The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE

Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.

3 0

3 years ago

When two waves collide and create a taller wave, what interference occurs?

oksano4ka [1.4K]

It would be Constructive interference. It occurs when two waves collide and make a taller wave.

3 0

3 years ago

An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 34.5

Inessa05 [86]

Answer:

the acceleration of the airplane is 5.06 x 10⁻³ m/s²

Explanation:

Given;

initial velocity of the airplane. u = 34.5 m/s

distance traveled by the airplane, s = 46,100 m

final velocity of the airplane, v = 40.7 m/s

The acceleration of the airplane is calculated from the following kinematic equation;

v² = u² + 2as

$2as= v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(40.7)^2 -(34.5)^2}{2 \times 46,100} \\\\a = 5.06 \ \times \ 10^{-3} \ m/s^2$

Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²

5 0

3 years ago