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crimeas [40]
3 years ago
8

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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A 5.00 kilogram block slides along a horizontal,frictionless surface at 10.0 meters per second. for 4.00 seconds. The magnitude
finlep [7]
In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
                   = (5 * 10) Kg m/s
                   = 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".
5 0
3 years ago
When doing yoga stretches, you are working on the cardiovascular endurance component of physical fitness.
Tomtit [17]
I believe it is true :D
6 0
3 years ago
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What is an expression for the vertical component of this vector?
grandymaker [24]
The correct choice is D .
4 0
3 years ago
Two positive charges of 6 µC are separated by a distance of 50 cm in air. What is the electric field strength at the midpoint of
attashe74 [19]

Answer:

The electric field strength at the midpoint of the line joining the charges is zero (0)

Explanation:

Given that the two charges are both positive (same charge) and are equal in magnitude that is 6uC. The electric field strength at the midpoint of the line joining the two charges will be equal and opposite in magnitude, therefore they will cancel each other out and the electric field strength at this point will be equal to zero.

4 0
3 years ago
Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.
SSSSS [86.1K]

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

\rho(r) = ar^2 - br^3

r is the radius of the spherical shell

dr is the thickness

volume of shell

dV = 4 \pi r^2 dr

mass of shell

dM = \rho(r)dV

\rho = \rho_0 - br

now,

dM = (\rho_0 - br)(4 \pi r^2)dr

integrating both side

M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr

M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})

M = \pi R^3(\dfrac{\rho_0}{3}+\rho)

we know,

a = \dfrac{GM}{R^2}

a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}

a =\pi RG(\dfrac{\rho_0}{3}+\rho)

a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)

a = 9.94 m/s²

7 0
3 years ago
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