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3 years ago

What is the biggest disadvantage of solar electricity, or at least, why is it not used much more than it currently is?

1 answer:

5 0

<span>The Solar electricity source of energy comes from the sun’s rays. It is more often in the form of a panel known as the solar panel. It has a photovoltaic system that generates and supplies energy in commercial and residential areas. Even though numerous countries are using this type of electricity to save electric-induced power, it is not used for a number of reasons. One is that 1 panel with a size of 6 by 10 solar cells is expensive. There are countries who cannot afford to buy them. Two, it requires a larger area to place the panel. Three, the efficiency is low. The smaller the panel is, the lower its efficiency and therefore produces lesser power to generate a specific area. </span>

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Một chất điểm khối lượng m=200g chuyển

MrMuchimi

Can you please translate to English?

8 0

2 years ago

A mass of 2.50 kg is in a gravitational field of 14.0 N/kg. What force acts on the mass?

Olenka [21]

To determine the force that acts on the mass, just multiply the mass by the gravitational field. Using the given data,
F = (2.50 kg)(14 N/kg) = 35 N
Therefore, the force that acts on the mass is equal to 35 N.

7 0

3 years ago

Read 2 more answers

1.) A stone falls from rest from the top of a cliff.

KengaRu [80]

Answer:

Explanation:

ignore air resistance

Let t be the time of fall for the dropped stone.

½(9.8)t² = 43.12(t - 2.2) + ½(9.8)(t - 2.2)²

4.9t² = 43.12t - 94.864 + 4.9(t² - 4.4t + 4.84)

4.9t² = 43.12t - 94.864 + 4.9t² - 21.56t + 23.716

0 = 21.56t - 71.148

t = 71.148/21.56 = 3.3 s

h = ½(9.8)3.3² = 53.361 = 53 m

h = 43.12(3.3 - 2.2) + ½(9.8)(3.3 - 2.2)² = 53.361 = 53 m

4 0

2 years ago

What must be the units for the gravitational constant G in order for gravitational force to have units of newtons?

babunello [35]

Answer:

m³/(kg⋅s²)

Explanation:

Hello.

In this case, since the involved formula is:

$F=G*\frac{m_1m_2}{r^2}$

By writing a dimensional analysis with the proper algebra handling, we obtain:

$N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}$

Thus, answer is:

m³/(kg⋅s²)

Note that the [=] is used to indicate the units of G.

Best regards

4 0

2 years ago

In a typical golf swing the club is in contact with the ball for about 0.0010 s. If the .045 kg ball experiences a force of 4000

Musya8 [376]

Answer:

v = 88.89 [m/s]

Explanation:

To solve this problem we must use the principle of conservation of momentum which tells us that the initial momentum of a body plus the momentum added to that body will be equal to the final momentum of the body.

We must make up the following equation:

$F*t = m*v$

where:

F = force applied = 4000 [N]

t = time = 0.001 [s]

m = mass = 0.045 [kg]

v = velocity [m/s]

$4000*0.001=0.045*v\\v=88.89[m/s]$

5 0

3 years ago