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3 years ago

7

What are advantages and disadvantages of apparent solar time? How is the situation improved by introducing mean solar time and s

tandard time?

1 answer:

Nutka1998 [239]3 years ago

3 0

Answer:

Advantage:

Apparent solar time gives the exact location of sun in the sky according to your precise location.

Disadvantage:

As the apparent solar time changes with the change in longitude. It is very difficult to track these changes in longitude. Hence, it is almost impossible to make plan for events.

Improvement in situation:

Situation can be improved using mean solar time because due to this people living in the longitude band agree upon a standard time. In this way, it is easy to plan for events.

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If each of the three rotor helicopter blades is 3.50 m long and has a mass of 120 kg , calculate the moment of inertia of the th

devlian [24]

Answer:

1470kgm²

Explanation:

The formula for expressing the moment of inertial is expressed as;

I = 1/3mr²

m is the mass of the body

r is the radius

Since there are three rotor blades, the moment of inertia will be;

I = 3(1/3mr²)

I = mr²

Given

m = 120kg

r = 3.50m

Required

Moment of inertia

Substitute the given values and get I

I = 120(3.50)²

I = 120(12.25)

I = 1470kgm²

Hence the moment of inertial of the three rotor blades about the axis of rotation is 1470kgm²

7 0

3 years ago

The FM radio band in most places goes from frequencies of about 89 MHz to 106 MHz. How long is the wavelength of the radiation a

ohaa [14]

Answer:

2.83 m

Explanation:

The relationship between frequency and wavelength for an electromagnetic wave is given by

$\lambda=\frac{c}{f}$

where

$\lambda$ is the wavelength

$c=3.0\cdot 10^8 m/s$ is the speed of light

$f$ is the frequency

For the FM radio waves in this problem, we have:

$f_1=89 MHz=89\cdot 10^6 Hz$ is the minimum frequency, so the maximum wavelength is

$\lambda_2=\frac{c}{f_1}=\frac{3\cdot 10^8}{89\cdot 10^6}=3.37 m$

The maximum frequency is instead

$f_2=106 MHz=106\cdot 10^6 Hz$

Therefore, the minimum wavelength is

$\lambda_1=\frac{c}{f_2}=\frac{3\cdot 10^8}{106\cdot 10^6}=2.83 m$

So, the wavelength at the beginning of the range is 2.83 m.

8 0

3 years ago

A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 781 K, and r

andreev551 [17]

Answer:2.517 J/K

Explanation:

Given

Reservoir 1 Temperature $T_1=781 K$

Reservoir 2 Temperature $T_2=335 K$

Let Q is the amount of heat Flows i.e. $Q=1477 J$

thus change in Entropy is given by $\frac{\sum Q}{T}$

$\Delta S=\frac{\sum Q}{T}=-\frac{Q}{T_1}+\frac{Q}{T_2}$

$\Delta S=\frac{\sum Q}{T}=-\frac{1477}{781}+\frac{1477}{335}$

$\Delta S=\frac{\sum Q}{T}=-1.891+4.4089$

$\Delta S=\frac{\sum Q}{T}=2.517 J/K$

6 0

3 years ago

Please answer all of these questions for brainly!

Nana76 [90]

Answer:

1. C
2. B
3. D
4. A

3 0

3 years ago

Objects with masses of 120 kg and a 420 kg are separated by 0.380 m. (a) Find the net gravitational force exerted by these objec

SOVA2 [1]

Answer:

$F_{net} = 6.879\times 10^{- 7}\ N$

Solution:

As per the question:

Mass of first object, m = 120 kg

Mass of second object, m' = 420 kg

Mass of the third object, M = 69.0 kg

Distance between the m and m', d = 0.380 m

Now,

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F = \frac{6.67\times 10^{-11}\times 120\times 0.69}{\frac({0.380}{2}^{2})} = 1.529\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F' = \frac{6.67\times 10^{-11}\times 420\times 0.69}{\frac({0.380}{2}^{2})} = 5.35\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':

$F_{net} = F + F'$

$F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N$

6 0

3 years ago