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pav-90 [236]
3 years ago
9

Which of Newton's laws explains why satellites need very little fuel to stay in oribit?

Physics
1 answer:
Angelina_Jolie [31]3 years ago
4 0

Sattelites don't need any fuel to stay in orbit. The applicable law is...."objects in motion tend to stay in motion". Having reached orbital velocity, any such object is essentially "falling" around the earth. Since there is no (or at least very little) friction in the vacuum of space, the object does not slow.... It simply continues.


Sattelites in "low" earth orbit do encounter some friction from the very thin upper atmosphere, and they will eventually "decay".

:)

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Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i
koban [17]

Answer:

A

   N  = 1340.86 \ slits  / cm

B

    \theta  = 15.7^o

Explanation:

From the question we are told that  

      The wavelength is  \lambda  =  650 \  nm  =  650  *10^{-9} \  m  

        The angle of  first bright fringe is  \theta  =  5^o  

        The order of the fringe considered is  n  =1

Generally the condition for constructive interference is  

       dsin (\theta ) = n * \lambda

=>    d =  \frac{1 *  650 *10^{-9 }}{ sin(5)}

=>    d = 7.458 *10^{-6} \  m

Converting to cm

           d = 7.458 *10^{-6} \  m = 7.458 *10^{-6}  * 100 =  0.0007458 \  cm

Generally the number of grating pre centimeter is  mathematically represented as

           N  =  \frac{1}{d}

=>         N  =  \frac{1}{0.0007458}

=>         N  = 1340.86 \ slits  / cm

Considering question B  

   From the question we are told that

     The first wavelength is  \lambda_1 =  650 \ nm  =  650 *10^{-9} \  m

     The second wavelength is  \lambda_2 = 429 \  m   =   420 *10^{-9 } \  m

      The order of the fringe is  n  =  2

       The grating is  N =  5000 \  slits / cm

Generally the slit width is mathematically represented as

              d =  \frac{1}{N  }

=>          d =  \frac{1}{ 5000  }

=>          d =   0.0002 \  c m  =  2.0 *10^{-6} \ m

Generally the condition for constructive interference for the first ray is mathematically represented as

         d sin(\theta_1) =  n *  \lambda_1

=>      \theta_1 = sin^{-1} [\frac{ 2 *  \lambda }{d}]

=>       \theta_1 = sin^{-1} [\frac{ 2 *   650 *10^{-9} }{ 2*10^{-6}}]

=>        \theta_1 = 40.5 ^o

Generally the condition for constructive interference for the second ray is mathematically represented as

         d sin(\theta_2) =  n *  \lambda_2

=>      \theta_2 = sin^{-1} [\frac{ 2 *  \lambda_1 }{d}]

=>       \theta_2 = sin^{-1} [\frac{ 2 *   420 *10^{-9} }{ 2*10^{-6}}]

=>        \theta_2 = 24.8  ^o

Generally the angular separation is mathematically represented as

            \theta  =  \theta_1 - \theta_1

=>          \theta  = 42.5^o -  24.8^o

=>          \theta  = 15.7^o

4 0
3 years ago
A 1210 kg rollercoaster car is
ratelena [41]

Answer: 4.98 m/s

Explanation:

You solve these kinetic energy, potential energy problems by using the fact P.E.+ K.E. = a constant as long as friction is ignored.

PEi = 0 in this case

KEi = ½mVi² = PEf+KEf = mghf + ½mVf²

½1210*8.31² = 1210*9.8*2.26 + ½1210*Vf²

½1210*Vf² = ½1210*8.31² - 1210*9.8*2.26

Vf² = 8.31² - 2*9.8*2.26 = 4.98² so Vf = 4.98m/s

3 0
3 years ago
 Why does a child in a wagon seem to fall backward when you give the wagon a sharp pull forward? 15. What force is needed to acc
slavikrds [6]

Answer:

Force, F = 77 N

Explanation:

A child in a wagon seem to fall backward when you give the wagon a sharp pull forward. It is due to Newton's third law of motion. The forward pull on wagon is called action force and the backward force is called reaction force. These two forces are equal in magnitude but they acts in opposite direction.

We need to calculate the force is needed to accelerate a sled. It can be calculated using the formula as :

F = m × a

Where

m = mass = 55 kg

a = acceleration = 1.4 m/s²

F=55\ kg\times 1.4\ m/s^2

F = 77 N

So, the force needed to accelerate a sled is 77 N. Hence, this is the required solution.

3 0
3 years ago
How much energy is needed to generate 0.71 x 10-16 kg of mass?
AleksAgata [21]

Answer:

6.39 J of energy is needed to generate 0.71 * 10⁻¹⁶ kg mass

Explanation:

According to the Equation: E = mc²

where the mass, m = 0.71 * 10⁻¹⁶ kg

the speed of light, c = 3 * 10⁸ m/s

The amount of energy needed to generate a mass of 0.71 * 10⁻¹⁶ kg is calculated as follows:

E = (0.71 * 10⁻¹⁶) (3 * 10⁸)²

E = 0.71 * 10⁻¹⁶ * 9 * 10¹⁶

E = 0.71 * 9

E = 6.39 J

6 0
3 years ago
A balloon rubbed against denim gains a charge of −7.0 µc. what is the electric force between the balloon and the denim when the
zubka84 [21]
The solution for this problem is computed by through this formula, F = kQq / d²Plugging in the given values above, we can now compute for the answer.
F = 8.98755e9N·m²/C² * -(7e-6C)² / (0.03m)² = -489N, the negative sign denotes attraction.
4 0
3 years ago
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