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likoan [24]
3 years ago
5

The force that keeps a satellite in earth’s orbit is always directed away from the center of earth. true or false.

Physics
1 answer:
pshichka [43]3 years ago
8 0
False............ It is in both directions in the form of centri-petal & fugal towards inwards and outwards respectively....
You might be interested in
What type of wave is produced when the frequency of a visible light wave is increased
SIZIF [17.4K]

-- If the frequency of a wave is too high for our eyes to detect it ... but not <u>too too</u> high ... we call it an <em>ultraviolet </em>wave.

-- If the wave's frequency is even higher than that, we call it an <em>X-ray</em> wave.

-- If the wave's frequency is even higher than that, we call it a <em>Gamma Ray </em>wave.

7 0
4 years ago
A projectile is fired straight upward with an initial veloc- ity of 100 m=s from the top of a building 20 m high and falls to th
SVETLANKA909090 [29]

Answer:

The answer to your question is:

a) h max = 529.7 m

b) t = 20.4 s

c) t = 20.6 s

Explanation:

a) h max = -(vo)² / 2g

              = 100² / 2(9.81)

              = 10000 / 19.62

              = 509.7 m

total height = 509.7 + 20 = 529.7 m

b)  

    h = gt² / 2

    t = √ 2h / g

    t = √ 2(509.7)/9.81

    t = √ 103.91

    t = 10.19 s  

    total time = 2 x t = 2 x 10.19 = 20.4 s

c)

   h = vot + 1/2gt²

   20 = 100t + 1/2(9.91) t²

  4.9t² + 100 t -20 = 0   quadratic equation

t = 0.19 s

Total time = 0.19 + 20.4 = 20.6 s

5 0
3 years ago
4. A chandelier brightens a ballroom after a waiter moves a switch
Leokris [45]

Answer:

because switch moves after a waiter

Explanation:

7 0
3 years ago
What is the energy (in joules) and the wavelength (in meters) of the line in the spectrum of hydrogen that represents the moveme
soldi70 [24.7K]

Answer:

The energy is 4.57x10^{-19} J and the wavelength is 4.34x10^{-7}m for the line in the spectrum of hydrogen that represents the movement of an electron from Bohr orbit with n = 2 to the orbit with n = 5.

<em>In what part of the electromagnetic spectrum do we find this radiation? </em>

In the Ultraviolet part of the electromagnetic spectrum.

Explanation:

The energy of the absorbed photon can be known by the difference in energy between the two states in which the transition is happening (In this case from n = 2 to n = 5):

E = E_{upper}-E_{lower}   (1)

The permitted energy for the atom of hydrogen, according with the Bohr's model, is defined as:

E_{n} = -\frac{13.606 eV}{n^{2}}   (2)

Or it can be expressed in Joules, since 1eV = 1.60x10^{-19}J

E_{n} = -\frac{2.18x10^{-18} J}{n^{2}}   (3)

Where the value -2.18x10^{-18} represents the energy of the ground state¹ and n is the principal quantum number.

For the case of n = 2:

E_{2} = -\frac{2.18x10^{-18} J}{(2)^{2}}

E_{2} = -5.45x10^{-19} J

For the case of n = 5:

E_{5} = -\frac{2.18x10^{-18} J}{(5)^{2}}

E_{5} = -8.72x10^{-20} J

Replacing those values in equation (1) it is gotten:

E = -8.72x10^{-20} J-(-5.45x10^{-19} J )

E = 4.57x10^{-19} J

The wavelength can be determined by means of the Rydberg formula:

\frac{1}{\lambda} = R(\frac{1}{n_{l}^{2}}-\frac{1}{n_{u}^{2}})  (4)

Where R is the Rydberg constant, with a value of 1.097x10^{7}m^{-1}

For this particular case n_{l} = 2 and n_{u} = 5:

\frac{1}{\lambda} = 1.097x10^{7}m^{-1}(\frac{1}{(2)^{2}}-\frac{1}{(5)^{2}})

\frac{1}{\lambda} = 1.097x10^{7}m^{-1}(0.21)

\frac{1}{\lambda} = 2303700m^{-1}

\lambda = \frac{1}{2303700m^{-1}}

\lambda = 4.34x10^{-7}m

So the energy is 4.57x10^{-19} J and the wavelength is 4.34x10^{-7}m for the line in the spectrum of hydrogen that represents the movement of an electron from Bohr orbit with n = 2 to the orbit with n = 5.

<em>In what part of the electromagnetic spectrum do we find this radiation? </em>

In the Ultraviolet part of the electromagnetic spectrum.

Key terms:

¹Ground state: State of minimum energy.  

8 0
3 years ago
A 120 N force, directed at an angle θ above a horizontal floor, is applied to a 28.0 kg chair sitting on the floor. If θ = 0°, w
ivanzaharov [21]

Answer:

See explanation

Explanation:

Given:-

- The applied force Fa = 120 N

- The angle made with Fa and horizontal =  θ

- The mass of chair m = 28.0 kg

Find:-

a) θ = 0 ,  the horizontal component Fh of the applied force and (b) the magnitude FN of the normal force of the floor on the chair?

If θ = 34.0°, what are (c)Fh and (d)FN?

If θ = 65.0°, what are (e)Fh and (f)FN?

Now assume that the coefficient of static friction between chair and floor is 0.410. What is the maximum force of static friction on the chair if θ is (g) 0°, (h)34.0°, and (i)65.0°?

Solution:-

- We will consider the applied force (Fa) and develop a general expression for Fh as function of θ. Using Trigonometry we have:

                                     Fh = Fa*cos ( θ )

- The normal contact force (N) can be expressed similarly by applying equilibrium conditions on the chair in vertical direction.

                                    N - mg - Fa*sin(θ) = 0

                                    N = m*g + Fa*sin(θ)  

For, θ = 0:

                                    Fh = Fa*cos(0) = Fa

                                    Fh = 120 N

                                    N = (28*9.81) + 120*0

                                    N = 274.68 N

For, θ = 34.0:

                                    Fh = Fa*cos(34) = (120)*(0.82903)

                                    Fh = 99.5 N

                                    N = (28*9.81) + 120*sin(34)

                                    N = 341.8 N

For, θ = 65.0:

                                    Fh = Fa*cos(65) = (120)*(0.42261)

                                    Fh = 50.7 N

                                    N = (28*9.81) + 120*sin(65)

                                    N = 383.4 N

- The maximum frictional force (Ff) is given by the following expression:

                                   Ff = N*us

Where,                        us = coefficient of static friction = 0.41

For, θ = 0:

                                   Ff = (274.68)*(0.41)

                                   Ff = 112.62 N

For, θ = 34:

                                   Ff = (341.8)*(0.41)

                                   Ff = 140.13 N

For, θ = 65:

                                   Ff = (383.4)*(0.41)

                                   Ff = 157.2 N

5 0
3 years ago
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