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Contact [7]
3 years ago
8

When a satellite is above earthâs atmosphere, is it also beyond the pull of earthâs gravity? defend your answer.when a satellite

is above earthâs atmosphere, is it also beyond the pull of earthâs gravity? defend your answer.no, gravity remains constant as altitude increases.no, gravity is an inverse-square law force and extends to infinite distance.no, gravity gets stronger as altitude increases.yes, gravity stops at the top of the atmosphere?
Physics
1 answer:
muminat3 years ago
8 0
No. Gravity is proportional to the inverse square of distance, and has no distance limit. A satellite that's, say, 400 miles above Earth's surface feels about 83% of the strength of gravity on the surface.
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Easy defimation of newtons second law of motion
shtirl [24]
Hi pupil here's your answer ::

_______________________________

Newton's Second Law  of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
 ie.,      F=ma

Where F is the force applied, m is the mass of the body, and a, the acceleration produced.

Or in simplest language it is the force applied to a particular object of particular mass multiplied by the acceleration caused by force .

______________________________

hope that it helps. . . . . .
8 0
3 years ago
A 0.1 kg toy contains a compressed spring. when the spring is released the toy fly 0.45 m upwards from ground level before falli
trapecia [35]

The speed of the toy when it hits the ground is 2.97 m/s.

The given parameters;

  • mass of the toy, m = 0.1 kg
  • the maximum height reached by the, h = 0.45 m

The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.

P.E = K.E

mgh_{max} = \frac{1}{2} mv_{max}^2\\\\gh_{max} = \frac{1}{2} v_{max}^2\\\\v_{max}^2= 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}}

Substitute the given values and solve the speed;

v_{max} = \sqrt{2\times 9.8 \times 0.45} \\\\v_{max} = 2.97 \ m/s

Thus, the speed of the toy when it hits the ground is 2.97 m/s.

Learn more here: brainly.com/question/7562874

7 0
2 years ago
How does an object's position and velocity change as the object accelerates?”
taurus [48]

Answer:

aerodynamics

Explanation:

if an object like a car is going 200 mph at max speed and then the car gets aerodynamic or smoothed to the point that air can get by the car it could end up going another 20 mph faster

7 0
3 years ago
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun
iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0
3 years ago
Monochromatic light passes through a double slit, producing interference, the distance between the slit centres is 1.2 mm and th
Alik [6]

Answer:

The wavelength of the light is 7200\ \AA.

Explanation:

Given that,

Distance between the slit centers d= 1.2 mm

Distance between constructive fringes \beta= 0.3\ cm

Distance between fringe and screen D= 5 m

We need to calculate the wavelength

Using formula of width

\beta=\dfrac{D\lambda}{d}

Put the value into the formula

0.3\times10^{-2}=\dfrac{5\times\lambda}{1.2\times10^{-3}}

\lambda=\dfrac{0.3\times10^{-2}\times1.2\times10^{-3}}{5}

\lambda=7.2\times10^{-7}\ m

\lambda=7200\ \AA

Hence, The wavelength of the light is 7200\ \AA.

8 0
2 years ago
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