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otez555 [7]
2 years ago
12

2) Why do satelites not Fall while revolving around the earth?

Physics
1 answer:
MrRissso [65]2 years ago
7 0

Answer:

Because they are moving to fast for Earth's gravity to pull it down.

You might be interested in
All electromagnetic waves travel in vacuum at the speed of c=3×10^8 m/s. Find the wavelength of microwaves of frequency 10^10 Hz
erica [24]
Recall the wave equation, c=f\lambda where c is the speed of the wave (m/s), f is the frequency of the wave (Hz) and λ is the wavelength of the wave (m).

c=f\lambda \Rightarrow \lambda =  \frac{c}{f} so \lambda =  \frac{3 \times 10^8}{10^{10}} = 0.03 \text{m}
5 0
3 years ago
The acceleration vector of a particle in projectile motion ________.
Alex73 [517]

Answer:

Points downward, and its magnitude is 9.8 m/s^2

Explanation:

The motion of a projectile consists of two independent motions:

- A uniform horizontal motion, with constant velocity and zero acceleration. In fact, there are no forces acting on the projectile along the horizontal direction (if we neglect air resistance), so the acceleration along this direction is zero.

- A vertical motion, with constant acceleration g = 9.8 m/s^2 towards the ground (downward), due to the presence of gravity wich "pulls" the projectile downward.

The total acceleration of the projectile is given by the resultant of the horizontal and vertical components of the acceleration. But we said that the horizontal component is zero, therefore the total acceleration corresponds just to its vertical component, therefore it is a vector with magnitude 9.8 m/s^2 which points downward.

4 0
2 years ago
Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value o
Elza [17]

Answer:

B or 2

Explanation:

5 0
2 years ago
A +2e charge is at the point (-1,0) mm in the x,y plane. A –e charge is at the point (0,1) mm. What is the electric field at the
Gennadij [26K]

Answer:

Let I and j be the unit vector along x and y axis respectively.

Electric field at origin is given by

E= kq1/r1^2 i + kq2/r2^2j

= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)

= (2.88i + 1.44j)*10^-3 N/C

Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3

F= (1.382 i + 0.691 j) *10^-21

 

 Goodluck

Explanation:

4 0
1 year ago
A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect
Stels [109]

Answer:

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta

sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}

r^2=d^2+x^2

So

E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}

Now by integrating above equation

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

4 0
3 years ago
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