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ira [324]
3 years ago
14

A 2 nC point charge is at the origin, and a second 5 nC point charge is on the x-axis at x = 8 m. Find the electric field (magni

tude and direction) at the point x = 2 m.
Physics
1 answer:
dimaraw [331]3 years ago
7 0

Answer:

The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

Explanation:

Given that,

Point charge at origin = 2 nC

Second charge = 5 nC

Distance at x axis = 8 m

We need to calculate the electric field at the point x = 2 m

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{q_{1}}{r_{1}^2}+\dfrac{q_{2}}{r_{2}^2})

Put the value into the formula

E=9\times10^{9}\times(\dfrac{2\times10^{-9}}{2^2}+\dfrac{5\times10^{-9}}{(8-2)^2})

E=5.75\ N/C

The direction is toward positive x- axis.

Hence, The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

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The mass of an object on earth is 20 kg what will be the mass of that object on moon? And why? ​
Alla [95]

Answer:

20kg

Explanation:

Mass is a measure of the amount of matter in an object. The mass of an object, the amount of matter inside it does not change based on location. E.g. Objects do not lose matter when they travel to the moon.

Weight, on the other hand is the downward force you exert on the ground. Weight is calculated by multiplying the mass by the gravitational field strength and changes in different places with different gravitational strength. E.g. The moon's gravitational strength is 1/5 of Earth's so the mass of the object would stay the same but the weight would be only 20% of the weight is had on earth.

Hope this helped!

7 0
3 years ago
Describe a well-known hypothesis that was discarded because it was found to be untrue.earth-centered model of the universe. the
polet [3.4K]
The widely accepted hypothesis before that turned out wrong was the Earth-Centered theory or the Geocentric Theory. This was proposed by the philosopher Ptolemy. He came about to this hypothesis from hi observation that from the Earth's perspective, the celestial bodies like the Sun, stars and the moon, look like they rotate around the Earth each day and night. However, this was disproved by Galileo Galelei by his Heliocentric Theory. He observed through the telescope that the Venus also changes phases like the moon. However, he deduced that this is not possible from the positions of the Venus, Earth, Moon and Sun. 
6 0
3 years ago
A very large sheet of insulating material has had an excess of electrons placed on it to a surface charge density of –3.00nC/m2
lys-0071 [83]

Answer: sheet of charge

Explanation:

a )

Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.

b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge.   The equi- potential surface is always perpendicular to electric field.

C ) Electric field which is almost uniform near the sheet of charge is equal t the following

E = σ / ε₀ where  σ is charge density of surface and  ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²

E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²

= .3389 x 10³

= 338.9 V / m

spacing between 1 V

= 1 / 338.9 m

= 2.95 X 10⁻3 m

= 2.95 mm.

3 0
3 years ago
A rock thrown with speed 8.50 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 19.0 m bef
frez [133]
Draw a diagram to illustrate the problem as shown below.

The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s

Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s

The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
   = 21.675 m

Answer: The height is 21.7 m (nearest tenth)

4 0
3 years ago
The space shuttle usually orbited Earth at altitudes of around 300.0 km. 1) Determine the time for one orbit of the shuttle abou
777dan777 [17]

Answer:

Explanation:

distance of shuttle from centre of the earth = radius of the orbit

= 6300 + 300 = 6600 km

= 6600 x 10³

Formula of time period of the satellite

T = 2π R /v₀ , v₀ is orbital velocity

v₀ = √gR , ( if height  is small with respect to radius )

T = 2π R /√gR

= 2π√ R /√g

= 2 x 3.14 x √ 6600 x 10³ / √9.8

= 2 x 3.14 x 256.9 x 10 / 3.13

= 5154.41 s

= 5154.41 / 60 minutes

= 85.91 m

85.9 minutes.

2 ) No of sunrise per day = no of rotation per day

= 24 x 60 / 85.9

= 16.76

or 17 sunrises.

3 0
4 years ago
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