For a projectile, why is the initial horizontal velocity equal to the final horizontal velocity?

1 answer:

4 0

I am assuming that the context is that of a thrown projectile (a ball, a bullet, etc.) in the gravitational field of the Earth.

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charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric fiel

IRISSAK [1]

Answer:

Explanation:

Electric field due to charge at origin

= k Q / r²

k is a constant , Q is charge and r is distance

= 9 x 10⁹ x 5 x 10⁻⁶ / .5²

= 180 x 10³ N /C

In vector form

E₁ = 180 x 10³ j

Electric field due to q₂ charge

= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²

= 30.33 x 10³ N / C

It will have negative slope θ with x axis

Tan θ = .5 / √.5² + .8²

= .5 / .94

θ = 28°

E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j

= 26.78 x 10³ i - 14.24 x 10³ j

Total electric field

E = E₁ + E₂

= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j

= 26.78 x 10³ i + 165.76 X 10³ j

magnitude

= √(26.78² + 165.76² ) x 10³ N /C

= 167.8 x 10³ N / C .

3 0

3 years ago

Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

Gennadij [26K]

Answer:

a. The station is rotating at $1.496 \frac{rev}{min}$

b. the rotation needed is $2.8502 \frac{rev}{min}$

Explanation:

We know that the centripetal acceleration is

$a_{c}= \omega ^2 r$

where $\omega$ is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).