Under what conditions will the projectile have the greatest velocity when it hits the ground?

1 answer:

8 0

I'm going without any details here.
The projectile would have the greatest velocity if thrown from a high distance, it would also help depending on what the gravity was which on earth would help increase the velocity since it's strong and pull more on it increasing weight but not mass.

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What happens to a current when the electric field in the wire increase

Gnom [1K]

Answer:

If the voltage is increased then the electric field is higher, and electron velocity (average) is proportional to this field. Then you have an increase in speed. And current is total charge passing per time unit, so current is proportional to velocity value of charge (and to voltage in resistors and wire).

Explanation:

5 0

4 years ago

A +26.3 uC charge qy is repelled by a force

Musya8 [376]

Answer:

+1.46×10¯⁶ C

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C

Force (F) = 0.615 N

Distance apart (r) = 0.750 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Charge 2 (q₂) =?

The value of the second charge can be obtained as follow:

F = Kq₁q₂ / r²

0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²

0.615 = 236700 × q₂ / 0.5625

Cross multiply

236700 × q₂ = 0.615 × 0.5625

Divide both side by 236700

q₂ = (0.615 × 0.5625) / 236700

q₂ = +1.46×10¯⁶ C

NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C

5 0

3 years ago

A particle executes linear harmonic motion about the point x = 0. At t = 0, it has displacement x = 0.37 cm and zero velocity. T

lukranit [14]

Answer:

(a) The period is 4s

(b) The angular frequency is pi/2 radians

(d) The displacement at time is (0.37 cm) cos((pi/2)*t)

(e) The Velocity at time t is v = (0.58 cm)(sin((pi/2)*t)

(f) The maximum speed is $v_{m} = -0.58 cm/s$

(g) The maximum acceleration is 0.91 cm/s^2

Explanation:

We have a particle which oscillates with frequency of f = 0.25 Hz about the point x = 0.At t = 0, the displacement of the particle is = 0.37 cm and its velocity is zero.

(a) The period of the oscillations is,

$T = 1/f$