Answer:
X(t) = 9.8 *t - 4.9 * t^2
Explanation:
We set a frame of reference with origin at the hand of the girl the moment she releases the ball. We assume her hand will be in the same position when she catches it again. The positive X axis point upwards.The ball will be subject to a constant gravitational acceleration of -9.81 m/s^2.
We use the equation for position under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a *t^2
X0 = 0 because it is at the origin of the coordinate system.
We know that at t = 2, the position will be zero.
X(2) = 0 = V0 * 2 + 1/2 * -9.81 * 2^2
0 = 2 * V0 - 4.9 * 4
2 * V0 = 19.6
V0 = 9.8 m/s
Then the position of the ball as a function of time is:
X(t) = 9.8 *t - 4.9 * t^2
Answer:
Hi, There!
O 10 m/s
O 10 yd/s
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Have a great day!
Answer:
Moving them farther apart
Explanation:
The electric field between the two charges Q and q separated by a distance r is given by
It shows that the electric field is inversely proportional to the square of the distance between two charges.
So, as the distance between two charges increases, the electric filed between the two charges decreases.
Answer:
Following are the solution to this question:
Explanation:
That light takes a very long time to hit the planet, and the object is far off the earth. The light of such an item near to the planet takes less time to enter it. The star is 2,5 million light-years from the Planet on the far side of the Andromeda Galaxy. But on the other hand, the moon is 15 crore miles from the earth, so sunlight is quickly reached on the ground as the other thing.
That milky way away from the earth is 66,500 light-years far, that distance between Earth and Orion nebula is 1,344 light-years, with such a distance of 4,367 light-years. The earth is 5.2261 trillion km apart from Pluto.
Answer:
Explanation:
Let m be mass of each sphere and θ be angle, string makes with vertex in equilibrium.
Let T be tension in the hanging string
T cosθ = mg ( for balancing in vertical direction )
for balancing in horizontal direction
Tsinθ = F ( F is force of repulsion between two charges sphere)
Dividing the two equations
Tanθ = F / mg
tan17 = F / (7.1 x 10⁻³ x 9.8)
F = 21.27 x 10⁻³ N
if q be charge on each sphere , force of repulsion between the two
F = k q x q / r² ( r is distance between two sphere , r = 2 x .7 x sin17 = .41 m )
21.27 x 10⁻³ = (9 X 10⁹ x q²) / .41²
q² = .3973 x 10⁻¹²
q = .63 x 10⁻⁶ C
no of electrons required = q / charge on a single electron
= .63 x 10⁻⁶ / 1.6 x 10⁻¹⁹
= .39375 x 10¹³
3.9375 x 10¹² .
