When t=2, the ball has fallen d(2) = 16 (2²) = 64 feet .
When t=5, the ball has fallen d(5) = 16 (5²) = 400 feet .
Distance fallen from t=2 until t=5 is (400 - 64) = 336 feet.
Time period between t=2 until t=5 is (5 - 2) = 3 seconds.
Average speed of the ball from t=2 until t=5 is
(distance covered) / (time to cover the distance)
= 336 feet / 3 seconds = 112 feet per second.
That's what choice-C says.
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I think it's 16.8 because you multiply 5.6x3=16.8
Answer:
5.77×10¹⁰ m
Explanation:
From the question,
Applying
Kepler's third law
P² = d³...................... Equation 1
Where P = Planet's period, d = distance between the center of the planet and the sun.
make d the subject of the formula n equation 1
d =
.................. Equation 2
Given: P = 0.24 sidereal.
Substitute the value of P into equation 2
d =
d = 0.386 Au
d = 0.386×1.496×10¹¹
d = 5.77×10¹⁰ m
Answer:
13 m/s east
Explanation:
We can solve the problem by using the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision:

where
m = 0.1 kg is the mass of each puck
u1 = +13 m/s is the initial velocity of puck 1
u2 = -18 m/s is the initial velocity of puck 2 (here I assume the west direction to be the negative direction, so I put a negative sign)
v1 = -18 m/s is the final velocity of puck 1
v2 = ? is the final velocity of puck 2
Simplifying m from the formula and substituting the data, we can find the final velocity of puck 2, v2:

And the positive sign means that puck 2 is moving east.