Answer:
a = 2.72 ms⁻²
32.83 s
Explanation:
By using the kinematic equations you get,
v² = u² +2as and v = u + at where all terms in usual meaning
Using 1st equation,
89.3² = 0² + 2a×1465 ⇒ a = 2.72 ms⁻²
By 2nd equation,
89.3 = 0 + 2.72×t ⇒ t = 32.83 s
A run though an open field during a thunderstorm is the answer
10.00 °C this is the right answer need more question feel free to post
consider the motion of the tennis ball in downward direction
Y = vertical displacement = 400 m
a = acceleration = acceleration due to gravity = 9.8 m/s²
v₀ = initial velocity of the ball at the top of building = 10 m/s
v = final velocity of the ball when it hits the ground = ?
using the kinematics equation
v² = v²₀ + 2 a Y
inserting the values
v² = 10² + 2 (9.8) (400)
v = 89.11 m/s
Answer:
The maximum data rate supported by this line is 39900 bps
Explanation:
The maximum data rate supported by this line can be obtained using the formula below
c = W*log2(S/N+1)
where;
c is the maximum data rate supported by the line
W is the bandwidth = 4kHz
S/N+1 is the signal to noise ratio = 1001
c = 4*log2(1001)
c = 39868.9 ≅ 39900 bps
Therefore, the maximum data rate supported by this line is 39900 bps
