Answer:
33.2 m
Explanation:
For the first object:
y₀ = 81.5 m
v₀ = 0 m/s
a = -9.8 m/s²
t₀ = 0 s
y = y₀ + v₀ t + ½ at²
y = 81.5 − 4.9t²
For the second object:
y₀ = 0 m
v₀ = 40.0 m/s
a = -9.8 m/s²
t₀ = 2.20 s
y = y₀ + v₀ t + ½ at²
y = 40(t−2.2) − 4.9(t−2.2)²
When they meet:
81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²
81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)
81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716
81.5 = 61.56t − 111.716
193.216 = 61.56t
t = 3.139
The position at that time is:
y = 81.5 − 4.9(3.139)²
y = 33.2
Speed=wavelength×frequency
speed= 0.005m × 10 Hz = 0.05m/s
Answer:
c. 65
Explanation:
The output is 65.
An array of length 10 is created first. Then, the first for-loop fill the array with different values; The array element now become: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. The array element are generated using the equation a[i] = i + 2; so when i is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. i must be less than the array length (10).
a[0] = 0 + 2 = 2
a[1] = 1 + 2 = 3
a[2] = 2 + 2 = 4
a[3] = 3 + 2 = 5
a[4] = 4 + 2 = 6
a[5] = 5 + 2 = 7
a[6] = 6 + 2 = 8
a[7] = 7 + 2 = 9
a[8] = 8 + 2 = 10
a[9] = 9 + 2 = 11
result variable is declared and initialized to 0.
The second for-loop goes through the array and add individual element to result.
Answer:
3.99*10^-3N/C
Explanation:
Using
Ep= kq/r²
Where r = 0.6mm = 0.6*10^-3m
K= 8.9*10^9 and q= 1.6*10^-19
So = 8.9*10^9 * 1.6*10^-19/0.6*10^-3)²
= 3.99*10^-3N/C
