Answer:
V = a * t = 9.8 m/s^2 * 2.3 s = 22.5 m/s velocity after 2.3 s
S = 1/2 g t^2 since initial speed is zero
S = 1/2 * 9.8 m/s^2 * 5.29 s^2 = 25.9 m
Answer:
Option C
Explanation:
Given:
- Clock speed f = 33 MHz
- The width of the bus w = 32 bits
Find:
what is the theoretical throughput in MiBytes?
Solution:
- First step is to convert the width of the bus to bytes as follows:
Bytes = 32 bits * (1 Bytes / 8 bits)
Bytes = 32 / 8
- Second step is to evaluate the time of the cycle:
Time period of clock T = 1 / f
Time period of clock T = 1 s / (33*10^6)
T = (1 / 33*10^6)
- Third step is to formulate the number of byte:
Number of byte = Bytes * T
= (32 / 8*33*10^6)
- Fourth step is to convert to Mi bytes:
Mibytes = Number of byte / 2^20
Mibytes = (32 / 8*33*10^6 * 2^20)
- The correct option is C
Answer:
U=α/(x+x_0)^2
v=1.88m/s
Explanation:
We have that
(a)
The potential is calculated by using
(b)
m=0.5kg
The acceleration can be obtained if we calculate the force for x=4, and after we compute the acceleration
and finally, we can use the equation for the final speed
I hope this is useful for you
regards