If I’m gonna keep it abuck with you I’m lost
Answer:
1. They both uses same energy
2. The 6 kg ball requires more power than 3kg ball
Explanation:
Sample 1
m = 3kg
g= 10m/s^2
h = 2m
t = 2secs
W = mgh = 3 x 10 x 2 = 60J
P= w/t = 60/2 = 30watts
Sample 2
m = 6kg
g= 10m/s^2
h = 1m
t = 1sec
W = mgh = 6 x 10 x 1 = 60J
P= w/t = 60/1 = 60watts
They both uses same energy but different power. The 6 kg ball requires more power than 3kg ball
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
First find Acceleration
- Initial velocity=u=0m/s
- Final velocity=v=42m/s
- Time=t=2s
- Distance=s
- Acceleration=a
Using second equation of kinematics