The acceleration of the ball is 5 m/s^2. This can be calculated using a formula that relates the change in velocity, acceleration, and time. This formula is:

Vf = Vi + at

where:
Vf = final velocity
Vi = initial velocity
a = acceleration
t = time

Substituting the values gives:

30 = 20 + a(2)
<span>a = 5 m/s^2 --> Final Answer</span>

6 0

3 years ago

Computers A and B implement the same ISA. Computer A has a clock cycle time of 200 ps and an effective CPI of 1.5 for some progr

lesya692 [45]

Answer:

Computer A is 1.41 times faster than the Computer B

Explanation:

Assume that number of instruction in the program is 1

Clock time of computer A is $CT_{A} =200 ps$

Clock time of computer B is $CT_{B} =250 ps$

Effective CPI of computer A is $CPI_{A} =1.5$

Effective CPI of computer B is $CPI_{B} =1.7$

CPU time of A is

$CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec$

CPU time of B is

$CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec$

Hence Computer A is Faster by $\frac{425}{300} =1.41$

Computer A is 1.41 times faster than the Computer B

4 0

3 years ago

Need help! Photo says all! Will mark brainliest :)

AVprozaik [17]

Answer:

C-D

Explanation:

As you can see from the graph, the distance from A to B was from 0 m to 6 m in a duration of 3 seconds.

Divide 6 meters by 3 seconds to find the speed:

6 ÷ 3 = 2 m/s

B-C is not moving due to a straight line as said in the graph, so speed is

0 m/s.

There is also C-D since the car traveled from a distance of 9 meters

(6 -(-3) = 9) in 3 seconds too. (NOTE: The graph line going down does not mean it is slowing down, but rather going to a certain distance like going backwards)

Divide 9 meters by 3 seconds to get the speed:

9 ÷ 3 = 3 m/s

Between A-B, B-C, and C-D, C-D has the fastest speed recorded with 3 m/s.

A-D does not count here as the line has no connection between point A and point D.

Cheers!

6 0

3 years ago