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expeople1 [14]
3 years ago
6

Swim 2km at 3km/h. Run at 12km/h for 1 ½ hours. Cycling 55km at 30 km/h. What is the mean overall speed.

Physics
1 answer:
exis [7]3 years ago
5 0

Answer:

t1 = S1 / V1 = 2 km / 3 km/hr = 2/3 hr        where S1 = 2 km

S2 = 12 km/hr * 3/2  hr = 18 km      where t2 = 3/2 hr

t3 = S3 / V3 = 55 km / 30 km/hr = 11/6 hr   where S3 = 55 km

T = t1 + t2 + t3 = 4/6 + 9/6 + 11 / 6 = 24/6 = 4 hrs

S = S1 + S2 + S3 = 2 + 18 + 55 = 75 km

V =  S / T = 75 km / 4 hrs = 18.75 km/hr     for the average speed

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A ___ force is not path-dependent and does not change the overall mechanical energy of an object.​
finlep [7]

Answer:

conservative

Explanation:

Nonconservative force is the force that depends on a path, however conservative does not depend on a path and it is not associated with the potential energy. When the work is done by an unconservative force, mechanical energy is added or removed. Friction is the best example for a non-conservative force. When these non-conservative forces are acting, the mechanical energy changes but these are not preserved.

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6 0
3 years ago
Read 2 more answers
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denis-greek [22]

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8 0
3 years ago
WILL GIVE YOU BRAINLIST IF YOU ANSWER Which of the following characteristics of the Arctic rabbit is specifically an adaptation
andrew-mc [135]

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5 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 isCPI_{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

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4 0
3 years ago
A red train traveling at 72 km/hr and a green train traveling at 144 km/hr are headed toward one another along a straight level
Temka [501]
First, convert all the km/hr into m/s

You will get that
initial speed = 20 m/s
Initial speed of Green train = 40 m/s
Initial separation = 950 m
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relative acceleration = -4 m/s^2

v = u + at
0 = 60 - 4t

t = 15s

s = ut + 1/2  *at * t

s = 60 * 15  - 1/2 *4 * 225
s = 900 - 450

Separation when they stop  = 450 m

hope this helps

5 0
4 years ago
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