Answer:
Option 3. The tennis ball began from rest and rolls at a rate of 14.7 m/s safer 1.5 seconds.
Explanation:
To know the the correct answer to the question, it is important that we know the definition of acceleration.
Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as:
a = (v – u) /t
Where
a => acceleration
v => final velocity
u => Initial velocity
t => time
With the above information in mind, let us consider the options given in the question above to know which conform to the difinition of acceleration.
For Option 1,
We were told that the tennis ball has the following:
Distance = 4 m
Time = 1.5 s
This talks about the speed and not the acceleration.
Speed = distance / time
For Option 2,
We were only told about the average speed and nothing else.
For Option 3,
We were told that the tennis ball have the following:
Initial velocity (u) = 0 m/s
Final velocity (v) = 14.7 m/s
Time = 1.5 s
This talks about the acceleration.
a = (v – u) /t
For Option 4,
We were only told that the tennis rolls to the right at an average speed. This talks about the average velocity. We need more information like time to justify the acceleration.
From the above illustrations, option 3 gives the correct answer to the question.
Answer: It would be 12 m/s.
Explanation: It would be this because If you go from rest to sprint it would be 12 m/s. Also, I did this the other day.
Refer to the diagram shown below.
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)
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Answer:
Therefore,
1/3 ≤ Re ≤ 10/3 ohms
The equivalent resistance CANNOT be any value outside the boundary above
That is
Re CANNOT be greater than 10/3 ohms and CANNOT be less than 1/3 ohms, given that R1,R2,R3 are all between 1 and 10ohms.
Explanation:
For a parallel resistor arrangement.
The equivalent resistance (Re) of the three resistors is given by;
1/Re = 1/R1 + 1/R2 + 1/R3
1/Re = (R2R3 + R1R3 + R1R2)/R1R2R3
Re = R1R2R3/(R2R3 + R1R3 + R1R2)
Therefore, if R1,R2,R3 are between 1-10ohms
We need to calculate the range of values of Re.
Taking the lower bound 1
R1= R2=R3= 1ohms
Re = 1/(1+1+1)
Re = 1/3 ohms
Taking the upper bound 10 ohms
R1= R2=R3= 10 ohms
Re = 1000/(100+100+100)
Re = 1000/300
Re = 10/3 ohms
Therefore,
1/3 ≤ Re ≤ 10/3 ohms
The equivalent resistance CANNOT be any value outside the boundary above
That is
Re cannot be greater than 10/3 ohms and cannot be less than 1/3 ohms, given that R1,R2,R3 are all between 1 and 10ohms.
