K.E =1/2mv2
M=6kg
V=3m/s
K.E=1/2 X 6 X 3 X 3
=1/2 X 6 X 9
=27 J
Answer:
Explanation:
Given that,
Current measure is
i=10±0.6 Amps
And also,
R=45.0±2.0 Ω
Power dissipated by
P=i²R
Then
P=(10±0.6)²(45.0±2.0)
P=10²×45
P=450Watts
Now, calculating the uncertainty
∆P=|P| • √(2(∆i/i)²+(∆R/R)²)
∆P=450√ (2×(0.6/10)²+(2/45)²)
∆P=450√(0.0072+0.001975)
∆P=450√0.009175
∆P=43.1
The uncertainty in power is 43.1
Then,
P=450 ± 43.1 Watts
Answer:
a) Acceleration of the car is given as
![a_{car} = -21 m/s^2](https://tex.z-dn.net/?f=a_%7Bcar%7D%20%3D%20-21%20m%2Fs%5E2)
b) Acceleration of the truck is given as
![a_{truck} = 10.15 m/s^2](https://tex.z-dn.net/?f=a_%7Btruck%7D%20%3D%2010.15%20m%2Fs%5E2)
Explanation:
As we know that there is no external force in the direction of motion of truck and car
So here we can say that the momentum of the system before and after collision must be conserved
So here we will have
![m_1v_1 + m_2v_2 = (m_1 + m_2)v](https://tex.z-dn.net/?f=m_1v_1%20%2B%20m_2v_2%20%3D%20%28m_1%20%2B%20m_2%29v)
now we have
![1400 (6.32) + 2900(0) = (1400 + 2900) v](https://tex.z-dn.net/?f=1400%20%286.32%29%20%2B%202900%280%29%20%3D%20%281400%20%2B%202900%29%20v)
![v = 2.06 m/s](https://tex.z-dn.net/?f=v%20%3D%202.06%20m%2Fs)
a) For acceleration of car we know that it is rate of change in velocity of car
so we have
![a_{car} = \frac{v_f - v_i}{t}](https://tex.z-dn.net/?f=a_%7Bcar%7D%20%3D%20%5Cfrac%7Bv_f%20-%20v_i%7D%7Bt%7D)
![a_{car} = \frac{2.06 - 6.32}{0.203}](https://tex.z-dn.net/?f=a_%7Bcar%7D%20%3D%20%5Cfrac%7B2.06%20-%206.32%7D%7B0.203%7D)
![a_{car} = -21 m/s^2](https://tex.z-dn.net/?f=a_%7Bcar%7D%20%3D%20-21%20m%2Fs%5E2)
b) For acceleration of truck we will find the rate of change in velocity of the truck
so we have
![a_{truck} = \frac{v_f - v_i}{t}](https://tex.z-dn.net/?f=a_%7Btruck%7D%20%3D%20%5Cfrac%7Bv_f%20-%20v_i%7D%7Bt%7D)
![a_{truck} = \frac{2.06 - 0}{0.203}](https://tex.z-dn.net/?f=a_%7Btruck%7D%20%3D%20%5Cfrac%7B2.06%20-%200%7D%7B0.203%7D)
![a_{truck} = 10.15 m/s^2](https://tex.z-dn.net/?f=a_%7Btruck%7D%20%3D%2010.15%20m%2Fs%5E2)
Answer:
μ = 0.604
Explanation:
For the cat to stay in place on the merry go round, the maximum static frictional force must be equal in magnitude to that of the centripetal force.
Now, Centripetal force is given as;
Fc = mv²/r
Where r is radius and v is tangential speed and m is mass.
We also know that maximum static frictional force is given by;
F_static = μmg
Where μ is coefficient of friction
Now, equating both forces, we have;
mv²/r = μmg
Divide through by m;
v²/r = μg
Now, tangential speed can be expressed as;
v = circumference/period
Thus, v = 2πr/T
Where T is period of rotation and
2πr is the circumference of the merry go round.
Thus,
v²/r = μg is now;
(2πr/T)²/r = μg
Making μ the subject, we have;
(2πr/T)²/rg = μ
μ = [(2π x 5.4)/6]²/(5.4 x 9.8)
μ = 0.604