Answer:
Yes, the errors are likely to be relevant
Explanation:
A systematic error occurs as a result of the instrument used in carrying out and experiment. These errors are a result of small fluctuations in the measurement properties of the instrument. This happens when the instrument departs from non-ideal situations, for example as a result of physical expansion or change in temperature. For instance, let the resistance be measured to be up to 10 Ω ± 1 Ω
The error of the resistance, ε = 0.01Ω
Answer:
2.87m
Explanation:
Using the law of gravitation to solve this question
F = GMm/r²
G is the gravitational constant
M and m are the masses
r is the distance between the masses
Substitute the given values
G = 6.67×10^-11 m³/kgs²
M =8.8 x 10^6 kg
m = 5.6 x 10^5 kg
F =440N
400 = 6.67×10^-11×8.8 x 10^6 ×5.6 x 10^5/r²
400r² = 328.698×10
400r² = 3286.98
r² = 3286.98/400
r² = 8.21745
r = √8.21745
r = 2.87m
Hence the distance of separation is 2.87m
Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2
Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.
We have for the car
distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s
the v car = distance/time= 81.1 m/11.6s= 7 m/s
In order to calculate the acceleration we have to use the kinematic equation for the train from the rest
distance train = (a* t^2)/2
distance train : distance travel by the car at constant speed
so distance train= (vcar*36.35)m=421 m
the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2
Answer:
a) 2.933 m
b) 4.534 m
Explanation:
We're given the equation
v(t) = -0.4t² + 2t
If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.
See attachment for the calculations
The conclusion of the attachment will be
7.467 - 2.933 and that is 4.534 m
Thus, The distance it travels in the second 2 sec is 4.534 m
