Answer:
displacement at 45 s = 30
65 s = 50
So the average speed over the interval from 45 s to 65 s is
(50 - 30) cm / 20 s = 1 cm / sec
As a check an average speed of 1 cm / sec for 20 sec will produce a
displacement of 1 cm / sec * 20 sec = 20 cm or from 30 to 50 cm
Answer:
7
Explanation:
A substance that is neither acidic nor basic is neutral. The pH scale measures how acidic or basic a substance is. The pH scale ranges from 0 to 14. A pH of 7 is neutral.
Answer:
![t_1 = \frac{v_i}{a_i}](https://tex.z-dn.net/?f=%20t_1%20%3D%20%5Cfrac%7Bv_i%7D%7Ba_i%7D)
![t_2 = \frac{v_i}{a_i}](https://tex.z-dn.net/?f=%20t_2%20%3D%20%5Cfrac%7Bv_i%7D%7Ba_i%7D)
Δd = ![v_it_1 = v_i^2/a_i](https://tex.z-dn.net/?f=%20v_it_1%20%3D%20v_i%5E2%2Fa_i)
Explanation:
As
, when the car is making full stop,
.
. Therefore,
![0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}](https://tex.z-dn.net/?f=0%20%3D%20v_i%20-%20a_it_1%5C%5Cv_i%20%3D%20a_it_1%5C%5Ct_1%20%3D%20%5Cfrac%7Bv_i%7D%7Ba_i%7D)
Apply the same formula above, with
and
, and the car is starting from 0 speed, we have
![v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}](https://tex.z-dn.net/?f=%20v_i%20%3D%200%20%2B%20a_it_2%5C%5Ct_2%20%3D%20%5Cfrac%7Bv_i%7D%7Ba_i%7D)
As
. After
, the car would have traveled a distance of
![s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}](https://tex.z-dn.net/?f=s%28t%29%20%3D%20s%28t_1%29%20%2B%20s%28t_2%29%5C%5Cs%28t_1%29%20%3D%20%28v_it_1%20-%20%5Cfrac%7Ba_it_1%5E2%7D%7B2%7D%29%5C%5C%20s%28t_2%29%20%3D%20%5Cfrac%7Ba_it_2%5E2%7D%7B2%7D)
Hence ![s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}](https://tex.z-dn.net/?f=%20s%28t%29%20%3D%20%28v_it_1%20-%20%5Cfrac%7Ba_it_1%5E2%7D%7B2%7D%29%20%2B%20%5Cfrac%7Ba_it_2%5E2%7D%7B2%7D%20)
As
we can simplify ![s(t) = v_it_1](https://tex.z-dn.net/?f=s%28t%29%20%3D%20v_it_1)
After t time, the train would have traveled a distance of ![s(t) = v_i(t_1 + t_2) = 2v_it_1](https://tex.z-dn.net/?f=%20s%28t%29%20%3D%20v_i%28t_1%20%2B%20t_2%29%20%3D%202v_it_1)
Therefore, Δd would be ![2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i](https://tex.z-dn.net/?f=%202v_it_1%20-%20v_it_1%20%3D%20v_it_1%20%3D%20v_i%5E2%2Fa_i)
Answer:
a) = 258352.5J
b) = 23.63 m/s
c) = 1.8m
Explanation:
Data;
Mass = 925kg
Distance (s) = 28.5m
Force constant (k) = 8.0*10⁴ N/m
g = 9.8 m/s²
a) = work = force * distance
But force = mass * acceleration
Force = 925 * 9.8 = 9065N
Work = F * s = 9065 * 28.5 = 258352.5J
b) acceleration (a) = (v² - u²) / 2s
a = v² / 2s
v² = a * 2s
v² = 9.8 * (2 * 28.5)
v² = 9.8 * 57
v² = 558.6
v = √(558.6)
V = 23.63 m/s
C). The work stops when the work done to raise the spring equals the work done to stop it by the spring
W = ½kx²
258352.5 = ½ * 8.0*10⁴ * x²
(2 * 258352.5) = 8.0*10⁴x²
516705 = 8.0*10⁴x²
X² = 516705 / 8.0*10⁴
X² = 6.46
X = √(6.46)
X = 2.54m
The compression was about 2.54m
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
![B=\frac{\mu_0 I}{2\pi r}](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B%5Cmu_0%20I%7D%7B2%5Cpi%20r%7D)
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:
![B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B%284%5Cpi%5Ccdot%2010%5E%7B-7%7D%29%2815%29%7D%7B2%5Cpi%280.0025%29%7D%3D1.2%5Ccdot%2010%5E%7B-3%7DT%20%3D%201.2%20mT)