Answer:
OA. . review all safety procedures and the lab activity procedure
Explanation:
Answer:
Explanation:
This is a problem based on time dilation , a theory given by Albert Einstein .
The formula of time dilation is as follows .
t₁ = ![\frac{t}{\sqrt{1-\frac{v^2}{c^2} } }](https://tex.z-dn.net/?f=%5Cfrac%7Bt%7D%7B%5Csqrt%7B1-%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%20%7D%20%7D)
t is time measured on the earth and t₁ is time measured by man on ship .
A ) Given t = 20 years , t₁ = ? v = .4c
![\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }](https://tex.z-dn.net/?f=%5Cfrac%7B20%7D%7B%5Csqrt%7B1-%5Cfrac%7B.16c%5E2%7D%7Bc%5E2%7D%20%7D%20%7D)
=1.09 x 20
t₁= 21.82 years
B ) Given t = 5 years , t₁ = ? v = .2c
![\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B%5Csqrt%7B1-%5Cfrac%7B.04c%5E2%7D%7Bc%5E2%7D%20%7D%20%7D)
=1.02 x 5
t₁= 5.1 years
C ) Given t = 10 years , t₁ = ? v = .8c
![\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }](https://tex.z-dn.net/?f=%5Cfrac%7B10%7D%7B%5Csqrt%7B1-%5Cfrac%7B.64c%5E2%7D%7Bc%5E2%7D%20%7D%20%7D)
=1.67 x 10
t₁= 16.7 years
D ) Given t = 10 years , t₁ = ? v = .4c
![\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }](https://tex.z-dn.net/?f=%5Cfrac%7B10%7D%7B%5Csqrt%7B1-%5Cfrac%7B.16c%5E2%7D%7Bc%5E2%7D%20%7D%20%7D)
=1.09 x 10
t₁= 10.9 years
E ) Given t = 20 years , t₁ = ? v = .8c
![\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }](https://tex.z-dn.net/?f=%5Cfrac%7B20%7D%7B%5Csqrt%7B1-%5Cfrac%7B.64c%5E2%7D%7Bc%5E2%7D%20%7D%20%7D)
=1.67 x 20
t₁= 33.4 years
Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:
![2gh = v_f^2 - v_i^2](https://tex.z-dn.net/?f=2gh%20%3D%20v_f%5E2%20-%20v_i%5E2)
where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,
![(2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\](https://tex.z-dn.net/?f=%282%29%289.8%5C%20m%2Fs%5E2%29h%20%3D%20%280%5C%20m%2Fs%29%5E2-%284%5C%20m%2Fs%29%5E2%5C%5C)
<u>h = 0.82 m</u>
Now, for the time in air during upward motion we use first equation of motion:
![v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s](https://tex.z-dn.net/?f=v_f%20%3D%20v_i%20%2B%20gt_1%5C%5C0%5C%20m%2Fs%20%3D%204%5C%20m%2Fs%20%2B%20%28-9.8%5C%20m%2Fs%5E2%29t_1%5C%5Ct_1%20%3D%200.41%5C%20s)
(c)
Now we will consider the downward motion and use the third equation of motion:
![2gh = v_f^2-v_i^2](https://tex.z-dn.net/?f=2gh%20%3D%20v_f%5E2-v_i%5E2)
where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,
![2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\](https://tex.z-dn.net/?f=2%289.8%5C%20m%2Fs%5E2%29%282.62%5C%20m%29%20%3D%20v_f%5E2%20-%20%280%5C%20m%2Fs%29%5E2%5C%5C)
<u>vf = 7.17 m/s</u>
Now, for the time in air during downward motion we use the first equation of motion:
![v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s](https://tex.z-dn.net/?f=v_f%20%3D%20v_i%20%2B%20gt_1%5C%5C7.17%5C%20m%2Fs%20%3D%200%5C%20m%2Fs%20%2B%20%289.8%5C%20m%2Fs%5E2%29t_2%5C%5Ct_2%20%3D%200.73%5C%20s)
(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
<u>t = 1.14 s</u>
The directions of magnetic force and magnetic field lines are shown in the figure.
The direction to find out the magnetic field lines is given by right hand curl rule. If the thumb shows the direction of current, then the curling fingers show the direction of magnetic field lines.
The direction of force can be given by right hand thumb rule, where
Thumb - Direction of magnetic field lines
Forefinger - Magnetic
force
Centre finger -
Current
Such that forefinger, centre finger and thumb must be at 90 degrees to each other.