In order to give a spaceship at rest in a specific reference frame s a speed increment of 0.500c, seven increments are required. Then, in this new frame, it receives an additional 0.500c increment.
The speed of an object, also known as v in kinematics, is a scalar quantity that refers to the size of the change in that object's position over time or the size of the change in that object's position per unit of time. The distance travelled by an object in a certain period of time divided by the length of the period gives the object's average speed in that period.
The spacecraft moves at v1 = 0.5c after the initial increment.The equation becomes V2 = V+V1/1+V*V1/c after the second one. 2 V2 = 0.5c+0.50c/1+(0.50c)^2/c^ 2 = 0.80c
Likewise, V3 = 0.929c
V4 = 0.976c
V5 = 0.992c
V6 = 0.99c
V7 = 0.999c
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Answer:
(a) The magnetic energy density in the field is 6.366 J/m³
(b) The energy stored in the magnetic field within the solenoid is 5 kJ
Explanation:
magnitude of magnetic field inside solenoid, B = 4 T
inner diameter of solenoid, d = 6.2 cm
inner radius of the solenoid, r = 3.1 cm = 0.031 m
length of solenoid, L = 26 cm = 0.26 m
(a) The magnetic energy density in the field is given by;
(b) The energy stored in the magnetic field within the solenoid
The sum is the result of adding 9260 and 3240 together. Each number can
be broken down into constituent parts in order to make addition easier.
Each place in the number represents its value, so a 2 in the hundreds
place represents 200.
You can separate numbers out this way to
make it easier to add them. 9260 can be broken down into 9000+200+60
while 3240 is 3000+200+40. You can then add these six numbers together.
60+40 = 100
200+200 = 400
9000+3000 = 12000
Then add your three partial results together to receive the final answer:
12000+400+100 = 12500
This being a perfect collision means no energy is lost during the collision. Because this question asks for speed and not velocity, the speed will be the same because the final energy is the same. The speed after the collision would therefore be 1.27 m/s.
Answer:
The tube should be held vertically, perpendicular to the ground.
Explanation:
As the power lines of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its potential.
And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.
