Answer:
v = 27.3 m/s
Explanation:
Given that
Acceleration ,a= 4.2 m/s²
Time ,t= 6.5 s
Lets take the maximum speed gain by Thomson's= v
We know that ,if acceleration is constant then the speed v is given as
v= u + a t
v=final speed
u=initial speed
a=acceleration
t=time
Here the initial speed of Thomson's ,u = 0 m/s
Now by putting the values in the above equation we get
v= 0 + 4.2 x 6.5 m/s
v = 27.3 m/s
Therefore the maximum speed gain by Thomson will be 27.3 m/s.
Answer:
120 km/s
Explanation:
Given data :
Radius of the star is r = 19 km
Rotational time period of the star is T = 1 s
Therefore, we know that the velocity of the star is given by :
V = 119380.52 m/s
Therefore, the velocity of the point on the equator of the star is = 120 km/s
Answer:
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Explanation:
<span>Crust. The thin solid outermost layer of Earth. ...Asthenosphere. The lower layer of the crust. ...Lithosphere.Plasticity: is solid but still being able to. flow without being a liquid.The cool, rigid outermost layer of the Earth. ...<span>the solid part of the earth consisting of the crust and outer mantle.</span></span>
The first right-hand rule determines the directions of magnetic force, conventional current and the magnetic field. Given any two of theses, the third can be found.
The second Right-Hand Rule determines the direction of the magnetic field around a current-carrying wire and vice-versa<span> </span>
So, assuming that a magnetic field <span>exists and its direction is known and assuming that a charged particle moves in a specific direction through that field with velocity (v(, to determine the direction of force on the particle we should use the second right-hand rule.</span>