Answer:
Maximum 13 m overhang
Explanation:
Since the brick is uniform. Its mass must be equally distributed along its length. So the center of mass is in geometric center, which is 26 / 2 = 13 m from each end of the brick.
For the brick to not tip over, its center of mass must NOT be overhung over the edge. It stays just on the edge. So the maximum overhang is 13 m.
Answer:
1.24Mev
Explanation:
Using
E= hc/lambda
= (6.62x10^-19) x(3x10^8m/s)/(1x10^-12) x 1.602x10^-9
= 1.24Mev
Answer:
transverse wave, motion in which all points on a wave oscillate along paths at right angles to the direction of the wave's advance. Surface ripples on water, seismic S (secondary) waves, and electromagnetic (e.g., radio and light) waves are examples of transverse waves.
Answer:
![v_S=\sqrt{2}v_L](https://tex.z-dn.net/?f=v_S%3D%5Csqrt%7B2%7Dv_L)
Explanation:
The acceleration experimented while taking a curve is the centripetal acceleration
. Since
, we have that: ![\frac{v_S^2}{r_S}=\frac{2v_L^2}{r_L}](https://tex.z-dn.net/?f=%5Cfrac%7Bv_S%5E2%7D%7Br_S%7D%3D%5Cfrac%7B2v_L%5E2%7D%7Br_L%7D)
They take the same curve, so we have: ![r_S=r_L=R](https://tex.z-dn.net/?f=r_S%3Dr_L%3DR)
Which means: ![v_S^2=2v_L^2](https://tex.z-dn.net/?f=v_S%5E2%3D2v_L%5E2)
And finally we obtain: ![v_S=\sqrt{2}v_L](https://tex.z-dn.net/?f=v_S%3D%5Csqrt%7B2%7Dv_L)
<em>as we know that P=mv .. where m is mass and v is velocity ..this relation shows that momentum has a direct relation with these two quantities .. so when velocity is increased twice
P' = mv
P'=m2v
P' = 2 mv here mv is P so
P'=2P
this shows that momentum will also be doubled by doubling the velocity :)</em>