Answer:
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Explanation:
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Answer:5.7m/s
Explanation:
Mass=1kg
Initial velocity=u=8m/s
height=h=1.6m
Final velocity =v
Acceleration due to gravity=g=9.8m/s^2
v^2=u^2-2xgxh
v^2=8^2-2x9.8x1.6
v^2=8x8-2x9.8x1.6
v^2=64-31.36
v^2=32.64
Take the square root of both sides
√(v^2)=√(32.64)
v=5.7
Speed at the height of 1.6m is 5.7m/s
Answer:
Explanation:
Our values are,
We have all the values to apply the law of linear momentum, however, it is necessary to define the two lines in which the study will be carried out. Being an intersection the vehicle of mass m_1 approaches through the X axis, while the vehicle of mass m_2 approaches by the y axis. In the collision equation on the X axis, we despise the velocity of object 2, since it does not come in this direction.
For the particular case on the Y axis, we do the same with the speed of object 1.
By taking a final velocity as a component, we can obtain the angle between the two by relating the equations through the tangent
Replacing in any of the two functions, given above, we will find the final speed after the collision,
Answer:
The P site
Explanation:
After the formation of peptide linkage, the trna of P site becomes free, more to the E site and then slip away. The A site trna has dipeptide instead of a single aminoacid.
In the prescence of translocase and energy from GTP, the ribosome move in such a way the peptidyl bearing trna of A site come to lie with P sit. This exposes next codon to A site. A new aminoacid trna complex reaches the fresh codon exposed at A site.
