They are all units of measure of length
Explanation:
Length is a scalar quantity representing a distance between two points, and it can be expressed in different units.
The SI units of the length is the metre (m), which is defined as the length of the path travelled by light in vacuum during a time interval of 1/299,792,458 of a second.
A unit which is common used is a multiple of the meter, the kilometre (km) which corresponds to 1000 metres:
1 km = 1000 m
Another unit used in the UK system is the mile (mi), where the conversion factor between miles and metres is
1 mi = 1609.34 m
Finally, these units are not suitable to be used to measure astronomical distances - such as those between stars and galaxies. For this, another unit is used, which is the light-year (ly), which corresponds to the distance travelled by the light in a vacuum in one year, and its conversion factor to the metre is:

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Answer:



Explanation:
g = Acceleration due to gravity = 
= Angle of slope = 
v = Velocity of child at the bottom of the slide
= Coefficient of kinetic friction
= Coefficient of static friction
h = Height of slope = 1.8 m
The energy balance of the system is given by

The speed of the child at the bottom of the slide is 
Length of the slide is given by


The force energy balance of the system is given by

The coefficient of kinetic friction is
.
For static friction

So, the minimum possible value for the coefficient of static friction is
.
Answer:
b) Cells will pass through the G1/S checkpoint even if conditions are not ideal for cell division.
Explanation:
In the given problem, if there exists a gain-of-function mutation for the given cell, there would not be the formation of cyclin E when there is the possibility of cells movement via the checkpoint of the G1/S, even when there are non-deal conditions for the division of cell. Thus, the correct option in the lists of options is the option b.
Explanation:
It is given that, the water from a fire hose follows a path described by equation :
........(1)
The x component of constant velocity, 
We need to find the resultant velocity at the point (2,3).
Let
and 
Differentiating equation (1) wrt t as,



When x = 2 and 
So,


Resultant velocity, 

v = 6.4 m/s
So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.