Answer: Option C. Is the correct answer.
Explanation: Wave length is the distance between the crest of each wave. If you were to decrease it you would be decreasing the distance between each wave therefore making it smaller and closer together as option C. shows.
Blood flows through the major artery at 1 m/s for 0.5 m then at a 0.6 m/s over a distance of another 0.5 m through the small artery the average speed of blood is 0.4 m/s.
We know that average speed = 
=0.4 m/s
Average speed is an important component in determining how long it takes to finish a journey. Average speed is simply a technique that assists us in calculating trip time and distance. It is obvious that the speed changes throughout the travel, making determining the average speed even more critical.
There are various methods for determining an object's or vehicle's average speed.
It is most desired when the speed of the object remains constant during the voyage, i.e. does not rise or decrease.
The approach for determining the average Speed is to divide. Divide the distance the vehicle travels by the time it travels to get the result.
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P= 10m = 10 x 9.10 = 91 N
Work is done lifting a 9.10-kg box straight up onto a shelf that is 1.80 m high :
A= Ph = 91 x 1.80 = 163.8 J
ok done. Thank to me :>
Half the potential difference of the the1-µF
A circuit must have a capacitance of 2 F across a 1 kV potential difference for an electrical technician. He has access to a sizable number of 1F capacitors, each of which can sustain a potential difference of no more than 400 V. Please suggest a configuration that uses the fewest capacitors possible.
The 2-mu F capacitor has the following characteristics: none of the aforementioned; half the charge of the 1-mu F capacitor; twice the charge of the 1-mu F capacitor; and half the potential difference of the 1-mu F capacitor.
Q = C V, C = Capacitance of the capacitor gives the charge stored by a capacitor with an applied voltage V. V is the applied voltage.
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Answer: 0.0180701 s
Explanation:
Given the following :
Length of string (L) = 10 m
Weight of string (W) = 0.32 N
Weight attached to lower end = 1kN = 1×10^3
Using the relation:
Time (t) = √ (weight of string * Length) / weight attached to lower end * acceleration due to gravity
g = acceleration due to gravity = 9.8m/s^2
Weight of string = 0.32N
Time(t) = √ (0.32 * 10) / [(1*10^3) * (9.8)]
Time = √3.2 / 9800
= √0.0003265
= 0.0180701s
