Explanation:it is beause they are sharper and also have less surface area and therefore more pressure
Transverse waves travel on a direction that is perpendicular to the motion of the particles (or whatever medium is waving) So the particles must be moving east to west, which is transverse to the north-south motion of the wave
Answer:
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
Explanation:
We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.
Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn
They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole
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L + W₂ L + W₁ L / 2 = 0
T_{y} = W₂ + W₁ / 2
T_{y} = 120 + 150/2
T_{y} = 195 lb
we use trigonometry to find the cable tension
sin 30 = T_{y} / T
T = T_{y} / sin 30
T = 195 / sin 30
T = 390 lb
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
T < 500 lb
Answer:
The centripetal force acting on the child is 39400.56 N.
Explanation:
Given:
Mass of the child is, 
Radius of the barrel is, 
Number of revolutions are, 
Time taken for 10 revolutions is, 
Therefore, the time period of the child is given as:

Now, angular velocity is related to time period as:

Now, centripetal force acting on the child is given as:

Therefore, the centripetal force acting on the child is 39400.56 N.
Answer: 14.52*10^6 m/s
Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.
the change in potential energy for the electron; e*ΔV is equal to energy kinetic gained for the electron so:
e*ΔV=1/2*m*v^2 v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s