Answer:
The velocity of Mosquito with respect to earth will be 0.302m/s
Explanation:
V(ma) = 1.10 m/s, east Velocity of mosquito with respect to air
V(ae) = 1.4 m/s at 35° Velocity of air with respect to Earth in west of south direction.
Velocity of Mosquito with respect to earth will be
V(me) = V(ma) + V(ae)
We need to find the mosquito’s speed with respect to Earth in the x direction.
V(x, me) = V(x, ma) + V(x, ae) = V(ma) + V(ae)(cos theta(ae) )
Angle (ae) = –90.0° − 35°=−125°
V(x, me) = 1.10 + (1.4)Cos(-125)
= 1.10 + 1.4(-0.57)
= 1.10 -0.798
= 0.302
So the velocity of Mosquito with respect to earth will be 0.302m/s
Answer:
6.86 meters
Explanation:
Let the compression of the string be represented by x, and the height of projection of the toy rocket be represented by h.
So that;
x = 9 cm = 0.09 m
In its rest position (i.e before the launch), the spring has a stored potential energy which is given as;
PE = 
K
= x 830 x 
= 415 x 0.0081
= 3.3615
The potential energy in the string = 3.36 Joules
Also,
PE = mgh
where: m is the mass, g is the gravitational force and h the height.
m = 50 g = 0.05 kg, g = 9.8 m
Thus,
PE = 0.05 x 9.8 x h
3.3615 = 0.05 x 9.8 x h
3.3615 = 0.49h
⇒ h = 
= 6.8602
The height of the toy rocket would be 6.86 meters.
I know what it is but its hard to see
It’s voltage can easily be modified and it allows power to be transmitted at a high voltage rate before being lowered to a smaller voltage rate to be safe.
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