Your velocity forward in would be 1.66 m/s. to solve you would use the velocity equation.
Answer:
a) 2Hz
b) 0.5seconds
c) 0.34m/s
Explanation:
a) Frequency is defined as the number of oscillations completed by a wave in one second.
If my finger is dipped into a pan of water twice each second, it means that my hand has made 2 oscillations through the water in one second. An oscillation is a to and fro movement of a particle, body or wave through a medium.
Based on the conclusion, the frequency of the waterwaves will be 2Hertz or 2cycles/sec.
2) Period T of a wave is defined as the time taken by a wave to complete one oscillation. It is the reciprocal of the frequency of a wave.
T = 1/F
Given frequency = 2Hertz
T = 1/2
T = 0.5seconds
Period of the water wave is 0.5seconds
c) speed of the wave v is expressed according to the relationship:
velocity = frequency × wavelength
Given:
Frequency = 2Hertz
Wavelength = 0.17m (Since it is the difference between two successive crest or trough of a wave)
Velocity = 2×0.17
Velocity = 0.34m/s
Answer:
A
Explanation:
the answer would be basicly A
Out of the following given choices;
a. decelerate; accelerate
b. accelerate; decelerate
c. stop; accelerate
d. decelerate; stop
<span>The answer is B. This is because
the cold front usually rotates around the warm front as cold air mass coverage
in the low-pressure system. This causes the cold air mass to accelerate to
catch up with the warm front. When an occluded front ( that is the boundary
that separates the older cool air mass already in place from new incoming cold
air mass), that the system decelerates. </span>
a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J
b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J
Applying Law of Energy conservation :
K 1+U 1
=K 2+U 2
⇒K 1− r 1GmM
=K 2− r 2 GmM
where M=5.0×10 23kg,r1
=> R=3.0×10 6m and m=10kg
(a) If K 1
=5.0×10 7J and r 2
=4.0×10 6 m, then the above equation leads to
K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J
(b) In this case, we require K 2
=0 and r2
=8.0×10 6m, and solve for K 1:K 1
=K 2 +GmM (r 11− r 21)=6.9×10 7 J
Learn more about Kinetic energy on:
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