Answer:
A transverse and D electromagnetic
on edg2020
Explanation:
I got it right on the test
good luck
Here we will the speed of seagull which is v = 9 m/s
this is the speed of seagull when there is no effect of wind on it
now in part a)
if effect of wind is in opposite direction then it travels 6 km in 20 min
so the average speed is given by the ratio of total distance and total time
now since effect of wind is in opposite direction then we can say
Part b)
now if bird travels in the same direction of wind then we will have
now we can find the time to go back
Part c)
Total time of round trip when wind is present
now when there is no wind total time is given by
So due to wind time will be more
Answer:
True
Explanation:
An electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.
Answer:
the volume is 0.253 cm³
Explanation:
The pressure underwater is related with the pressure in the surface through Pascal's law:
P(h)= Po + ρgh
where Po= pressure at a depth h under the surface (we assume = 1atm=101325 Pa) , ρ= density of water ,g= gravity , h= depth at h meters)
replacing values
P(h)= Po + ρgh = 101325 Pa + 1025 Kg/m³ * 9.8 m/s² * 20 m = 302225 Pa
Also assuming that the bubble behaves as an ideal gas
PV=nRT
where
P= absolute pressure, V= gas volume ,n= number of moles of gas, R= ideal gas constant , T= absolute temperature
therefore assuming that the mass of the bubble is the same ( it does not absorb other bubbles, divides into smaller ones or allow significant diffusion over its surface) we have
at the surface) PoVo=nRTo
at the depth h) PV=nRT
dividing both equations
(P/Po)(V/Vo)=(T/To)
or
V=Vo*(Po/P)(T/To) = 0.80 cm³ * (101325 Pa/302225 Pa)*(277K/293K) = 0.253 cm³
V = 0.253 cm³
Answer:
Moving a unit "positive" test charge from A to B will result in a reduction in potential
V = K Q / R potential at a point
V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q
V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8
V2 - V1 = -4.17 * 562.5 J/C
V = - 2346 Volts
