Answer: summer and greater than 66°N
Explanation:
Summer is the most appropriate season in which the sun does not set. It is a typical situation for the Arctic Circle. It is the latitude in which the sun does not set in the day and typically during the month of July. The north of this circle is a period of sunshine which last for a period of about six months in the North Pole. Summer never rises in the winter solstice (December).
Answer:
a,V=311.15m/s
b.246Hz
c.245Hz
d. 1.4m
Explanation:
One of the 63.5-cm-long strings of an ordinaryguitar is tuned to produce the note {\rm B_3}(frequency 245 Hz) when vibrating inits fundamental mode.
A.Find the speed of transverse waves on this string.B.If the tension in this string is increased by1.0%, what will be the new fundamental frequency ofthe string?C.If the speed of sound in the surrounding air is 344m/s, find the frequency of the sound wave produced in theair by the vibration of the {\rm B_3} string.D.If the speed of sound in the surrounding air is 344m/s, find the wavelength of the sound wave produced in theair by the vibration of the {\rm B_3} string
NOTE
speed is distance by the wave per time
frequency is the number of oscillation the wave front makes in one seconds
the wave speed is given by

recall also that the wave speed is v=f lambda
for a standing wave , we know the fundamental frequency of a string is
f1=v/2L
L=length of the string
f1=245Hz
V=?
L=0.635m
V=245*2*0.635
V=311.15m/s
b. tension in the string is increased by 1%
F2=F+1%
f2=101F%
substituting for F2 into this equation


v2=
v2=1.01^0.5*311.15m/s
v2=312.7m/s
for the new fundamental frequency we have
f2=312.7/2*0.635
f2=246Hz
c. the frequency of the sound wave equal the frequency of the string that created it
c. fs=245Hz
d. speed of sound in air344m/s
v=344m/s
v=f*lambda
lambda is the wavelength
344=245*lambda
344/245=1.40m
wavelength of string B3 is 1.4m
Answer:
V = (5.8cm/s)i, (4.7cm/s)j
Explanation:
Given :
r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i^+( 4.70 cm/s )tj^
To obtain the average velocity (V)
V = (r2 - r1) / (t2 - t1)
To obtain r1 and r2, substitute t1 = 0 and t2 = 2 respectively in the equation above
r1 = [ 4.50 cm +( 2.90 cm/s2 ) 0]i^+( 4.70 cm/s )0 j
r1 = 4.50 cm + 0 + 0 = (4.50cm)i + 0j
r2 = [ 4.50 cm +( 2.90 cm/s2 )2²]i^+( 4.70 cm/s )2 j
r2 = 4.50cm + (2.90 × 4)i + (4.70 × 2)j
r2 = (16.1cm)i + (9.4cm)j
V = [(16.1 - 4.50)i - (9.4 - 0)j] / 2 - 0
V = 11.6i / 2 ; 9.4j / 2
V = (5.8cm/s)i, (4.7cm/s)j
Steam<span> is water in the gas phase, which is </span>formed<span> when water boils. </span>Steam<span> is invisible; however, "</span>steam<span>" often refers to wet </span>steam<span>, the visible mist or aerosol of water droplets </span>formed<span> as this water vapour condenses.
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The momentum before is zero because nothing is moving. Note that because momentum is a conserved vector, the total momentum after the shot is also zero. You just have to add the products of mass and velocity in opposite directions. So the answer is zero