Electrons flow from the positive end of a source towards the negative end
Answer:
1. Speed and velocity both involve a numeric rate describing the distance traveled by a body in a unit of time. However, speed describes the rate of a body traveling in any direction in a unit of time, while velocity describes the rate of a body traveling in a particular direction in a unit of time.
2. Answers may vary, but should resemble the following:
Average velocity explains the velocity the body traveled overall, not taking into consideration each spot in the trip. If a car moves at 65 km/h on average, it may have slowed down for some parts and sped up for others. Overall though, it would have made a certain distance of travel within a specified unit of time that totals the average velocity of 65 km/h.
Instantaneous velocity explains the velocity of a body at a particular instant of the trip. The instantaneous velocity of a car stopped at a stop sign would be 0 m/s even if it was moving before and will continue to move after this stop. The velocity at that particular instant is the instantaneous velocity.
Uniform velocity is when the distance being covered is changing uniformly with time. For example, if a car moves 20 km every 30 minutes and continues to do so in the same direction, it's traveling with a uniform velocity.
3. a=v2−v1t
a=20 m/s−60 m/s6 s
a=−406
a = –6.7 m/s2
4. v2 = v1 + at
v2 = 14 m/s + (3 m/s2 × 6 s)
v2 = 14 + 18
v2 = 32 m/s
5. v=st
v=375 km5 h
v = 75 km/h
6. First, convert the minutes to seconds. Since there are 60 seconds in one minute, multiply:
60 × 15 (minutes) = 900 seconds
s = v × t
s = 6 m/s × 900 s
s = 5,400 m
7. t=sv
t=80 km35 km/hr
t = 2.29 hr
8. a=v2−v1t
a=50 m/s−15 m/s4 s
a=35 m/s4 s
a = 8.75 m/s2
9. vav=v1+v22
vav=15 m/s+50 m/s2
vav=65 m/s2
vav = 32.5 m/s
10. a=v2−v1t
a=0 m/s−11.5 m/s3.5 s
a = –3.29 m/s2
Explanation:
Answer:
B. over the symbol.
Explanation:
vectors are represented with a symbol carrying an arrow head with also indicates direction
The work done by gas is 0.753 J and change in internal energy is 4.247J
So we are given that mass is 2kg , radius 1 cm and the amount of heat is 5 cm
The piston raised by 2.4cm
As we know that Work done is PΔV
Where ΔV is change in volume
Therefore ΔV = πr^2 h = π x (.01)^2 x .024 =7.53×10^(-6)m^3
Here pressure is 10^5 pa
So W = 
Therefore W = 0.753 J
Now coming to change in internal energy
Change in Internal Energy = Heat Added - Energy lost in work
∴ 5J - 0.753 J = 4.247J
Hence the change in internal energy is 4.247 J
Learn more about Work done here
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