Answer:
C!
Explanation:
Cold fronts generally advance at average speeds of 20 to 25 mph. toward the east — faster in the winter than summer — and are usually oriented along a northeast to southwest line.
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:
False
Explanation:
The depending on what type of pressure is enclosed in a object, the smaller the object the more pressure because that pressure has limited room to escape.
B) the mass of the reactant equals the mass of the product.
The object must be placed at the distance greater than the focal length of the lens
