Question: In which situation would a space probe most likely experience centripetal acceleration?
as it revolves around a planet
as it flies straight past a moon
as it is pulled in a line toward the Sun
as it lifts off from Earth
Answer:
When "space probe revolves around a planet" most likely to experience centripetal acceleration
Explanation:
Centripetal acceleration defined as the rate in change of tangential velocity. Also, as per Newton's second law, any kind of force will be directly proportional to the acceleration attained by the object. So, for centripetal acceleration, the force will be the centripetal force. The centripetal force will be acting on an object rotating in a circular motion with its direction of force towards the center. Thus, centripetal acceleration will be experienced by an object or a space probe when it is in a circular motion. It means the space probe is revolving around a planet.
Answer:
35 m/s^2
Explanation:
Decceleration = change in velocity / change in time
= ( 70 m/s ) / 2 seconds = 35 m/s^2
DEcceleration = 35 m/s^2
Jason's speed changes by - 35 m/s^2
You probably do this as a DC circuit which is not quite correct, but it will get you an answer. The study is a great deal more complicated.
Problem One: Secondary Power.
W = E * I
E = 10 A
I = 24V
W = 24 * 10
W = 240 Watts.
Here's the thing you have to know. These transformers are 100% efficient (or are assumed so). So whatever wattage is in the secondary, it is the same as that in the primary.
Primary Power = Secondary Power
Secondary Power = 240 Watts
Primary Power = 240 Watts
W = E * I
E = 240 volts
W = 240 watts
I = W/E = 240 / 240
I = 1 Amp. Answer Part One
Part Two
Answered above. 240 watts.
Part Three
Answered above. 240 watts.
Secondar
B
HOPE IT HELPS LET ME KNOW IF U NEED EXPLANATION
Upstream speed = S - 1
Downstream speed = S + 1
Average speed = total distance / total time
Average speed = (S - 1) + (S + 1) / 2
= S
S = 6 miles / 4 hours
S = 1.5 miles per hour
