Let's use ' t ' to represent half of the time, in hours.
The distance traveled in the first half of the time is (80 t) km.
The distance traveled in the last half of the time is (40 t) km.
The total distance covered is (80t + 40t) = (120t) km.
You said that the total distance covered was 60 km,
so ...
120 t = 60 km
Divide each side by 120 : t (half of the time) = 0.5 hour
Average speed = (total distance covered) / (time to cover the distance)
= (60 km) / (1 hour)
= 60 km/hr .
You've failed because you failing becomes a statement rather than it becoming fact or what actually happened.
An arrow which shows the direction that the probe should be moving in order for it to enter the orbit is X.
<h3>What is an orbit?</h3>
An orbit can be defined as the curved path through which a astronomical (celestial) object such as planet Earth, in space move around a Moon, Sun, planet or star.
In this scenario, if the scientists want the probe to enter the orbit they should ensure that probe moves in direction X. This ultimately implies that, the probe must move in the same direction as the orbit, in order to enter it.
Read more on orbit here: brainly.com/question/18496962
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Answer:
= 4.86 s
= 1.98 s
Explanation:
<u><em>Given:</em></u>
Length = l = 1 m
Acceleration due to gravity of moon = 
= 1.67 m/s²
Acceleration due to gravity of Earth = 
= 10 m/s²
<u><em>Required:</em></u>
Time period = T = ?
<u><em>Formula:</em></u>
T = 2π 
<u><em>Solution:</em></u>
<u>For moon</u>
<em>Putting the givens,</em>
T = 2(3.14) 
T = 6.3 
T = 6.3 × 0.77
T = 4.86 sec
<u>For Earth,</u>
<em>Putting the givens</em>
T = 2π 
T = 2(3.14) 
T = 6.3 × 0.32
T = 1.98 sec
Answer:
C. amount of charge on the source charge.
Explanation:
Electric field lines can be defined as a graphical representation of the vector field or electric field.
Basically, it was first introduced by Michael Faraday and it is typically a curve drawn to the tangent of a point is in the direction of the net field acting on each point.
The number, or density, of field lines on a source charge indicate the amount of charge on the source charge. Therefore, the density of field lines on a source charge is directly proportional to quantity of charge on the source.
