From the picture, I see that you had no trouble at all with #4.
Well, #5, 6, and 7 are easily handled in exactly the same way.
Just as you did with #4, please sketch these on paper
as I walk you through the solutions. That'll help you see
immediately what's going on.
#5.b).
Traveling east at 3 m/s for 4 seconds,
he covers (3 m/s) x (4 sec) = 12 meters.
Traveling south at 5 m/s for 2 seconds,
he covers (5 m/s) x (2 sec) = 10 meters.
The total distance he covers is (12m + 10m) = 22 meters.
#5.c).
Average speed (scalar)
= (distance covered)/(time to cover the distance)
= (22 meters)/(6 sec) = 3-2/3 m/s .
#5.d).
Displacement (vector)
= distance between the start-point and the end-point,
regardless of the route traveled,
in the direction from the start-point to the end-point.
Distance from the start-point to the end-point =
√(12² + 10²) = √(144 + 100) = √(244) = 15.62 meters
in the direction of arctan(10/12) south of east
= 39.8° south of east.
#5.e).
Average velocity (vector) =
(displacement vector) / (time)
= 15.62 meters directed 39.8° south of east / 6 seconds
= 2.603 m/s directed 39.8° south of east.
#6).
Magnitude = √(5.2² + 2.1²) = √(27.04 + 4.41) = √31.45 = 5.608 km.
Direction = arctan(5.2/2.1) south of east
= 68° south of east = 158° bearing .
#7).
Magnitude = √(39² + 57²) = √(1521 + 3249) = √( 4770)
= 69.07 m/s .
Direction = arctan (57/39) south of west
= 55.6° south of west
Bearing = 214.4°
Compass: 0.65° past "southwest by south".
I'm grateful for the privilege and opportunity to practice my math,
and I shall cherish the bounty of 5 points that came with it.
________b_____ 7 km east
|
| 2km north.
|a
|
°
pythagorean theorem : ✓a² + b² = c²
c² = a² + b² = 4 + 49 = 53
c = ✓53 km
displacement = c = ✓53 km
distance = 10 + 3 + 2 = 15 km
Answer:
The velocity of the ball when its hit the ground will be 54.22 m/sec
Explanation:
We have given height from which ball is dropped h = 150 m
Acceleration due to gravity 
As the ball is dropped so initial velocity will be zero so u = 0 m/sec
According to third equation of motion we know that 
So the velocity of the ball when its hit the ground will be 54.22 m/sec
Explanation:
Given that,
Charge on a spherical drop of water is 43 pC
The potential at its surface is 540 V
(a) The electric potential on the surface is given by :
r is the radius of the drop
(b) Let R is the radius of the spherical drop, when two such drops of the same charge and radius combine to form a single spherical drop. ATQ,
Now the charge on the new drop is 2q. New potential is given by :
Hence, the radius of the drop is 
and the potential at the surface of the new drop is 856.79 V.
