Answer:
19.62 ms
Explanation:
t = Time taken = 2 s
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s² (we take downward direction as positive)
Using the equations of motion
The pebble's speed upon contact with the water is 19.62 ms
Response:
The horizontal span of Sosa is 276.526 ft or 84.28 meters.
Explanation:
As illustrated in the diagram, let point O denote Sosa's starting position. She travels 361 ft at a 50-degree angle relative to the horizontal.
sin 50 =
0.7660 = h / 361
h = 276.526 ft
h = 84.28 meters
The horizontal distance of Sosa is 276.526 ft or 84.28 meters.
Answer:
(a). The distance traveled is 7.06 km towards the west.
(b). Their displacement magnitude is 7.51 km.
Explanation:
The information given states that,
The geese initially move 4.0 km directly west, then alter course to the north at a 40° angle, covering an additional 4.0 km.
Based on the diagram,
(a). To determine the distance
We will apply the distance formula
Insert the values into the equation
The resultant distance is 7.06 km westward.
(b). We will find the total displacement's magnitude
Using the displacement formula
Insert the values into the equation
The total displacement magnitude is 7.51 km.
In conclusion, (a). The traveled distance is 7.06 km towards the west.
(b). The magnitude of their total displacement is 7.51 km.
Answer:
The density comes out to be Mg/µL
Explanation:
Given data:
The density of nuclear matter is approximately kg/m³
1 ml corresponds to 1 cm³
To determine:
The density of nuclear matter in Mg/µL
Solution:
We recognize that:
1 Mg equals 1000 kg
Thus, 1 m³ is equal to cm³
Moreover, 1 cm³ is equivalent to 1 mL
Thus, we can conclude that 1 mL is equal to 10³ µL
With this, we convert the density as follows:
Density = kg/m³
Density = kg/m³ ×
Mg/µL
Density = Mg/µL