4.2 liters..... there are 1,000 mL in a liter and there is a total of 4200 mL in this case which is divided by 1000 which gives you 4.2 liters.
Answer:
865.08 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 243 m/s
Height (h) of the cliff = 62 m
Horizontal distance (s) =?
Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:
Height (h) of the cliff = 62 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
62 = ½ × 9.8 × t²
62 = 4.9 × t²
Divide both side by 4.9
t² = 62/4.9
Take the square root of both side.
t = √(62/4.9)
t = 3.56 s
Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:
Initial velocity (u) = 243 m/s
Time (t) = 3.56 s
Horizontal distance (s) =?
s = ut
s = 243 × 3.56 s
s = 865.08 m
Thus, the cannon ball will impact the ground 865.08 m from the base of the cliff.
Answer:
0.12
Explanation:
The acceleration due to gravity of a planet with mass M and radius R is given as:
g = (G*M) / R²
Where G is gravitational constant.
The mass of the planet M = 3 times the mass of earth = 3 * 5.972 * 10^24 kg
The radius of the planet R = 5 times the radius of earth = 5 * 6.371 * 10^6 m
Therefore:
g(planet) = (6.67 * 10^(-11) * 3 * 5.972 * 10^24) / (5 * 6.371 * 10^6)²
g(planet) = 1.18 m/s²
Therefore ratio of acceleration due to gravity on the surface of the planet, g(planet) to acceleration due to gravity on the surface of the planet, g(earth) is:
g(planet)/g(earth) = 1.18/9.8 = 0.12
Answer:
C sedimentary rocks is the answer.
Explanation:
Answer:
Power, P = 600 watts
Explanation:
It is given that,
Mass of sprinter, m = 54 kg
Speed, v = 10 m/s
Time taken, t = 3 s
We need to find the average power generated. The work done divided by time taken is called power generated by the sprinter i.e.

Work done is equal to the change in kinetic energy of the sprinter.


P = 900 watts
So, the average power generated by the sprinter is 900 watts. Hence, this is the required solution.