Answer:
The gravity on this planet is stronger than that of earth.
Explanation:
First we need to find the acceleration due to gravity value of this planet to compare its gravity force with that of the earth. Hence, we will use second equation of motion:
h = Vi t + (0.5)gt²
where,
h = height or depth of crater = 100 m
Vi = Initial Velocity of rock = 0 m/s
t = time = 4 s
g = acceleration due to gravity on this planet = ?
Therefore,
100 m = (0 m/s)(4 s) + (0.5)(g)(4 s)²
g = (200 m)/(16 s²)
g = 12.5 m/s²
on earth:
ge = 9.8 m/s²
Since,
ge < g
Therefore,
<u>The gravity on this planet is stronger than that of earth.</u>
Answer:
I only speak English
Explanation:
I'm sorry can you type it in English
The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m
s = ut + 1 / 2 at²
s = Distance
u = Initial velocity
t = Time
a = Acceleration
Vertically,
s = 15.4 m
u = 0
a = 9.8 m / s²
15.4 = 0 + ( 1 / 2 * 9.8 * t² )
t² = 3.14
t = 1.77 s
Horizontally,
u = 3.01 m / s
a = 0 ( Since there is no external force )
s = ( 3.01 * 1.77 ) + 0
s = 5.34 m
Therefore, the distance travelled by the ball before hitting the ground is 5.34 m
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Answer:
v = 5.9 x 10⁷ m/s
Explanation:
The kinetic energy of the electron in terms of potential difference is given as:
--------------- equation (1)
where,
e = charge on electron = 1.6 x 10⁻¹⁹ C
V = Potential Difference = 9.9 KV = 9900 Volts
The kinetic energy in general is given as:
--------- equation (2)
where,
m = mass of electron = 9.1 x 10⁻³¹ kg
v = speed of electron = ?
Therefore, comparing equation (1) and equation (2), we get:

<u>v = 5.9 x 10⁷ m/s</u>
The displacement of the train after 2.23 seconds is 25.4 m.
<h3>
Resultant velocity of the train</h3>
The resultant velocity of the train is calculated as follows;
R² = vi² + vf² - 2vivf cos(θ)
where;
- θ is the angle between the velocity = (90 - 51) + 37 = 76⁰
R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)
R² = 129.75
R = √129.75
R = 11.39 m/s
<h3>Displacement of the train</h3>
Δx = vt
Δx = 11.39 m/s x 2.23 s
Δx = 25.4 m
Thus, the displacement of the train after 2.23 seconds is 25.4 m.
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