(1) The ball is in the air for <u>1.4 seconds.</u>
(2) The horizontal velocity of the ball as it rolls off the table is<u> 6.32 m/s.</u>
(3) The vertical velocity of the ball right before it hits the ground is <u>13.72 m/s.</u>
(4) The horizontal velocity of the ball right before it hits the ground is<u> 6.32 m/s.</u>
(5) The initial vertical velocity as soon as the ball comes of the cliff is <u>13.72 m/s.</u>
<h3>What is the time of motion of the ball?</h3>
The time of motion of the ball is calculated by applying the following equation.
t = √(2h/g)
where;
- h is the height of the cliff
- g is acceleration due to gravity
t = √(2h/g)
t = √(2 x 9.63 / 9.8)
t = 1.4 seconds
The horizontal velocity of the ball is calculated as follows;
v = d/t
where;
- d is the horizontal distance travelled by the ball = 8.85 m
v = 8.85 m / 1.4 s
v = 6.32 m/s
The vertical velocity of the ball before it hits the ground is calculated as;
vf = vi + gt
vf = 0 + 9.8 x 1.4
vf = 13.72 m/s
The horizontal velocity of the ball right before it hits the ground is calculated as;
the initial velocity of a projectile = final horizontal velocity
vxf = vxi = 6.32 m/s
The initial vertical velocity as soon as the ball comes off the cliff = final vertical velocity = 13.72 m/s
Learn more about horizontal velocity here: brainly.com/question/24949996
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Answer:
The velocity of the players will be <u>2.88 m/s</u> in the <u>east</u> direction.
Explanation:
Let 'v' be the velocity of the players after collision.
Consider the east direction as positive direction.
Given:
Mass of the first player is, 
kg
Initial velocity of the first player is, 
m/s
Mass of the second player is, 
kg
Initial velocity of the second player is, 
m/s
In order to solve this problem we use the law of conservation of momentum which says that the total momentum must be conserved before and after the collision. So we can write:
Solving for v, we get:
Therefore, their velocity after the collision is 2.88 m/s.
The sign of the velocity after collision is positive. So, the players will move in the east direction only after collision.
At the phase of D. Exhaust Stroke, the waste gases are removed during a four stroke engine. It is one of the phases of the four stroke engine. The cylinder from the fuel ignited during the compression step and removed from the cylinder.
Answer:
+1.46×10¯⁶ C
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C
Force (F) = 0.615 N
Distance apart (r) = 0.750 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Charge 2 (q₂) =?
The value of the second charge can be obtained as follow:
F = Kq₁q₂ / r²
0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²
0.615 = 236700 × q₂ / 0.5625
Cross multiply
236700 × q₂ = 0.615 × 0.5625
Divide both side by 236700
q₂ = (0.615 × 0.5625) / 236700
q₂ = +1.46×10¯⁶ C
NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C
Co carbon monoxide
sorry friend i don't know other ones
