The horizontal speed of the object 1.0 seconds later is 1) 5.0 m/s.
Explanation:
The motion of an object thrown horizontally off a cliff is a projectile motion, which follows a parabolic path that consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
This means that the horizontal speed of an object in projectile motion does not change, and remains constant during the whole motion.
Since in this case the object has been launched with a horizontal speed of
v = 5.0 m/s
this means that this speed will remain constant during the motion, so its horizontal speed 1.0 s later is also 5.0 m/s.
Learn more about projectile motion:
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The speed of an object is measured by calculating the distance it travels per unit time.
The formula in Physics for calculating speed is = distance/ time.
Since an object at rest is not covering any distance, the speed is always assumed to be 0 m/s.
The net force is 12 N to the left.
Answer:
an ellipse with the Sun at one focus or D
Explanation:
Answer for edgenuity
Answer:
1. 2.5×10¯⁹ N
2. 3.33×10¯¹¹ m/s²
Explanation:
1. Determination of the force of attraction.
Mass of astronaut (M₁) = 75 Kg
Mass of spacecraft (M₂) = 125000 Kg
Distance apart (r) = 500 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force of attraction (F) =?
The force of attraction between the astronaut and his spacecraft can be obtained as follow:
F = GM₁M₂ /r²
F = 6.67×10¯¹¹ × 75 × 125000 / 500²
F = 2.5×10¯⁹ N
Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N
2. Determination of the acceleration of the astronaut.
Mass of astronaut (m) = 75 Kg
Force (F) = 2.5×10¯⁹ N
Acceleration (a) of astronaut =?
The acceleration of the astronaut can be obtained as follow:
F = ma
2.5×10¯⁹ N = 75 × a
Divide both side by 75
a = 2.5×10¯⁹ / 75
a = 3.33×10¯¹¹ m/s²
Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²
