Answer: the speed at which it falls toward the Earth.
Explanation:
A bullet travelling across Earth's surface with some horizontal velocity is classical example of projectile motion.
Projectile motion is an idealization of the motion under the action of gravity neglecting the influence of the air (no drag force nor friction).
This kind of motion is the result of two independent motions: vertical motion and horizontal motion.
The observed net velocity is the vectorial sum of the vertical and horizontal velocities.
The horizontal velocity is constant, since there is not any force acting in the horizontal axis. Thi is, the object, following the first Law of Newton (inertia law) tends to continue in uniform rectilinear movement (with zero acceleration).
The vertical velocity, this is the velocity at which the bullet falls toward the Earth, is influenced (accelerated) by the action of the gravity of the Earth. So, the vertical velocity is accelerated by the pull of the Earth.
Vertical and horizontal velocities are independent of each other, which means that the speed or the magnitude of the horizontal velocity does not affect the speed at which an object (the bullet) falls toward the Earth.
Answer:
Explanation:
We shall apply Pascal's Law in fluid mechanics
According to it , pressure is transmitted in liquid from one point to another without any change .
25 cm diameter = 12.5 x 10⁻² m radius
Area = 3.14 x (12.5 x 10⁻²)²
= 490.625 x 10⁻⁴ m²
Pressure by vehicle
Force / area
13000 / 490.625 x 10⁻⁴
= 26.497 x 10⁴ Pa
5 cm diameter = 2.5 x 10⁻² radius
area = 3.14 x (2.5 x 10⁻²)²
= 19.625 x 10⁻⁴ m²
If we assume required force F on this area
Pressure = F / 19.625 x 10⁻⁴ Pa
According to Pascal Law
F / 19.625 x 10⁻⁴ = 26.497 x 10⁴
F = 19.625 x 26.497
= 520 N
Answer:
The acceleration of the cart is 1.0 m\s^2 in the negative direction.
Explanation:
Using the equation of motion:
Vf^2 = Vi^2 + 2*a*x
2*a*x = Vf^2 - Vi^2
a = (Vf^2 - Vi^2)/ 2*x
Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.
Let x = Xf -Xi
Where Xf is the final position of the cart and Xi the initial position of the cart.
x = 12.5 - 0
x = 12.5
The cart comes to a stop before changing direction
Vf = 0 m/s
a = (0^2 - 5^2)/ 2*12.5
a = - 1 m/s^2
The cart is decelerating
Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.
