The force exerted by a pressure of any gas over a surface its given by the formula P=F/S (where P is pressure, F force and S surface).
We can multiply both sides of the formula by S to obtain the force.
P*S=(F*S)/S
P*S=F
Solve for P=1.80*10^5 Pa and S=4.10*10^-4 m^2 ([Pa] =[N/m^s])
(1.80*10^5 N/m^s) * (4.10*10^-4 m^2) =F
73.8 N =F
The time the truck must apply the given force to increase its speed to given value is 5 s.
The given parameters;
- <em>applied force, F = 600 N</em>
- <em>mass of the truck, m = 1,500 kg</em>
- <em>speed of the truck, v = 2 m/s</em>
The force applied to the truck is determined by Newton's second law of motion; <em>which states that the force applied to an object is directly proportional to the product of mass and acceleration of the object.</em>
F = ma
Thus, the time the truck must apply the given force to increase its speed to given value is 5 s.
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Answer:
the shooting angle ia 18.4º
Explanation:
For resolution of this exercise we use projectile launch expressions, let's see the scope
R = Vo² sin (2θ) / g
sin 2θ = g R / Vo²
sin 2θ = 9.8 75/35²
2θ = sin⁻¹ (0.6)
θ = 18.4º
To know how for the arrow the tree branch we calculate the height of the arrow at this point
X2 = 75/2 = 37.5 m
We calculate the time to reach this point since the speed is constant on the X axis
X = Vox t
t2 = X2 / Vox = X2 / (Vo cosθ)
t2 = 37.5 / (35 cos 18.4)
t2 = 1.13 s
With this time we calculate the height at this point
Y = Voy t - ½ g t²
Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²
Y = 6.23 m
With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch
A rotating disc supplied with constant power where the relationship of the angular velocity of the disc and the number of rotations made by the disc is governed by Newton's second law for rotation. This law is specially made for rotating bodies which is extracted from Newton's second law of motion.
Example of surface events are erosion and weathering. Erosion is the carrying of a particle from one place to the other and weathering is the breaking down of particles. These processes help in rock formation because this allows physical changes (grouping together or breaking down) on a certain substance. Subsurface events are those which happened underground such as the flow of underground water which subsequently allow the deposition of minerals, etc.
