The north and west components of this displacement are 9.19 km and -7.71 km respectively.
<h3>
North component of the displacement</h3>
The north component of the displacement is calculated as follows;
Dy = D sinθ
Dy = 12 km x sin50
Dy = 9.19 km
<h3>West component of the displacement</h3>
Dx = D x cosθ
Dx = - 12 km x cos(50)
Dx = -7.71 km
Learn more about components of displacement here: brainly.com/question/13416288
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Answer:
a) v_{p} = 2.83 m / s
, b) 50.5º north east
Explanation:
This is a vector problem.

The speed of the ball with respect to the ground is the speed of the ball with respect to Mia plus the speed of Mia with respect to the ground
To make the sum we decompose the speed of the ball in its components
The angle of 30 east of the south, measured from the positive side of the x axis is
θ = 30 + 270 = 300
=
cos 300
= v_{b} sin. 300
v_{bx} = 3.60 cos 300 = 1.8 m / s
v_{by} = 3.60 sin 300 = -3,118 m / s
Let's add speeds on each axis
X axis
vₓ = v_{bx}
vₓ = 1.8 m / s
Y Axis
= v1 - vpy
v_{y} = 5.30 - 3.118
v_{y} = 2.182 m / s
The magnitude of the velocity can be found using the Pythagorean theorem
= √ (vₓ² + v_{y}²)
v_{p} = √ (1.8² + 2.182²)
v_{p} = 2,829 m / s
v_{p} = 2.83 m / s
b) for direction use trigonometry
tan θ =
/ vₓ
θ = tan ⁺¹ v_{y} / vₓ
θ = tan⁻¹ 2.182 / 1.8
Tea = 50.48º
This address is 50.5º north east
<span>So we want to know what is the direct effect of the gas particles if the force of their collision with the walls of the container is increased. Pressure, microscopically, is defined as the number of collisions per unit area. If we increase the force of the collision, the pressure increases so that is the direct effect. </span>
Answer:
110 mL
Explanation:
Ideal gas law:
PV = nRT
Assuming the container isn't rigid, and the pressure is constant, then:
V/T = V/T
Plug in values (remember to use absolute temperature).
V / 293 K = 150 mL / 393 K
V = 110 ml
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Answer:
The formula of energy is m
. So, energy is M
. Since velocity = displacement/time.
Explanation:
If you have any questions feel free to ask in the comments - Mark