Answer: magnitude of resultant = 2.811km, direction of resultant = 209.6°
Explanation: The vector ( in this case displacement) that lies on the y axis is 3.25km north and 1.50km due south.
Their resultant is gotten below
Magnitude of resultant = 3.25 - 1.50 = 1.75km
Direction of resultant = north (direction of the bigger vector)
2.20km is the only vector acting on the x axis due east, combining this vector with the resultant of the vectors above, we realize that 2.2km west is perpendicular to 1.75km due north. Since 1.75km us due north, it implies that it is the vector on the positive y axis (vy) and 2.20km due west implies that it is the vector on the negative x axis(vx), thus their resultant is gotten using phythagoras theorem.
R = √ vx² + vy²
R = √ 2.20² + 1.75²
R = √ 7.9025
R= 2.1811km.
θ = tan^-1 (vy/vx)
θ = tan ^-1 (1.25/2.20)
θ = tan ^-1 (0.5618)
θ = 29.6°.
The direction of the vector is south west which implies the third quadrant of the trigonometric quadrant which implies 180 + θ
Thus the direction of the vector is 180 + 29.60 = 209.6°
From the picture attached to this answer we can see roughly that the magnitude of resultant is longer than it component and the vector is placed on the 3rd quadrant which verifies our quantitative claim.
A2 = 16 m^2
Explanation:
Application of Pascal's law:
F1/A1 = F2/A2
Given:
F1 = 50 N. A1 = 1 m^2
F2 = 800 N A2 = ?
A2 = (F2/F1)A1 = (800 N/50 N)(1 m^2)
= 16 m^2
Answer:
According to Newton's first law of motion, an object will experience a change in motion if the forces are unbalanced.
Answer:
F = 5226.6 N
Explanation:
To solve a lever, the rotational equilibrium relation must be used.
We place the reference system on the fulcrum (pivot point) and assume that the positive direction is counterclockwise
F d₁ = W d₂
where F is the applied force, W is the weight to be lifted, d₁ and d₂ are the distances from the fulcrum.
In this case the length of the lever is L = 5m, t the distance desired by the fulcrum from the weight to be lifted is
d₂ = 200 cm = 2 m
therefore the distance to the applied force is
d₁ = L -d₂
d₁ = 5 -2
d₁= 3m
we clear from the equation
F = W d₂ / d₁
W = m g
F = m g d₂ / d₁
we calculate
F = 800 9.8 2/3
F = 5226.6 N
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