Take east to be the positive direction. Then the resultant force from adding <em>F</em>₁ and <em>F</em>₂ is
<em>F</em>₁ + <em>F</em>₂ = (-45 N) + 63 N = 18 N
which is positive, so it's directed east.
To this we add a third force <em>F</em>₃ such that the resultant is 12 N pointing west, making it negative, so that
18 N + <em>F</em>₃ = -12 N
<em>F</em>₃ = -30 N
So <em>F</em>₃ has a magnitude of 30 N and points west.
Answer:
the distance traveled by the car is 42.98 m.
Explanation:
Given;
mass of the car, m = 2500 kg
initial velocity of the car, u = 20 m/s
the braking force applied to the car, f = 5620 N
time of motion of the car, t = 2.5 s
The decelaration of the car is calculated as follows;
-F = ma
a = -F/m
a = -5620 / 2500
a = -2.248 m/s²
The distance traveled by the car is calculated as follows;
s = ut + ¹/₂at²
s = (20 x 2.5) + 0.5(-2.248)(2.5²)
s = 50 - 7.025
s = 42.98 m
Therefore, the distance traveled by the car is 42.98 m.
Answer:
1. True WA > WB > WC
Explanation:
In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.
A) Work is the product of force by distance and the cosine of the angle between them
WA = W h cos 0
WA = mg h
B) On a ramp without rubbing
Sin30 = h / L
L = h / sin 30
WB = F d cos θ
WB = F L cos 30
WB = mf (h / sin30) cos 30
WB = mg h ctan 30
C) Ramp with rubbing
W sin 30 - fr = ma
N- Wcos30 = 0
W sin 30 - μ W cos 30 = ma
F = W (sin30 - μ cos30)
WC = mg (sin30 - μ cos30) h / sin30
Wc = mg (1 - μ ctan30) h
When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum
Let's review the claims
1. True The work of gravity is the greatest and the work where there is friction is the least
2 False. The job where there is friction is the least
3 False work with rubbing is the least
4 False work with rubbing is the least
Explanation :
Distance is total path travelled by an object during its entire journey. It is a scalar quantity i.e only magnitude.
Displacement is the shortest distance covered by an object. It is basically the change in position of object. It is a vector quantity i.e direction as well as magnitude.
When an object is travelling in a straight line and stops at the end point, then both distance and displacement are same.
When an object is travelling in a straight line and then changes its direction or we can say come backwards then the magnitude of distance and displacement are different.
