Answer:
Explanation:
As we know that train is initially moving with the speed
now we know that
now the final speed of the train when it crossed the crossing
now we can use kinematics here
Now the time to cross that junction is given as
Answer:
33.2 m
Explanation:
For the first object:
y₀ = 81.5 m
v₀ = 0 m/s
a = -9.8 m/s²
t₀ = 0 s
y = y₀ + v₀ t + ½ at²
y = 81.5 − 4.9t²
For the second object:
y₀ = 0 m
v₀ = 40.0 m/s
a = -9.8 m/s²
t₀ = 2.20 s
y = y₀ + v₀ t + ½ at²
y = 40(t−2.2) − 4.9(t−2.2)²
When they meet:
81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²
81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)
81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716
81.5 = 61.56t − 111.716
193.216 = 61.56t
t = 3.139
The position at that time is:
y = 81.5 − 4.9(3.139)²
y = 33.2
Answer: 4.7m/s²
Explanation:
According to newton's first law,
Force = mass × acceleration
Since we are given more the one force, we will take the resultant of the two vectors.
Mass = 2.0kg
F1+F2 = (3i-8j)+(5i+3j)
Adding component wise, we have;
F1+F2 = 3i+5i-8j+3j
F1+F2 = 8i-5j
Resultant of the sum of the forces will be;
R² = (8i)²+(-5j)²
Since i.i = j.j = 1
R² = 8²+5²
R² = 64+25
R² = 89
R = √89
R = 9.4N
Our resultant force = 9.4N
Substituting in the formula
F = ma
9.4 = 2a
a = 9.4/2
a = 4.7m/s²
Therefore, magnitude of the acceleration of the particle is 4.7m/s²
Answer:
Coefficient of friction.
Explanation:
The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :
N is normal force.
= coefficient of friction
