A block lies on a frictionless floor. a force of 5 n pulls toward the east while a force of 4 n pulls toward the north. what is
the magnitude of the net force on the block?
1 answer:
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Explanation:
It is given that,
Length of the rod, l = 14 cm = 0.14 m
Charge on the rod,
We need to find the magnitude and direction of the net electric field produced by the charged rod at a point 36.0 cm to the right of its center along the axis of the rod, z = 36 cm = 0.36 m
Electric field at the axis of the rod is given by :
Where
is the linear charge density of the rod,
E = -7493170.57 N/C
or
Negative sign shows that the electric field is acting in inwards direction. Hence, this is the required solution.
Answer:
a) t = 19.6 s, b) fr = 1.274 10⁴ N
Explanation:
This is a Newton's second law problem
Y Axis
for the cabin
N₁-W₁ = 0
N₁ = W₁
for the trailer
N₂- W₂ = 0
N₂ = W₂
X axis
for the cabin plus trailer, where friction is only in the cabin
fr = (m₁ + m₂) a
the friction force equation is
fr = μ N
we substitute
μ N₁ = (m₁ + m₂) a
μ m₁ g = (m₁ + m₂) a
a = μ g
let's calculate
a = 0.65 9.8
a = 1,274 m / s²
a) to find the stopping distance we can use kinematics
Let's slow down the sI system
v₀ = 90 km / h (1000 m / 1km) (1h / 3600s) = 25 m / s
v = v₀ - a t
when it is stopped its speed is zero
0 = v₀ - at
t = v₀ / a
t = 25 / 1.274
t = 19.6 s
b) the friction force is
fr = 0.65 2000 9.8
fr = 1.274 10⁴ N
This is the braking force and also the forces that couple the cars.