Answer:
4960 N
Explanation:
First, find the acceleration.
Given:
v₀ = 6.33 m/s
v = 2.38 m/s
Δx = 4.20 m
Find: a
v² = v₀² + 2aΔx
(2.38 m/s)² = (6.33 m/s)² + 2a (4.20 m)
a = -4.10 m/s²
Next, find the force.
F = ma
F = (1210 kg) (-4.10 m/s²)
F = -4960 N
The magnitude of the force is 4960 N.
Answer:
option (b)
Explanation:
Let the resistance of each resistor is R.
In series combination,
The effective resistance is Rs.
rs = r + R + R + .... + n times = NR
Let V be the source of potential difference.
Power in series
Ps = v^2 / Rs = V^2 / NR ..... (1)
In parallel combination
the effective resistance is Rp
1 / Rp = 1 / R + 1 / R + .... + N times
1 / Rp = N / R
Rp = R / N
Power is parallel
Rp = v^2 / Rp = N V^2 / R ..... (2)
Divide equation (1) by equation (2) we get
Ps / Pp = 1 / N^2
"6.5 km/hr" is not a velocity. It's just a speed, so
we don't know what direction he's walking.
If he happens to be walking north, then it takes him
(12 km) / (6.5 km/hr) = 1.846... hours (rounded) .
If he's walking in any other direction, it takes him longer than that.
If the angle between north and the direction he's walking is
90 degrees or more, then he can never cover any northward
distance, no matter how long he walks.
Answer:
56.7°
Explanation:
Imagine a rectangle triangle.
The triangle has 3 sides.
One side is the height of the tower, let's name it A.
Another side is the distance from the base of the tower to the point where the waire touches the ground. Let's name that B.
Sides A and B are perpendicular.
The other side is the length of the wire. Let's name it C.
From trigonometry we know that:
cos(a) = B / C
Where a is the angle between B anc C, between the wire and the ground.
Therefore
a = arccos(B/C)
a = arccos(552/1005) = 56.7°
Answer:
a) α = 0.338 rad / s² b) θ = 21.9 rev
Explanation:
a) To solve this exercise we will use Newton's second law for rotational movement, that is, torque
τ = I α
fr r = I α
Now we write the translational Newton equation in the radial direction
N- F = 0
N = F
The friction force equation is
fr = μ N
fr = μ F
The moment of inertia of a saying is
I = ½ m r²
Let's replace in the torque equation
(μ F) r = (½ m r²) α
α = 2 μ F / (m r)
α = 2 0.2 24 / (86 0.33)
α = 0.338 rad / s²
b) let's use the relationship of rotational kinematics
w² = w₀² - 2 α θ
0 = w₀² - 2 α θ
θ = w₀² / 2 α
Let's reduce the angular velocity
w₀ = 92 rpm (2π rad / 1 rev) (1 min / 60s) = 9.634 rad / s
θ = 9.634 2 / (2 0.338)
θ = 137.3 rad
Let's reduce radians to revolutions
θ = 137.3 rad (1 rev / 2π rad)
θ = 21.9 rev
