1) The net force is 16 N to the right
2) The net force is 98 N to the left
3) The net force is 0.5 N downward
4) The net force is 170 N to the right
5) The net force is 175 N to the right
Explanation:
1)
To find the net force, we have to analyze all the forces acting on the box.
We have:
Therefore, the horizontal net force is
(to the right)
While the vertical force is
So the net force is 16 N to the right.
2)
In this case, we have the following forces:
Therefore, the horizontal net force is
(to the left)
While the vertical force is
(downward)
And so, the net force is 98 N to the left.
3)
The force acting on the squirrel in this problem are:
Both forces act vertically and there are no other forces acting in other directions, therefore the net force on the squirrel is simply equal to the net force on the vertical direction, which is:
And since the weight is larger than the air resistance, the direction of the net force is downward.
4)
The forces acting on Monkey are:
So, the net force in the horizontal direction is
(to the right)
While the net force in the vertical direction is
And therefore the net force is 170 N to the right
5)
The forces acting on Deer are:
So the net horizontal force is
to the right
While the net vertical force is
So the net force is 175 N to the right.
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Answer:
Kinetic energy, E = 133.38 Joules
Explanation:
It is given that,
Mass of the model airplane, m = 3 kg
Velocity component, v₁ = 5 m/s (due east)
Velocity component, v₂ = 8 m/s (due north)
Let v is the resultant of velocity. It is given by :
Let E is the kinetic energy of the plane. It is given by :
E = 133.38 Joules
So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.
Answer:
Yes, it will produce a diffraction pattern.
a. 3.9 mm b. 1.95 mm
Explanation:
The light shined from a single slit will produce a diffraction pattern because, the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.
Given
wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m
width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m
Distance of slit from central maximum , D = 1.65 m
Distance of first minimum from central maximum, y = ?
a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...
The relationship between y and D is given by
Since θ is small, sinθ ≈ θ ≈ tanθ
so, dθ = mλ ⇒ θ = mλ/d = y/D
Therefore, y = mλD/d
Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ = λD/d and y₋₁ = -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d
Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ = 3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm
b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ = λD/d and y₂ = 2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have
λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm