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1 answer:
Answer:
<em>The car took 3 hours to travel 180 miles</em>
Explanation:
<u>Constant Speed Motion</u>
An object travels at constant speed if the ratio of the distance traveled by the time taken is constant.
Expressed in a simple equation, we have:
Where
v = Speed of the object
d = Distance traveled
t = Time taken to travel d.
From the equation above, we can solve for t:
The car travels at v=60 mi/h for a distance of d=180 miles. The time taken is:
t = 3 hours
The car took 3 hours to travel 180 miles
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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
(a) Zero
When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.
This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy
where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:
where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.
(b) 9.8 m/s upward
We can find the velocity of the ball 1 s before reaching its highest point by using the equation:
where
a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward
v = 0 is the final velocity (at the highest point)
u is the initial velocity
t = 1 s is the time interval
Solving for u, we find
and the positive sign means it points upward.
(c) -9.8 m/s
The change in velocity during the 1-s interval is given by
where
v = 0 is the final velocity (at the highest point)
u = 9.8 m/s is the initial velocity
Substituting, we find
(d) 9.8 m/s downward
We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:
where this time we have
a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative
v is the final velocity (1 s after reaching the highest point)
u = 0 is the initial velocity (at the highest point)
t = 1 s is the time interval
Solving for v, we find
and the negative sign means it points downward.
(e) -9.8 m/s
The change in velocity during the 1-s interval is given by
where here we have
v = -9.8 m/s is the final velocity (1 s after reaching the highest point)
u = 0 is the initial velocity (at the highest point)
Substituting, we find
(f) -19.6 m/s
The change in velocity during the overall 2-s interval is given by
where in this case we have:
v = -9.8 m/s is the final velocity (1 s after reaching the highest point)
u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)
Substituting, we find
(g) -9.8 m/s^2
There is always one force acting on the ball during the motion: the force of gravity, which is given by
where
m is the mass of the ball
g = -9.8 m/s^2 is the acceleration due to gravity
According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so
which means that the acceleration is
and the negative sign means it points downward.