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dmitriy555 [2]
4 years ago
13

What is the symbol for angle?

Physics
1 answer:
zlopas [31]4 years ago
4 0

Answer:

angle is denoted by /_ , see in explanation

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In a Young's double-slit experiment the separation distance y between the second-order bright fringe and the central bright frin
Natasha2012 [34]

Answer:

y = 0.0233 m

Explanation:

In a Young's Double Slit Experiment the distance between two consecutive bright fringes is given by the formula:

Δx = λL/d

where,

Δx = distance between fringes

λ = wavelength of light

L = Distance between screen and slits

d = Slit Separation

Now, for initial case:

λ = 425 nm = 4.25  x 10⁻⁷ m

y = 2Δx = 0.0177 m => Δx = 8.85 x 10⁻³ m

Therefore,

8.85 x 10⁻³ m = (4.25 x 10⁻⁷ m)L/d

L/d = (8.85 x 10⁻³ m)/(4.25 x 10⁻⁷ m)

L/d = 2.08 x 10⁴

using this for λ = 560 nm = 5.6 x 10⁻⁷ m:

Δx = (5.6 x 10⁻⁷ m)(2.08 x 10⁴)

Δx = 0.0116 m

and,

y = 2Δx

y = (2)(0.0116 m)

<u>y = 0.0233 m</u>

3 0
3 years ago
An increase in the magnitude of velocity is??
bazaltina [42]
It means the speed of the object is increasing
and
there is a positive acceleration in the direction of the velocity
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8 0
3 years ago
Please need help fast
iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) 1.5 m/s^2

The average acceleration of the horse can be calculated as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

a=\frac{15-0}{10}=1.5 m/s^2

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance travelled

u is the initial velocity

t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

a=1.5 m/s^2

Substituting,

s=0+\frac{1}{2}(1.5)(10)^2=75 m

(d) See attached graphs

In a uniformly accelerated motion:

- The distance travelled (x) follows the equation mentioned in part c,

x=ut+\frac{1}{2}at^2

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

a=1.5 m/s^2 (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

6 0
3 years ago
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