4. The fastest runner in a track and field event is the one with the<br>

Estimate the distance (in cm) between the central bright region and the third dark fringe on a screen 5.00 m from two double sli

Artist 52 [7]

Answer:

y = 1.75 cm

Explanation:

In the double-slit experiment the equation for destructive interference is

d sin tea = (m + ½)

λ

let's use trigonometry to find the angle

tan θ = y / L

as all the experiment does not occur at small angles

tan θ = sin θ / cos θ = sin θ = y / L

we substitute

y = (m + 1/2 ) λ L / d

we calculate

y = (3 + ½) 500 10⁻⁹ 5.00 / 0.5 10⁻³

y = 1.75 10⁻² m

y = 1.75 cm

5 0

3 years ago

Please help on this one?

SSSSS [86.1K]

I think it would be near the sun due.to high rays of sun

8 0

3 years ago

Read 2 more answers

If the magnitude of a positive charge is tripled, what is the ratio of the original value of the electric field at a point to th

ipn [44]

Answer:

b)1 :3

Explanation:

Lets that

The value of a positive charge = q

As we know that electric filed on a point charge given as

$E=\dfrac{Kq}{r^2}$

Where ,K=Constant

q=Charge ,r=Distance

If the value of the charge gets tripled ,q'= 3 q

Then electric filed E'

$E'=\dfrac{Kq'}{r^2}$

$E'=\dfrac{3Kq}{r^2}$

E' = 3 E

Therefore we can say that

$\dfrac{E}{E'}==\dfrac{1}{3}$

therefore the answer will be --

b)1 :3

3 0

3 years ago

A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a

katen-ka-za [31]

Answer:b

Explanation:

Given

mass of first cart $m_1=6 kg$

mass of second cart $m_2=3 kg$

velocity of first cart $v_1=3 m/s$

conserving momentum

$m_1v_1+m_2v_2=(m_1+m_2)v$

$6\times 3+3\times 0=(9)\cdot v$

$v=2 m/s$

Initial kinetic Energy $K.E._1=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$

$K.E._1=\frac{1}{2}\cdot 6\cdot 3^2+0$

$K.E._1=27 J$

Final Kinetic Energy

$K.E._2=\frac{1}{2}(m_1+m_2)v^2$

$K.E._2=\frac{1}{2}(6+3)\cdot 2^2=18 J$

Ratio of initial Kinetic Energy to the Final Kinetic Energy

$=\frac{27}{18}=1.5$

6 0

4 years ago

a farmer lifts a bale of hay that is 12m off the ground . If he uses a force of 350 n, how much work does the farmer do?

hjlf

Answer:

4200 Joules

Explanation:

Work done =force x distance

From the question , we’re given f =350N ,

d = 12m

Using the above formula, we have

Workdone = 350 x 12

= 4200 Joules

5 0

3 years ago

4. The fastest runner in a track and field event is the one with the​

4. The fastest runner in a track and field event is the one with the