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kondor19780726 [428]
2 years ago
11

In a double-slit experiment, it is observed that the distance between adjacent maxima on a remote screen is 1.0 cm. what happens

to the distance between adjacent maxima when the slit separation is cut in half?
Physics
1 answer:
Studentka2010 [4]2 years ago
3 0
As the displacement of the maxima is calculated with the equation
(displacement)y=(m<span>λD)/slit separation (d)
halving the split separation doubles the displacement</span>
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Becky has a shower and uses 20,000g of water with a specific heat capacity of 4200J/KG. When the water is supplied with 336,000J
lara [203]

Answer:

46°C

Explanation:

q = mCΔT

336,000 J = (20 kg) (4200 J/kg/°C) (50°C − T)

T = 46°C

7 0
3 years ago
A bicycle with an initial velocity of +6 m/s accelerates at a rate of +2 m/s2 for 3 seconds. What distance does the bicycle trav
irga5000 [103]

Answer:

27 m

Explanation:

Given:

v₀ = 6 m/s

a = 2 m/s²

t = 3 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (6 m/s) (3 s) + ½ (2 m/s²) (3 s)²

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3 0
3 years ago
What is the gravitational force of attraction between a planet and a 17-kilogram mass that is falling freely toward the surface
PolarNik [594]

Answer:

a. 150 N

Explanation:

Gravitational Force: This is the force that act on a body under gravity.

The gravitational force always attract every object on or near the earth's surface. The earth therefore, exerts an attractive force on every object on or near it.

The S.I unit of gravitational force is Newton(N).

Mathematically, gravitational force of attraction is expressed as

(i) F = GmM/r² ........................ Equation 1 ( when it involves two object of different masses on the earth)

(ii) F = mg ............................... Equation 2 ( when it involves one mass and the gravitational field).

Given: m = 17 kg, g = 8.8 m/s²

Substituting into equation 2,

F = 17(8.8)

F = 149.6 N

F ≈ 150 N.

Thus the gravitational force = 150 N

The correct option is a. 150 N

5 0
2 years ago
The cannon on a battleship can fire a shell a maximum distance of 26.0 km.
Alla [95]
It is possible to demonstrate that the maximum distance occurs when the angle at which the projectile is fired is \theta = 45^{\circ}.
In fact, the laws of motions on both x- and y- directions are
S_x(t)= v_0 cos \theta t
S_y(t)= v_0 \sin \theta t -  \frac{1}{2} gt^2
From the second equation, we get the time t at which the projectile hits the ground, by requiring S_y(t)=0, and we get:
t= \frac{2 v_0 \sin \theta}{g}
And inserting this value into Sx(t), we find
S_x(t) = 2  \frac{v_0^2}{g}  \sin \theta \cos \theta= \frac{v_0^2}{g} \sin (2\theta)
And this value is maximum when \theta=45^{\circ}, so this is the angle at which the projectile reaches its maximum distance.

So now we can take again the law of motion on the x-axis
S_x(t)=  \frac{v_0^2}{g} \sin (2\theta)
And by using S_x = 26 km=26000 m, we find the value of the initial velocity v0:
v_0 =  \sqrt{ \frac{S_x g}{\sin (2\theta)} } = \sqrt{ \frac{(26000m)(9.81m/s^2)}{\sin (2\cdot 45^{\circ})} } =505 m/s
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sergey [27]
None of the above!!!!!!!!!??????????$
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