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timama [110]
3 years ago
6

Giving brainliest to whoever can solve!! Light of wavelength 675 nm passes through a double slit , and makes its first minimum (

m = 1) at an angle of 0.111 deg. What is the separation of the slits , d , in MILLIMETERS ?

Physics
1 answer:
Aloiza [94]3 years ago
5 0

Answer:

0.174

Explanation:

Using the equation with both the minimum and where you're solving for d, (1-1/2) 675/sin(0.111), you get 174210.2 so you convert that into millimeters by moving the decimal point up 6 (the difference of nm 10^-9 and millimeters 10^-3),leaving you with 0.174.

I hope this helped!

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A balloon contains 0.075 m^3 of
inessss [21]

Answer:

600 KPa.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V1) = 0.075 m³

Final volume (V2) = 0.45 m³

Final pressure (P2) = 100 KPa

Initial pressure (P1) =?

Temperature = constant

The initial pressure can be obtained by using the Boyle's law equation as shown below:

P1V1 = P2V2

P1 × 0.075 = 100 × 0.45

P1 × 0.075 = 45

Divide both side by 0.075

P1 = 45 / 0.075

P1 = 600 KPa.

Thus, the initial pressure in the balloon is 600 KPa.

7 0
2 years ago
What is the minimal mass of helium (density 0.18 kg/m3) needed to lift a balloon carrying two people in a basket, if the total m
Sergio039 [100]

Answer:   
 M[min] = M[basket+people+ balloon, not gas] * ΔR/R[b] 
 Î”R is the difference in density between the gas inside and surrounding the balloon. 
 R[b] is the density of gas inside the baloon.   
 ==================================== 
 Let V be the volume of helium required. 
 Upthrust on helium = Weight of the volume of air displaced = Density of air * g * Volume of helium = 1.225 * g * V   
 U = 1.225gV newtons 
 ---- 
 Weight of Helium = Volume of Helium * Density of Helium * g 
 W[h] = 0.18gV N   
 Net Upward force produced by helium, F = Upthrust - Weight = (1.225-0.18) gV = 1.045gV N  -----

 
 Weight of 260kg = 2549.7 N 
 Then to lift the whole thing, F > 2549.7 
 So minimal F would be 2549.7 
 ---- 
 1.045gV = 2549.7 
 V = 248.8 m^3   
 Mass of helium required = V * Density of Helium = 248.8 * 0.18 = 44.8kg (3sf)   
 =====   
 Let the density of the surroundings be R 
 Then U-W = (1-0.9)RgV = 0.1RgV   
 So 0.1RgV = 2549.7 N 
 V = 2549.7 / 0.1Rg   
 Assuming that R is again 1.255, V = 2071.7 m^3 
 Then mass of hot air required = 230.2 * 0.9R = 2340 kg   
 Notice from this that M = 2549.7/0.9Rg * 0.1R so   
 M[min] = Weight of basket * (difference in density between balloon's gas and surroundings / density of gas in balloon)   
 M[min] = M[basket] * ΔR/R[b]
3 0
3 years ago
I NEED HELP ASAP WHAT IS THE DISTANCE MOVED IN 15 SECONDS
Oduvanchick [21]

Answer:

5 I think

Explanation:

8 0
3 years ago
Read 2 more answers
The driver of a 1750 kg car traveling on a horizontal road at 110 km/h suddenly applies the brakes. Due to a slippery pavement,
Alexeev081 [22]

Answer: a=-2.4525 m/s^2

d=s=190.3 m

Explanation:The only force that is stopping the car and causing deceleration is the frictional force Fr

Fr = 25% of weight

W=mg

W=1750*9.81

W=17167.5

Hence

Fr=\frac{25}{100} * -17167.5\\\\Fr=-4291.875 N

Frictional force is negative as it acts in opposite direction

According to newton second law of motion

F=ma

hence

a=Fr/m

a=-4291.875/1750\\a=-2.4525

given

u= 110 km/h

u=110*1000/3600

u=30.55 m/s

to get t we know that final velocity v=0

v^2=u^2+2as\\0=30.55^2-2*2.4525*s\\s=190.34m

3 0
3 years ago
A crane lifts an air conditioner to the top of a building. If the building is 12 m high, and the air conditioner has a mass of 2
andrey2020 [161]

Work needed = 23,520 J

<h3>Further explanation</h3>

Given

height = 12 m

mass = 200 kg

Required

work needed by the crane

Solution

Work is the transfer of energy caused by the force acting on a moving object  

Work is the product of force with the displacement of objects.  

Can be formulated  

W = F x d  

W = Work, J, Nm  

F = Force, N  

d = distance, m  

F = m x g

Input the value :

W = mgd

W = 200 kg x 9.8 m/s²x12 m

W = 23520 J

5 0
3 years ago
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