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Sidana [21]
2 years ago
10

How can gravity be simulated in an orbiting space station?.

Physics
1 answer:
Alex777 [14]2 years ago
6 0

Answer:

A spinning space station will have centrifugal force acting on the inhabitants which if adjusted right can simulate the force of gravity on Earth

Explanation:

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Light of wavelength 630 nm falls on two slits and produces an interference pattern in which the third-order bright red fringe is
goblinko [34]

Answer:

d \frac{x}{l} = m×  λ⇒ d = λ ×m×l / x

= 630×10^{-9} m × 3×3m/ 45×10^{-3} m

= 1.26×10^{-4}m

Explanation:

the above calculation is based on Young’s double slit experiment where the two slits provide two coherent light sources which results either constructive interference or destructive interference when passing through a double slit.

6 0
3 years ago
How do you find the average speed of an object?
Aleonysh [2.5K]

Answer:

Average speed is total distance divided by total time.

v = d / t

5 0
3 years ago
Read 2 more answers
Two simple pendulum of slightly different length , are set off oscillating in step is a time of 20s has elasped , during which t
adell [148]

Answer:

Length of longer pendulum = 99.3 cm

Length of shorter pendulum = 82.2 cm

Explanation:

Since the longer pendulum undergoes 10 oscillations in 20 s, its period T = 20 s/10 = 2 s.

From T = 2π√(l/g), the length of the pendulum. l = T²g/4π²

substituting T = 2s and g = 9.8 m/s² we have

l = T²g/4π²

= (2 s)² × 9.8 m/s² ÷ 4π²

= 39.2 m ÷ 4π²

= 0.993 m

= 99.3 cm

Now, for the shorter pendulum to be in step with the longer pendulum, it must have completed some oscillations more than the longer pendulum. Let x be the number of oscillations more in t = 20 s. Let n₁ = number of oscillations of longer pendulum and n₂ = number of oscillations of longer pendulum.

So, n₂ = n₁ + x. Also n₁ = t/T₁ and n₂ = t/T₂ where T₂ = period of shorter pendulum.

t/T₂ = t/T₁ + x

1/T₂ = 1/T₁ + x  (1)

Also, the T₂ = t/n₂ = t/(n₁ + x)  (2)

From (1) T₂ = T₁/(T₁ + x) (3)

equating (2) and (3) we have

t/(n₁ + x) = T₁/(T₁ + x)

substituting t = 20 s and n₁ = 10 and T₁ = 2s, we have

20 s/(10 + x) = 2/(2 + x)

10/(10 + x) = 1/(2 + x)

(10 + x)/10 = (2 + x)

(10 + x) = 10(2 + x)

10 + x = 20 + 10x

collecting like terms

10x - x = 20 - 10

9x = 10

x = 10/9

x = 1.11

x ≅ 1 oscillation

substituting x into (2)

T₂ = t/n₂ = t/(n₁ + x)

= 20/(10 + 1)

= 20/11

= 1.82 s

Since length l = T²g/4π²

substituting T = 1.82 s and g = 9.8 m/s² we have

l = T²g/4π²

= (1.82 s)² × 9.8 m/s² ÷ 4π²

= 32.46 m ÷ 4π²

= 0.822 m

= 82.2 cm

6 0
3 years ago
As more babies are piled into a shopping cart, and the velocity stays the same, what happens to its momentum?
vagabundo [1.1K]
The momentum would increase assuming the velocity stays the same. P=Mv
7 0
2 years ago
A 2 kg blob of putty moving at 3 m/s slams into a 3 kg blob of putty at rest. Which blob has the most momentum prior to the coll
Ray Of Light [21]

initial momentum of 2 kg blob is given as

P_1 = m_1v_1

here we have

m_1 = 2kg

v_1 = 3m/s

P_1 = 3(2) = 6 kg m/s

initial momentum of 3 kg blob is given as

P_2 = m_2v_2

here we have

m_2 = 3kg

v_2 = 0m/s

P_2 = 3(0) = 0 kg m/s

So initially 2 kg Blob has most momentum before they collide

5 0
3 years ago
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