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PIT_PIT [208]
3 years ago
6

What is superstring theory?

Physics
1 answer:
Stels [109]3 years ago
4 0
Superstring theory is an attempt to explain all of the particles and fundamental forces of nature in one theory by modeling them as vibrations of tiny supersymmetric strings.
You might be interested in
Why current remains same in series combination of resistors in all resistors and p.d. remains different?
IgorC [24]
Current at all points of a series circuit must be the same, because there's no place in the circuit where electrons are being manufactured, and no place where they're leaking out and falling on the floor. The nimber of electrons that leaves the loop is the same number that entered it. I'm not sure what is nmeant by "p.d. remains different" .
4 0
3 years ago
A cricketer throws a ball sideways with an initial velocity of 30 m/s. She releases the ball from a height of 1.3m. Calculate ho
ioda

Answer:

78.34

Explanation:

1.3/30=78.3m

8 0
2 years ago
I have to write something here, so like hello and please help​
ExtremeBDS [4]

Answer:

the extension would be less the new extension might be 3 cm

Explanation:

5 0
3 years ago
A ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally aw
pshichka [43]

Answer:

44.3 m/s

Explanation:

Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.

What is the magnitude of its velocity just before it strikes the ground ?

The parameters given are:

Height H = 100m

Since the ball is thrown from a top of a building, initial velocity U = 0

Let g = 9.8m/s^2

Using third equation of motion

V^2 = U^2 + 2gH

Substitute all the parameters into the formula

V^2 = 2 × 9.8 × 100

V^2 = 200 × 9.8

V^2 = 1960

V = 44.27 m/s

Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately

6 0
3 years ago
*8–52. Beam AB has a negligible mass and thickness, and supports the 200-kg uniform block. It is pinned at A and rests on the to
denis23 [38]

Answer:

μ₁ = 0.1048

μ₂ = 0.1375

Explanation:

Using  static equation can find in both point the moment and the forces so:

∑ M = F *d  , ∑ F = 0

∑ M A = 0

N₁ * 3 - 200 * 9.81 * 1.5 = 0

N₁ = 981  

∑ M y = 0

N₂ + 300 * ³/₅ - 981 - 20 * 9.81 = 0

N₂ = 997.2 N

∑ M C = 0

F₁ * 1.75  - 300 * ⁴/₅  * 0.75 = 0

F₁ = 102.86

∑ M B = 0

300 * ⁴/₅ * 1 - F₂ * 1.75 = 0

F₂ = 137.14 N

The Force F1 and F2 related the coefficients of static friction

F₁ = μ₁ * N₁   ⇒  102.86 N = μ₁ * 981 ⇒ μ₁ = 0.1048

F₂= μ₂ * N₂  ⇒  137.14 N = μ₂ * 997  ⇒ μ₂ = 0.1375

8 0
3 years ago
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