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3 years ago

8

What is a string theory ?

2 answers:

Serggg [28]3 years ago

8 0

It is a unified theory than can describe whole world.

Dovator [93]3 years ago

7 0

Answer:

String theory is a set of attempts to model the four known fundamental interactions—gravitation, electromagnetism, strong nuclear force, weak nuclear force—together in one theory. ... Einstein had sought a unified field theory, a single model to explain the fundamental interactions or mechanics of the universe

Explanation:

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A rope is wrapped around a pulley many times. The pulley can be modeled as a solid disk of radius R and mass M, and a mass mA ha

aleksandr82 [10.1K]

The magnitude of the tangential acceleration of the hanging mass is 2mg/MR

<h3>Tangential acceleration of the hanging mass</h3>

The tangential acceleration of the hanging mass around the pulley is determined from the principle of conservation of angular momentum as shown below;

τ = Iα

Where;

I is the moment of inertia
α is the angular velocity

$\alpha = \frac{\tau}{I} \\\\\alpha = \frac{mgR}{3/2MR^2} \\\\\alpha = \frac{2mgR}{3MR^2} \\\\\alpha = \frac{2mg}{3MR}$

Where;

m is the hanging mass
M is the mass of solid disk

The tangential acceleration is calculated as follows;

$a = \alpha R\\\\a = \frac{2mg}{3MR} \times R\\\\a = \frac{2mg}{3M}$

Thus, the magnitude of the tangential acceleration of the hanging mass is 2mg/MR

Learn more about tangential acceleration here: brainly.com/question/11476496

4 0

3 years ago

A force of 5 N accelerates an object. The object’s mass is 50 g. What is the acceleration of the object?

lana [24]

50 g = 0.5 kg. a = 5 / 0.5 = 10 m/s^2.

5 0

3 years ago

What is the name for the area of dust and debris which orbited the young sun and eventually became planets and moons?

alexandr402 [8]

I believe the answer is C. protoplanetary disc. Let me know if this helps

5 0

3 years ago

Read 2 more answers

Please help with physics 2 part question giving 50 points

AveGali [126]

Answer:

2.88×10⁻⁹ s

2.40×10¹⁵ m/s²

Explanation:

Given:

v₀ = 12300 m/s

v = 6.92×10⁶ m/s

Δx = 0.997 cm = 0.00997 m

Part 1) Find: t

Δx = ½ (v + v₀)t

0.00997 m = ½ (6.92×10⁶ m/s + 12300 m/s) t

t = 2.88×10⁻⁹ s

Part 2) Find: a

(6.92×10⁶ m/s)² = (12300 m/s)² + 2a (0.00997 m)

a = 2.40×10¹⁵ m/s²

5 0

4 years ago

Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge th

Tresset [83]

Given Information:

Kinetic energy of proton = KE = 5 MeV

Radius = R = 0530 m

Speed of light = c = 3x10⁸ m/s

Permittivity of free space = ε = 8.854×10⁻¹² F/m

Mass of proton m = 1.67x10⁻²⁷ kg

Required Information:

Energy radiated per second = dE/dt = ?

Answer:

$\frac{dE}{dt} = 7.346x10^{14}$ $eV/s$

Explanation:

The rate at which energy is emitted from an accelerating charge with charge q and acceleration a is given by

$\frac{dE}{dt} =\frac{q^{2}a^{2} }{6\pi{\epsilon}c^{3}}$

Where ε is the permittivity of free space and c is the speed of light

We know that centripetal acceleration is given by

$a = \frac{v^2}{R}$

Whereas kinetic energy is given by

$KE = \frac{1}{2}mv^{2}$

$\frac{2KE}{m} = v^2$

Substitute in the equation of acceleration

$a = \frac{2KE}{mR}$

$a=1.13x10^{34}$ $eV/s^{2}$

Therefore, the corresponding energy radiated per second is

$\frac{dE}{dt} =\frac{(1.60x10^{-19})^{2}(1.13x10^{34})^{2} }{6\pi{\epsilon}c^{3}}$

$\frac{dE}{dt} =\frac{3.27x10^{30} }{6\pi8.854x10^{-12}(3x10^{8}^)^{3}}$

$\frac{dE}{dt} = 7.346x10^{14}$ $eV/s$

4 0

3 years ago