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3 years ago

What is superstring theory?

1 answer:

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.

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Suppose we have two planets with the same mass, but the radius of the second one is twice the size of the first one. How does th

bogdanovich [222]

The free-fall acceleration on the second planet is one-fourth the value of the first planet.

Calculation:

Consider the mass of planet A to be, M

the mass of planet B to be, Mₓ = M

the radius of planet A to be, R₁

the radius of planet B to be, R₂

The acceleration due to gravity on planet A's surface is given as:

g = GM/R₁² - (1)

Similarly, the acceleration due to gravity on planet B's surface is given as:

g' = GM/R₂² [where, R₂ = 2R₁]

= GM/4R₁² -(2)

From equation 1 & 2, we get:

g/g' = GM/R₁² ÷ GM/4R₁²

g/g' = 4/1

Thus we get,

g' = 1/4 g

Therefore, the free-fall acceleration on the second planet is one-fourth the value of the first planet.

Learn more about free-fall here:

<u>brainly.com/question/13299152</u>

#SPJ4

6 0

2 years ago

A canon is tilled back 30.0 degrees and shoots a cannon ball at 155 m/s. What is the

Umnica [9.8K]

Answer:

$= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m$

Therefore, highest point that the cannon ball reaches is 168.7m

Explanation:

the cannon is fired at an angle 30 o to the horizonatal with a speed of 155 m/s

highest point that the cannon ball reaches?

$H_{max}=\frac{V^2\sin ^2 \theta}{2g}$

g = 9.8m/s2

$= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m$

Therefore, highest point that the cannon ball reaches is 168.7m

6 0

4 years ago

HURRY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Nata [24]

When the child is moving fastest

6 0

4 years ago

Read 2 more answers

A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di

Lyrx [107]

Answer:

Explanation:

From the information given:

The angular speed for the block $\omega = 50 \ rad/s$

Disk radius (r) = 0.2 m

The block Initial velocity is:

$v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s$

Change in the block's angular speed is:

$\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s$

However, on the disk, moment of inertIa is:

$I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2$

The time t = 10s

∴

Frictional torques by the wall on the disk is:

$T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m$

Finally, the frictional force is calculated as:

$F = \dfrac{T}r{}$

$F= \dfrac{0.6}{0.2} \\ \\ F = 3N$

8 0

3 years ago

A sprinter who is running a 200-m race travels the second 100 m in much less time than the first 100m because

Harrizon [31]

Answer:

He is warmed up now

Explanation:

His muscles are better and stretched now

3 0

3 years ago

Read 2 more answers