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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.
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A ship sets sail from Rotterdam, The Netherlands, heading due north at 8.00 m/s relative to the water. The local ocean current i
Nuetrik [128]

Answer:

The velocity of the ship relative to the earth V = 9.05 \frac{m}{s}

Explanation:

The local ocean current is  = 1.52 m/s

Direction \theta = 40°

Velocity component in X - direction V_{x} = 1.52 \cos 40°

V_{x} = 1.164 \frac{m}{s}

Velocity component in Y - direction V_{y} = 8 + 1.52 \sin 40°

V_{y} = 8.97 \frac{m}{s}

The velocity of the ship relative to the earth

V = \sqrt{V_{x}^{2} + V_{y} ^{2}  }

Put the values of V_{x} and V_{y} we get,

⇒ V = \sqrt{1.164^{2} + 8.97 ^{2}  }

⇒ V = 9.05 \frac{m}{s}

This is the velocity of the ship relative to the earth.

7 0
3 years ago
Name the fundamental units involved in pascal<br>​
gtnhenbr [62]

Explanation:

Hey there!

Here,

Pascal is a unit of pressure.

pressure =  \frac{f}{a}

Now, As per the formula the units are:

kg, m and s^2.

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

5 0
3 years ago
The kinetic energy of a body of mass 15 kg is 30 joule. What is its momentum?
lys-0071 [83]

This problem is a piece o' cake, IF you know the formulas for both kinetic energy and momentum.  So here they are:

Kinetic energy = (1/2) · (mass) · (speed²)

Momentum = (mass) · (speed)

So, now ... We know that

==> mass = 15 kg,  and

==> kinetic energy = 30 Joules

Take those pieces of info and pluggum into the formula for kinetic energy:

Kinetic energy = (1/2) · (mass) · (speed²)

30 Joules = (1/2) · (15 kg) · (speed²)

60 Joules = (15 kg) · (speed²)

4 m²/s² = speed²

Speed = 2 m/s

THAT's all you need !  Now you can find momentum:

Momentum = (mass) · (speed)

Momentum = (15 kg) · (2 m/s)

<em>Momentum = 30 kg·m/s</em>

<em>(Notice that in this problem, although their units are different, the magnitude of the KE is equal to the magnitude of the momentum.  When I saw this, I wondered whether that's always true.  So I did a little more work, and I found out that it isn't ... it's a coincidence that's true for this problem and some others, but it's usually not true.)</em>

8 0
3 years ago
Applying newton's version of kepler's third law (or the orbital velocity law) to the a star orbiting 40,000 light-years from the
Ugo [173]

Applying Newtons version of Kepler's third law or the orbital velocity law to the star orbiting 40000 light years from the center of the Milky Way Galaxy allows us to determine the mass of the Milky Way Galaxy that lies within 40000 light years in the galactic center.

<h3></h3><h3>What is orbital velocity law?</h3>

The orbital velocity law states that, the orbital velocity is directly proportional to the mass of the body for which it is being calculated and inversely proportional to the radius of the body. Earths orbital velocity near its surface is around 8km/sec if the air resistance is disregarded.

In space exploration, orbital velocity is a crucial topic. Space authorities heavily rely on it to comprehend how to launch satellites. It aids scientists in figuring out the velocities at which satellites must orbit a planet or other celestial body to prevent collapsing into it. The speed at which one body orbits the other body is known as the orbital velocity. The term "orbit" refers to an object's consistent circular motion around the Earth. The distance between the object and the earth's centre determines the orbit's velocity.

To know more about orbital velocity law, refer brainly.com/question/11353717

#SPJ4

5 0
2 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
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