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Naily [24]
3 years ago
8

Matter comprises all of them and among them are independent. what are they

Physics
1 answer:
telo118 [61]3 years ago
8 0

<em>Matter is composed of elementary particles i.e. quarks and leptons.</em>

<em>Matter is composed of elementary particles i.e. quarks and leptons.Matter is composed of elementary particles which is called quarks and leptons. Quarks consist of protons, neutrons and electrons. All observable matter is made up of up quarks, down quarks and electrons.</em>

<em>Matter is composed of elementary particles i.e. quarks and leptons.Matter is composed of elementary particles which is called quarks and leptons. Quarks consist of protons, neutrons and electrons. All observable matter is made up of up quarks, down quarks and electrons.Lepton is an elementary particle consist of half-integer spin that does not undergo strong interactions. Leptons exist on two main classes i.e. charged leptons, and neutral leptons. Electron, electron neutrino, muon, muon neutrino, tau and tau neutrino are the six types of leptons.</em>

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Which of the following is an example of chemical change?
Maru [420]

Answer:

c

Explanation:

all the others r physical

8 0
3 years ago
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A merry-go-round with a a radius of R = 1.9 m and moment of inertia I = 209 kg-m2 is spinning with an initial angular speed of ω
RideAnS [48]

Answer:

340.67 kgm²/s

Explanation:

R = Radius of merry-go-round = 1.9 m

I = Moment of inertia = 209 kgm²

\omega_i = Initial angular velocity = 1.63 rad/s

m = Mass of person = 73 kg

v = Velocity = 4.8 m/s

Initial angular momentum is given by

L=I\omega_i\\\Rightarrow L=209\times 1.63\\\Rightarrow L=340.67\ kgm^2/s

The initial angular momentum of the merry-go-round is 340.67 kgm²/s

8 0
3 years ago
We have an Atwood device, two blocks connect by a string strung over a pulley, but the twist this time is that both blocks are o
Zanzabum

The Acceleration of the system is 6.41 m/s².

Given,

α= 15°, m₁ = 7kg

β= 65°, m₂ = 11 kg

Let, a be the acceleration and T is the tensions at the end it's the cord.

Let, the mass m₂ be coming down along the inclined plane along the inclined surface towards downward m₂g sin β and the tension in the upward direction,

Resultant force, m₂a=m₂g sin β -T

11a=((11) ×g sin 65°) -T  ...(i)

Now, considering the motion of m₁ which moves downwards, the forces are m₁g sinα, and T both are acting downwards.

Resultant force m₁a = m₁g sin α+T

7a =7g sin 15°+T  ...(ii)

Solving both the equations by adding them,

18a=11gsin 65°+7g sin 15°-T+T

18a=11gsin 65°+7g sin 15°=115.45

a=115.45/18=6.41 m/s²

Hence, the Acceleration of the system is 6.41 m/s².

Learn more about the acceleration here:

brainly.com/question/22048837

#SPJ10

6 0
2 years ago
If you lived on the Moon, would you see Earth go through phases? If so, would the sequence of phases be the same as those of the
marissa [1.9K]

Answer:

yes; yes

Explanation:

Phases of the moon refers to the shapes of the moon due to the lit part of it visible from the Earth. On a new moon day, the moon comes between the sun and the earth such that the lit portion is not visible from the Earth. On a full moon day, the earth comes between the sun and the moon and the whole lit part is visible.

When one would view the earth from the moon, the earth would also be visible as going through the phases. The order would be reversed. Understand this with the following example, On a new moon day, the Earth would be visible completely lit from the moon. So it will be full Earth day on the moon. On a full moon day, the lit side of the Earth would be completely away and hence, from the moon, new earth would be there.

8 0
3 years ago
A circular coil 14.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this locati
sp2606 [1]

Answer:

a)T = 2.9*10^{-5} N-m

b) north edge will rise up

Explanation:

torque on the coil is given as

T = NIABsin\theta

where N is number of loop =  9 loops

i is current = 7.80 A

-B -earth magnetic field = 5.00*10^{-5} T

A- area of circular coil

A = \frac{\pi d^{2}}{4}

A =\frac{\pi .14^{2}}{4}

A =0.015 m2

PUTITNG ALL VALUE TO GET TORQUE

T = 9*7.8*0.015*5*10^{-5} sin{90-56}

T = 2.9*10^{-5} N-m

b) north edge will rise up

3 0
3 years ago
Read 2 more answers
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