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loris [4]
2 years ago
12

If 11,550 J heat is absorbed by a copper object when it is heated from 20.0°C to 30.0°C,

Physics
1 answer:
RSB [31]2 years ago
5 0

Answer:

The mass of the copper is 3.00kg

Explanation:

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The throwing back by a wall or barrier of a sound wave without absorbing<br> it. *<br> 1 point
Semenov [28]

Answer:Reflection

Explanation:

The throwing back of a sound wave without absorbing it is called reflection

In acoustic reflection of sound is termed as echo i.e. sound arrived at the listener after a particular delay depending upon the position of barrier to the observer.

The reflection of sound is used in many devices like megaphone, trumpets, etc. It is also used in auditorium such that the ceiling of the auditorium is curved for multiple reflections of sound so that sound can be reached at every corner of the auditorium.

8 0
3 years ago
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Nana76 [90]

Answer:

Electron shell

Nucleus

Neutrons

Explanation:

An atom is made up of three fundamental subatomic particles which are the protons, neutrons and electrons.

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5 0
3 years ago
Apply a force of 50N to the left describe the motion of the box
GenaCL600 [577]
When a force is applied to the box , this will cause an acceleration to the box.
(force =mass×acceleration)

So the box has a constant acceleration and a changing velocity.
4 0
3 years ago
What is the term used to label the energy levels of electrons?
mel-nik [20]
Electron volts...........
4 0
3 years ago
A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a
IgorLugansk [536]

Answer:1.5

Explanation:

Given

mass of first  cart m_1=6 kg

initial Velocity u_1=3 m/s

mass of second cart m_2=3 kg

u_2=0 m/s

In the absence of External Force we can conserve momentum

m_1u_1+m_2u_2=(m_1+m_2)v

v=\frac{m_1u_1+m_2u_2}{m_1+m_2}

v=\frac{6\times 3+3\times 0}{6+3}

v=2 m/s

Final kinetic Energy of two masses

K.E._2=\frac{1}{2}(m_1+m_2)v^2

K.E._2=\frac{1}{2}\cdot (3+6)\cdot (2)^2

K.E._2=18 J

Initial Kinetic Energy

K.E._1=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2

K.E._1=\frac{1}{2}6\times 3^2+0

K.E._1=27 J

ratio =\frac{K.E._1}{K.E._2}=\frac{27}{18}=1.5

5 0
3 years ago
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