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The bodies in this universe attract one another name the scientist who propounded this statement

Colt1911 [192]

<h2><em>Answer: According to wikipedia Sir Isaac Newton has told this statement.</em></h2><h2><em>hope its helps you.</em></h2><h2><em>have a great day.</em></h2><h2><em> keep smiling be happy stay safe. </em></h2><h2><em /></h2>

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7 0

3 years ago

A circular horizontal surface having an area of 1020.5 mm2that is completely immersed in a fluid experiences a uniform force of

Bingel [31]

Answer:

Faya vei Abhikesh brass hahahahahaha

Explanation:

3 0

3 years ago

A 1.00 kg object moving in the + x direction at 10.0 m/s collides with a 1.50 kg object traveling at 5.00 m/s in the - x directi

marysya [2.9K]

Answer:

The amount of kinetic energy lost during the collision is 60.75 J

Explanation:

The given parameters are;

The mass of the object moving in the +x direction, m₁ = 1.00 kg

The velocity of the object moving in the +x direction, v₁ = 10.0 m/s

The mass of the object moving in the -x direction, m₂ = 1.50 kg

The velocity of the object moving in the +x direction, v₂ = 5.00 m/s

The final velocity of the 1.00 kg mass after collision, v₃ = 4.00 m/s

The direction of motion of the 1.00 kg mass after collision = -x direction

The total initial kinetic energy of the system, $K.E._{total \ initial}$ = 1/2·m₁·v₁² + 1/2·m₂·v₂²

∴ $K.E._{total \ initial}$ = 1/2×1.00×10² + 1/2×1.50×5² = 68.75 J

The final kinetic energy of the system, $K.E._{final}$ = 1/2·m₁·v₃²

$K.E._{final}$ = 1/2×1.00×4² = 8 J

The amount of kinetic energy lost during the collision, $\Delta K.E._{system}$ is given as follows;

$\Delta K.E._{system}$ = $K.E._{final}$ - $K.E._{total \ initial}$ = 8 J - 68.75 J = -60.75 J

The amount of kinetic energy lost during the collision = 60.75 J.

7 0

3 years ago

Human perception of the frequency of sound waves is called:

Olin [163]

Answer:

The study of the human perception of sound is called psychoacoustics.

Explanation:

5 0

3 years ago

A clay ball (mass = 0.25kg) has a rightward momentum of +1.75 kg∙m/s. A second clay ball (mass = 0.25 kg) has a leftward momentu

julia-pushkina [17]

1)

$\begin{gathered} E1=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_1v_2^2 \\ where: \\ m_1=m_2=0.25kg \\ v_1=7m/s \\ v_2=-7m/s \\ so: \\ E1=\frac{1}{2}0.25(7^2)+\frac{1}{2}0.25(7^2) \\ E1=6.125+6.125 \\ E1=12.25J \end{gathered}$

Answer:

d. 12.25J

------------------------

2)

According to the conservation of energy:

$\begin{gathered} E1=E2 \\ so: \\ E2=12.25J \end{gathered}$

Answer:

b. 12.25

-------------------------------

3)

$P1=m_1v_2+m_2v_2=+1.75-1.75=0$

Answer:

d. 0 kg∙m/s

-----------------------------------------

4)

Using conservation of momentum:

$\begin{gathered} P1=P2 \\ so: \\ P2=0 \end{gathered}$

Answer:

b. 0 kg•m/s

3 0

1 year ago