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elena-14-01-66 [18.8K]
3 years ago
8

According to principle of conservation of energy, the total momentum of a system of masses in any direction remains constant unl

ess acted upon by an external force in that direction. A. True B. False
Physics
1 answer:
Sergeu [11.5K]3 years ago
6 0
The principle of conservation of energy states that energy can neither be created nor destroyed, but can be changed from one form to another. The law of inertia on the other hand states that a body remains at a state of rest or motion on a straight line unless compelled by an external force. So i think that the answer is B, which is FALSE because the question is more related to the law of inertia(Newtons first law) than to the law of conservation of energy. Hope i helped.  
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The mass of an electron is A) equal to the mass of a proton B) less than the mass of a neutron C) greater than the mass of a pro
julsineya [31]

Answer:

B

Explanation:

Atomic structure contains electrons, protons and neutrons.

Electron is very light compared to proton and neutrons.

Given that the mass of an electron is

A) equal to the mass of a proton

B) less than the mass of a neutron

C) greater than the mass of a proton

D) equal to the mass of a neutron

The correct answer is B which is less than the mass of the neurons.

4 0
3 years ago
Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.
andreev551 [17]

Answer:

the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = m_A

then, the mass of object B = m_B = \frac{m_A}{4}

work done on object A = -18 J

work done on object B = -18 J

let v_i be the initial speed

let v_f be the final speed

For object A;

K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2  = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ -  v_i^2 )\ =- 18\\\\v_f^2 \ -  v_i^2  = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ -  v_i^2  = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\

v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\

Thus, the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

To obtain the change in the final speed of object B, apply the following equations.

K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B =  \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\

\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i

Thus, the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

3 0
3 years ago
Which best describes the current atomic model?
Lorico [155]
<span>A: It is not an exact representation of the atom, but is close enough to be very useful.

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3 0
3 years ago
Read 2 more answers
The position of a 55 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds. determine the velocity at t=0.4
NNADVOKAT [17]
The position of the mass is given by (in cm):
x(t)=2 \cos (10 t)
The velocity is the derivative of the position:
v(t) =  \frac{dx(t)}{dt} =-10\cdot 2 \sin (10t)=-20 \sin (10t)
Substituting t=0.40 s, we can find the velocity at this time:
v(0.40 s)= -20 \sin (10 \cdot 0.4)=15 cm/s=15 \cdot 10^{-2}m/s
5 0
3 years ago
Read 2 more answers
All of the following are factors affecting flow rate except what?
faust18 [17]

Answer:

c. Vessel side holes

Explanation:

  • The "Poiseuille formula" which is given by \\\begin{aligned} \small Q& =  \small \frac{\pi r^4}{8 \eta}.\frac{\Delta P}{\Delta L}\\\end{aligned} describes the volumetric flow rate (\small Q) through tubular sections.
  • Here, \Delta P,\,\, \Delta L,\,\, r,\,\, \eta represent the injection pressure difference, the length of the section, the radius of the section and the viscosity index of the fluid that flows through the section respectively.
  • With this, one can confirm that all the factors except the vessel side holes affect the flow rate.
  • Side holes, however, are a factor that could give a measure of how much volume would flow to a particular location. In such a situation the flow rate remains unchanged and one location would receive a lower volume (not the whole) as some volume would spill out at the side holes.

#SPJ4

5 0
2 years ago
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