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strojnjashka [21]
3 years ago
8

Suppose Car 1 collides with Car 2; Car 2 was not moving. In an ideal situation with no friction, according to the Law of Conserv

ation of Momentum, what happens to the momentum within this system after the collision?
A) It remains the same.
B) It is greatly reduced.
C) It is slightly reduced.
D) It is slightly increased.
Physics
2 answers:
Jobisdone [24]3 years ago
6 0

Answer: The answer is A.

Explanation:i just took the test

insens350 [35]3 years ago
4 0
Your answer is C it is slightly reduced
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If a scientist unknowingly breaks the law, he is guilty of _____.
Margaret [11]

Answer:

noncompliance

Explanation:

If a scientist unknowingly breaks the law, he is guilty of <u>noncompliance</u>

8 0
3 years ago
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According to the law of conservation of energy,
vodka [1.7K]

According to the law of conservation of energy,  

A. an object loses most of its energy as friction

<u>B. the total amount of energy for a system stays the same</u>

Energy is never lost due to the law of conservation

C. the potential energy of an object is always greater than its kinetic energy

D. the kinetic energy of an object is always greater than its potential energy​

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Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i
SCORPION-xisa [38]

Answer:

The value is  r =  5.077 \  m

Explanation:

From the question we are told that

   The  Coulomb constant is  k =  9.0 *10^{9} \  N\cdot  m^2  /C^2

   The  charge on the electron/proton  is  e =  1.6*10^{-19} \  C

    The  mass of proton m_{proton} =  1.67*10^{-27} \  kg

    The  mass of  electron is  m_{electron } =  9.11 *10^{-31} \ kg

Generally for the electron to be held up by the force gravity

   Then    

       Electric force on the electron  =  The  gravitational Force

i.e  

            m_{electron} *  g  = \frac{ k *  e^2  }{r^2 }

         \frac{9*10^9 *  (1.60 *10^{-19})^2  }{r^2 }  =     9.11 *10^{-31 }  *  9.81

         r =  \sqrt{25.78}

         r =  5.077  \  m

7 0
3 years ago
A tank having a volume of 0.85 m 3 initially contains water as a two-phase liquid-vapor mixture at 260 o C and a quality of 0.7.
Keith_Richards [23]

Answer:=14,160 kJ

Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below

Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase

C MPa m^3/kg kJ/kg kJ/kg

1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture

2 260 4.689 0.0422 2599 2797 1 Saturated Vapor

The mass initially contained in the tank is m1 = V/v1

m1 =0.85 m^3 /0.02993 m^3 /kg

= 28.4 kg

The mass finally contained in the tank is

m2 =V2/v

= 0.85 m^3 /0.0422 m^3 /kg

= 20.14 kg

The heat transfer is then

Qcv = m2u2 − m1u1 − he(m2 − m1)

Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ

4 0
3 years ago
Is magnitude a<br> vector<br> scalars<br> both<br> neither
Doss [256]

Answer:

scalars

Explanation:

7 0
3 years ago
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