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amm1812
3 years ago
10

Consider a system to be one train car moving toward another train car at rest. When the train cars collide, the two cars stick t

ogether.
What is the total momentum of the system after the collision?
800 kg • m/s
1,600 kg • m/s
2,400 kg • m/s
4,000 kg • m/s

Physics
2 answers:
sweet [91]3 years ago
5 0

Answer:

C. 2,400 kg • m/s

Explanation:

Doing it now! Good luck

oee [108]3 years ago
3 0

Answer:

Momentum after = 2400 kgm/s

Explanation:

Momentum after = momentum before

Momentum after = m1v1 +m2v2

Momentum after = (600)(4) + (400)(0)

Momentum after = 2400 kgm/s

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A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a
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     Considering the unknown resistence as R and using the Ohm's First Law, we have:

i= \frac{V}{R_{eq}}  \\ 0.5= \frac{12}{R+10}  \\ R+10=24 \\ R=14-Ohm
 
     The equivalent resistence is given by the resistor series with the lamp resistence.

R_{eq}=R+10 \\ R_{eq}=14+10 \\ \boxed {R_{eq}=24-Ohm}

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6 0
3 years ago
Kepler's First Law states that the shape of planetary orbits is a/an ________________ ?
harkovskaia [24]

Answer:

elliptical orbit

Explanation:

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First Law : It states that all the planets revolve around the sun in an elliptical path and the sun is at one focus of that elliptical path.

4 0
3 years ago
A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
3 years ago
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