Answer:A)3074m,3074.37m/s
B)12905.4m,19100.54m/s
Explanation:the formula for electric field =V/d
Where V=velocity of the particle,d=distance covered from the electric field source
A)E=V/d
1.48=4.55×10^5/d
d=4.55×10^5/1.48=3074m
Distance of B from centre of the field=3074+0.370=3074.370m
B)d=V/E=1.91×10^4/1.48
d=12905.4m
At point B
Distance=12905.4+0.370=12905.770
V=Ed=12905.770×1.48=19100.54m/s
Answer:
physical property
Explanation: Some substances exist as gases at room temperature (oxygen and carbon dioxide), while others, like water and mercury metal, exist as liquids.
Use the definition of average acceleration:
<em>a</em> = ∆<em>v</em> / <em>t</em>
If <em>v</em> is the starting speed, then ∆<em>v</em> = 0 - <em>v</em>, so solve for <em>v</em> :
-6.42 m/s² = (0 - <em>v</em>) / (2.85 s)
<em>v</em> = (6.42 m/s²) (2.85 s)
<em>v</em> ≈ 18.3 m/s
Answer:A force can be described as a push or a pull. Pushes and pulls can be seen to act on objects when they begin to move, speed up, slow down or change direction.
The image gallery on this page shows children at play using push and pull actions to move objects. Children use push and pull all the time while at play. While playing, students learn to manipulate objects and materials and make observations about their actions.
Teachers may use this teaching resource over a number of lessons to explore and develop their students' understanding that a push or a pull affects how an object moves or changes shape.
Explanation:
Answer:
ΔS=2*m*Cp*ln((T1+T2)/(2*(T1*T2)^1/2))
Explanation:
The concepts and formulas that I will use to solve this exercise are the integration and the change in the entropy of the universe. To calculate the final temperature of the water the expression for the equilibrium temperature will be used. Similarly, to find the change in entropy from cold to hot water, the equation of the change of entropy will be used. In the attached image is detailed the step by step of the resolution.