The antacid is basic so it neutralizes acidity or lowers it. Then if it goes into the esophagus, it's not as strong and it doesn't hurt, and it also calms your stomach because the acidity in your stomach is also lower. Antacids are therefore taken by many people, especially as they grow older and things like heartburn become more common.
Answer:C) car X
Explanation:
Given
All the cars have identical Engine thus Force Produced by car X will be equal to Y and Z
and 
Since Car X is most massive so acceleration associated with it will be minimum
acceleration of car X is minimum thus it will travel farthest
Answer: B
Explanation:
It's not the time it took to heat the substance, so that rules out A and C.
This means that we only have to choose between
B. the area of contact
D. the area of the substances
(since everything else in each of those answers are the same)
Area of contact matters more (e.g. an object with greater surface area is exposed to the air more, will lose/gain heat quicker than an object with less surface area).
Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region?
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres
Answer:

Explanation:
Given data:
Momentum of moving model train, 
Mass of the stationary model train, 
Initial speed of the stationary model train, 
Assume there is no external force is acting on the given train system.
In this case, the total linear momentum of the trains would be conserved.
Let the final linear momentum of the trains be 
Thus,




