In trajectory, acceleration vector is not the same with the velocity vector but it depends on the change in velocity. The acceleration path is perpendicular to the velocity vector path. The Correct answers are:
a.) D
b.) B
c.) C
d.) A
A Vector is a quantity that has both magnitude and direction.
Given that a ball is fired from a cannon at point 1 and follows the trajectory shown in the figure.
a.) The vector best represents the ball's velocity at position 2 is D
b.) The vector best represents the ball's acceleration at point 2 is B
c.) Which vector best represents the ball's velocity at position 3 = C
d.) Which vector best represents the ball's acceleration at point = A
In trajectory, acceleration vector is not the same with the velocity vector but depends on the change in velocity. The acceleration path is perpendicular to the velocity vector path.
At point 3, since the ball path is parabolic, the velocity at that point is not equal to zero but will tend to take a horizontal path.
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a gas increases in pressure from 2.00 atm to 6.00 atm at a constant volume of 1.00 m3 and then expands at constant pressure to a volume of 3.00 m3 before returning to its initial state as shown in the figure below. how much work is done in one cycle?
The work done on the given gas in one cycle is -405.3 kJ.
The given parameters;
initial pressure of the gas, P₁ = 2 atmfinal pressure of the gas, P₂ = 6 atminitial volume of the gas, V₁ = 1 m³final volume of the gas, V₂ = 3 m³Convert the pressure to Pascal (N/m²);
1 atm = 101325
The work done in one cycle is the area of the triangle and it is calculated as follows
[tex]Area = \frac{1}{2} \times base \times height\\\\Area = \frac{1}{2} \times (3 \ m^3\ -\ 1 \ m^3)\times (6 \ atm \ - \ 2 \ atm)\\\\Area = 4 \ atm -m^3\\\\Area = 4 \ atm -m^3 \ \times \frac{101325 \ Pa}{1 \ atm} \\\\Area = 405,300 \ m^3.Pa\\\\Area = 405,300 \ m^3. (N/m^2)\\\\Area = 405,300 \ Nm\\\\Area = 405,300 \ J\\\\Area = 405.3 \ kJ[/tex]
the net work done on the gas = - 405.3 kJ
Thus, the work done on the given gas in one cycle is -405.3 kJ.
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what do u guys think of my drawing?
Answer: yess sirrrrrrrr
Explanation:
Answer:
oou I like it! u did a good job ^^
How to find density of a gas.
Answer:
Subtract the mass of the cylinder from the mass of the cylinder when it contains gasoline. This is the mass of the gasoline. Divide this figure by the volume, 100 ml, to get the density.
What is the amount of charge when 13.5a is flowing for 2 1/2 hours
Answer:
Quick Answer:
a. 8.9Ω
b. 1.2×104 C
Explanation:
Answer:
Answer provided in explanation
Explanation:
why do air waves move faster than earthquake waves?
Answer:
P-waves and S-waves are body waves that propagate through the planet. P-waves travel 60% faster than S-waves on average because the interior of the Earth does not react the same way to both of them. ... The energy is thus less easily transmitted through the medium, and S-waves are slower.
Explanation:
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Can u plz help me with my hw
Answer:
1) 4 x 25m = 100m
2) 0 because after 4 lengths, he's back at the starting block.
3) speed is distance over time so speed here is 100m/125s = 0.8m/s
4) ...
5) 100m / 1.25m/s = 80s
6.a) 100m / 0.5m/s = 200s
7.b) ... can't draw this now..
If a train is moving south at 50 m/s and a person is running SOUTH on the train at 5 m/s, what is the velocity of the person as seen by someone not moving (watching the train going by)
A-45 m/s North
B-50 m/s South
C-55 m/s North
D-100 m/s South
The velocity of the person as seen by someone not moving (watching the train going by) will be 55 m/s South. I can therefore conclude that none of the option given is correct.
Given that a train is moving south at 50 m/s and a person is running SOUTH on the train at 5 m/s.
That means the person is running relatively to the velocity of the train. Since the train and the person are moving in the same direction, the relative velocity seen by an observer watching the train going will be the addition of both the train and the person in the direction of the train.
That is,
Relative velocity = 50 + 5
Relative velocity = 55 m/s South.
Therefore, the velocity of the person as seen by someone not moving (watching the train going by) will be 55 m/s South.
I can therefore conclude that none of the option given is correct.
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True or False: Objects fall at a constant speed.
Answer:
Explanation:
true