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Sidana [21]
3 years ago
7

A car slowing down for a stop sign is an example of.....?

Physics
2 answers:
Elza [17]3 years ago
7 0
D-negative Acceleration
maxonik [38]3 years ago
3 0
The answer would be D, negative acceleration
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The distance and displacement of a object in motion can be the same (true or false)
Rasek [7]

Aswer:

False, the values ​​of the distance traveled and the displacement only coincide when the trayectorie is a straight line. Otherwise, the distance will always be greater than the offset.

Although these terms are used synonymously in other cases, they are totally different. Since the distance that a mobile travels is the equivalent of the length of its trajectory. Whereas, the displacement will be a vector magnitude.

<u>xXCherryCakeXx</u>.

4 0
2 years ago
A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net w
tatyana61 [14]
(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s
And so, the kinetic energy of the object is
K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s
And so the new kinetic energy is
K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J

So, the work done on the object is the variation of kinetic energy of the object:
W=\Delta K=120 J-60 J=60 J
7 0
2 years ago
which of the following are vector quantities? check all that apply. a. force b. acceleration c. displacement d. mass
Ratling [72]
Force, acceleration, and Displacement are all vector quantities.
4 0
3 years ago
A 235 kg motorcycle moves with a velocity of 7 m/s. What's it's kenetic energy?
antiseptic1488 [7]
The non-relativistic formula for low speed  v < 0.1c is:

K.E = 0.5mv^2 = 0.5 * 235 * (7)^2 = 5757.5 J
7 0
3 years ago
A frictionless piston–cylinder device contains 7 kg of nitrogen at 100 kPa and 250 K. Nitrogen is now compressed slowly accordin
lora16 [44]

Answer: -1038.8 kJ

Explanation:

From the question, we can see that PV^n = constant. And as such, we can deduce that it is a polytropic process. Thus, we can use the polytropic work equation to calculate the needed work input.

from the question we were given

Mass of nitrogen, m = 7kg

initial temperature, T1 = 250k

Final temperature, T2 = 450k

Polytropic index, n = 1.4

Specific gas constant, R = 0.2968kJ/kgK

W = [p2 * v2 - p1 * v1] / 1 - n

W = [m * R * T2 - T1] / 1 - n

W = 7*0.2968*(450 - 250)] / 1 - 1.4

W = [7*0.2968*200] / -0.4

W = 415.52 / -0.4

W = -1038.8 kJ

5 0
2 years ago
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