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3 years ago

14

If the velocity of the particle is uniform, then the path of the particle must be a

1 answer:

lianna [129]3 years ago

6 0

Answer:

write yoir question carefully i am not understanding it

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Help me to answer my questions please

Andrei [34K]

2 is c
3 is a
4 is b
5 is c

3 0

3 years ago

Two identical balls move directly toward each other with equal speeds. how will the balls move if they collide and stick togethe

Citrus2011 [14]

The momentum of both the identical balls would eventually be transferred to one another when it comes to a point wherein they will collide. In addition, the phenomenon is called an elastic collision wherein both the momentum and energy of the system would considered to be conserved.

7 0

3 years ago

38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

7 0

3 years ago

The kinetic energy of a moving object is E=12mv2. A 61 kg runner is moving at 10kmh. However, her speedometer is only accurate t

jek_recluse [69]

Answer:

$e=3367.2J$

% $e=1.43$ %

Explanation:

From the exercise we know two information. The real speed and the experimental measured by the speedometer

$v_{r}=10km/h=2.77m/s$

Since the speedometer is only accurate to within 0.1km/h the experimental speed is

$v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s$

Knowing that we can calculate Kinetic energy for the real and experimental speed

$E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J$

$E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J$

Now, the potential error in her calculated kinetic energy is:

$e=E_{r}-E_{e}=(234023-230656)J=3367.2J$

% $e=\frac{E_{r}-E_{e}}{E_{r}}x100=\frac{(234023-230656)J}{234023J}x100=1.43$ %

4 0

3 years ago

A certain car traveling in a straight line path has a velocity of +10.0 m/s at some instant. after 3.00 s, its velocity is +6.00

makvit [3.9K]

The formula is:
v = v o + a t
6 = 10 + 3 * a
3 a = 10 - 6
a = 4 : 3
a = - 1.33 m/s² ( because the car slows down )
Answer: The average acceleration of the car is - 1.33 m/s²

3 0

3 years ago