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Ymorist [56]
3 years ago
6

the gas in a cylinder has a volume of 4 liters at a pressure of 123 kPa. the pressure of the gas is increased to 204 kpa. assumi

ng the temperature remains constant, what would the new volume be?
Physics
1 answer:
Marat540 [252]3 years ago
7 0
According to boyle's law, pressure is inversely proportional to volume. Therefore (PV) initial = (PV) final. 123x4 = 204xV. V=123x4/204 = 2.41L.
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What does Atomic Collisions mean
stiks02 [169]

Answer:

Atomic and molecular collision processes are the physical interactions of atoms and molecules when they are brought into close contact with each other and with electrons, protons, neutrons or ions. This includes energy-conserving elastic scattering and inelastic scattering.

4 0
3 years ago
A man stands at the edge of a cliff and throws a rock downward with A speed of 12.0 m s . Sometimes later it strikes the ground
kvasek [131]

B) 48.0 m/s

We can actually start to solve the problem from B for simplicity.

The motion of the rock is a uniformly accelerated motion (free fall), so we can find the final speed using the following suvat equation

v^2 -u^2 = 2as

where

v is the final velocity

u = 12.0 is the initial velocity (positive since we take downward as positive direction)

a=g=9.8 m/s^2 is the acceleration of gravity

s = 110 m is the vertical displacement

Solving for v, we find the final velocity (and so, the speed of the rock at impact):

v = \sqrt{u^2+2as}=\sqrt{12^2+2(9.8)(110)}=48.0 m/s

A) 3.67 s

Now we can find the time of flight of the rock by using the following suvat equation

v=u+at

where

v = 48.0 is the final velocity at the moment of impact

u = 12.0 is the initial velocity

a=g=9.8 m/s^2 is the acceleration of gravity

t is the time it takes for the rock to reach the ground

And solving for t, we find

t=\frac{v-u}{a}=\frac{48.0-12.0}{9.8}=3.67 s

6 0
3 years ago
Two resistors ( 3 ohms & 6 ohms) in a Parallel circuit with a power supply = 12 volts. The current through resistor 3 ohms i
balu736 [363]

Answer:

4 A

Explanation:

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Explanation:

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2 years ago
1. If an object that stands 3 centimeters high is placed 12 centimeters in front of a plane
igor_vitrenko [27]

Answer:

1. 12 cm

2. 0.133 m

3. 0.03 m

4. Plane mirror

Virtual image

Upright

Behind the mirror

The same size as the object

Concave mirror when the object is located a distance greater than the focal length from the mirror's surface

Real image

Inverted image

In front of the the mirror

Diminished when the object is beyond the center of curvature

Same size as object when the object is placed at the center of curvature

Enlarged when the object is placed between the center of curvature of the mirror and the focus of the mirror

Concave mirror when the object is located a distance less than the focal length from the mirror's surface

Virtual image

Upright image

Behind the the mirror

Enlarged

Convex mirror

Type = Virtual image

Appearance = Upright image

Placement = Behind the mirror

Size = Smaller than the object

Explanation:

1. For plane mirror, since there is no magnification, the virtual image distance from the mirror = object distance from the mirror = 12 cm behind the mirror

2. The height of the object = 0.3 m

The distance of the object from the mirror = 0.4 meters

Height of image formed = 0.1 meter

We have;

Magnification, \ m = \dfrac{Image \ height }{Object \ height } = \dfrac{Image \ distance \ from \ mirror }{Object\ distance \ from \ mirror }

m = \dfrac{0.1}{0.3 } = \dfrac{Image \ distance \ from \ mirror }{0.4 }

Image distance from the mirror = 0.1/0.3×0.4 = 2/15 = 0.133 m

Image distance from the mirror = 0.133 m

3. m = \dfrac{Image \ height}{0.10 } = \dfrac{0.06 }{0.20 }

The image height = 0.06/0.2×0.1 = 3/100 = 0.03 meter

The image height = 0.03 meter

4. Plane mirror

Type = Virtual image

Appearance = Upright image with the left transformed to right

Placement = Behind the mirror

Size = The same size as the object

Concave mirror when the object is located a distance greater than the focal length from the mirror's surface

Type = Real image

Appearance = Inverted image

Placement = In front of the the mirror

Size = Diminished when the object is beyond the center of curvature

Same size as object when the object is placed at the center of curvature

Enlarged when the object is placed between the center of curvature of the mirror and the focus of the mirror

Concave mirror when the object is located a distance less than the focal length from the mirror's surface

Type = Virtual image

Appearance = Upright image

Placement = Behind the the mirror

Size = Enlarged

Convex mirror

Type = Virtual image

Appearance = Upright image

Placement = Behind the mirror

Size = Smaller than the object.

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4 years ago
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