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3 years ago

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As the resistance increased under 25 V, the current blank. Compared to calculated currents, experimental currents proved to be b

lank . Currents varied proportionally according to resistance: 10 Ω of resistance produced a current of blank A, while 100 Ω of resistance produced a current of blank A.

1 answer:

sukhopar [10]3 years ago

4 0

Answer:

Decreased, The Same, 2.5 and .25

Explanation:

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CClassify each characteristic of sound waves.

Alex787 [66]

Explanation : Explain each characteristic of sound waves.

Intensity : the intensity of the sound wave is understand as the power carry by sound wave per unit area in the direction perpendicular to that area.

Loudness : loudness is the quality of the loud and soft of the sound wave.

Frequency : Human normal hear sound frequency between 20 Hz to kHz.

Pitch : Pitch is the quality of low and high of sound wave . pitch relates to the frequency of the slowest vibration in the sound wave for simple sound.

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3 years ago

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A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in 1.5 seconds

horrorfan [7]

Answer:

The force has been reduced by 8018 N

Explanation:

The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:

$F\Delta t = m\Delta v$

where:

F is the force exerted on the car

$\Delta t$ is the duration of the collision

m = 1400 kg is the mass of the car

$\Delta v=-27 m/s$ is the change in velocity of the car

We can re-write the equation as

$F=\frac{m\Delta v}{\Delta t}$

In the 1st collision, the time is 1.5 seconds, so the force is

$F_1=\frac{(1400)(-27)}{1.5}=-25,200 N$

In the 2nd collision, the time is increased to 2.2 seconds, so the force is

$F_2=\frac{(1400)(-27)}{2.2}=-17,182 N$

Therefore, the force has been reduced by:

$F_2-F_1=-17,182-(-25,200)=8018 N$

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3 years ago

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Which of the following elements is in the same period as phosphorus?

Arte-miy333 [17]

The answer is B. magnesium I am pretty sure

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3 years ago

A team exerts a force of 10 N towards the south in a tug of war. The other, opposite team, exerts a force of 17 N towards the no

ikadub [295]

Answer:

Option C

Explanation:

According to the question:

Force exerted by the team towards south, F = 10 N

Force exerted by the opposite team towards North, F' = 17 N

Net Force, $\vec{F_{net}} = \vec{F'} - \vec{F}$

$\vec{F_{net}} = \vec{F'} - \vec{F} = 17 - 10 = 7 N$

Thus the force will be along the direction of force whose magnitude is higher

Therefore,

$\vec{F_{net}} = 7 N$ towards North

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2 years ago

An insect 5.25 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

Zigmanuir [339]

Answer:

(A) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

Explanation:

height of the insect (h) = 5.25 mm = 0.525 cm

distance of the insect (s) = 25 cm

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)

index of refraction (n) = 1.7

(A) we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{∞} -\frac{1}{-12.5} )$

$\frac{1}{f} =(0.7)(0 + \frac{1}{12.5} )$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) if the lens is reversed, the radius of curvatures would be interchanged

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = 12.5 cm

we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{12.5} -\frac{1}{∞} )$

$\frac{1}{f} =(0.7)( \frac{1}{12.5} - 0)$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

7 0

3 years ago