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3 years ago

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Starting with a constant velocity of 45 km/h, a car accelerates for 35 seconds at an acceleration of 0.45 m/s2 . What is the vel

ocity of the car at the end of the period of 35 seconds of acceleration?

1 answer:

DENIUS [597]3 years ago

5 0

Answer:

28.3 m/s

Explanation:

Vi = 45 Km/h = 12.5 m/s

Vf - Vi = at

Vf -12.5 = 0.45(35)

Vf= 28.3 m/s

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A 50-cm wire placed in an east-west direction is moved horizontally to the north with a speed of 2.0 m/s. the horizontal compone

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When a wire is moved inside uniform magnetic field then its free electrons will experience magnetic force on it due to which wire will have potential difference at its ends.

Now here we will have magnetic field due to earth and wire is moving in this constant field so induced emf is given by formula

$EMF = v.(B x L)$

given that

$B = 25\mu Tj - 50\mu Tk$

$v = 2 m/s j$

$L = 0.50 m (-i)$

now by using the above formula we will have

$EMF = 2(j) .(25\mu j - 50\mu k) x (-0.50 i)$

$EMF = 2(j) .(12.5\mu k + 25\mu j)$

$EMF = 50 \mu Volts$

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3 years ago

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If vector A ⃗ has components A x and A y and makes an angle θ with the +x axis, then

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Then the tangent of angle-Θ is (Ay / Ax).

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3 years ago

Issac and Blaise decide to race. They both start at the same position at the same time. Issac runs at 2m/s but decides to take a

FromTheMoon [43]

Let the Blaise runs for time "t" to complete the race

so the total distance he moved is given by

$d_1 = 1* t$

Now Issac runs for time t = "t - 2*60"

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now it is given that Blaise wins by 10 m distance

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3 years ago

what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe

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Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

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Time, t = 7.4 s

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$a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2$

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2$

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s$

Acceleration,

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3 years ago

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adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

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2 years ago