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3 years ago

A 1000 kg car drives off of u 30m cliff of

Physics

1 answer:

Inessa05 [86]3 years ago

5 0

The mechanical energy of the car is $1.54\cdot 10^6 J$

Explanation:

The mechanical energy of an object is equal to the sum of its gravitational potential energy (GPE) and its kinetic energy (KE):

E = GPE + KE

The GPE of the car is given by:

$GPE = mgh$

where

m = 1000 kg is the mass of the car

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

h = 30 m is the height of the car above the ground

Substituting,

$GPE=(1000)(9.8)(30)=2.94\cdot 10^5 J$

The kinetic energy of the car is given by:

$KE=\frac{1}{2}mv^2$

where

m = 1000 kg

v = 50 m/s is the speed of the car

Substituting,

$KE=\frac{1}{2}(1000)(50)^2=1.25\cdot 10^6 J$

Therefore, the mechanical energy is

$E=GPE+KE=2.94\cdot 10^5 + 1.25\cdot 10^6 = 1.54\cdot 10^6 J$

Learn more about potential energy and kinetic energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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Unpolarized light with an intensity of 22.4 ????ux passes through a polarizer whose transmission axis is vertically oriented. (a

irina1246 [14]

Answer:

a) I₁ = 11.2 Lux , vertical direction , b) I₂ = 1.44 Lux

Explanation:

a) A polarized is a system that absorbs light that is not polarized in the direction of its axis, therefore half of the non-polarized light must be absorbed

consequently the above the processed light has half of the incident intensity and the directional of the polarized

I₁ = I₀ / 2

I₁ = 22.4 / 2

I₁ = 11.2 Lux

is polarized in the vertical direction

b) The polarized light falls on a second polarizer, therefore it must comply with the law of Malus

I₂ = I₁ cos² θ

I₂ = 11.2 cos² 69

I₂ = 1.44 Lux

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3 years ago

A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c

ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

$\int _{0}^{i}di =9\int _{0}^{t}dt$

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

$B = \frac{\mu_{0}i}{2\pi r}$

$B = \frac{\mu_{0}\times 9t}{2\pi r}$

(b)

Magnetic flux,

$\phi=\int B\times a dr$

$\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr$

$\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )$

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$\phi = 1.89 \times 10^{-7}t$

(c)

R = 3 ohm

$e = -\frac{d\phi}{dt}$

magnitude of voltage is

e = 1.89 x 10^-7 V

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3 years ago

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Explanation:

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Serga [27]

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