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Delicious77 [7]

2 years ago

11

Assuming 84.0% efficiency for the conversion of electrical power by the motor, what current must the 13.0-V batteries of a 716 k

g electric car be able to supply to climb a 3.00 x 102 m high hill in 2.00 min at a constant 22.0 m/s speed while exerting 7.00 x 102 N of force to overcome air resistance and friction

1 answer:

tino4ka555 [31]2 years ago

8 0

Answer:

$\mathbf{ current(I) =1766.67 \ A}$

Explanation:

Given that:

The air resistance and friction = 700 N

The gravity caused force = 716 × 9.8 = 7016.8

Total force = (7016.8 + 700) N

Total force = 7716.8 N

∴

$13 \times current(I) \times 0.84 = \dfrac{7716.8 \times 300}{2 \times 60}$

$current(I) \times 10.92= 19292$

$current(I) = \dfrac{19292}{10.92}$

$\mathbf{ current(I) =1766.67 \ A}$

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An astronaut sitting on the launch pad on Earth's surface is 6,400 kilometers from Earth's center and weighs 400 newtons. Calcul

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Answer:

weight at height = 100 N .

Explanation:

The problem relates to variation of weight due to change in height .

Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .

At the surface :

Applying Newton's law of gravitation

mg₀ = G Mm / R²

At height h from centre

mg₁ = G Mm /h²

Given mg₀ = 400 N

400 = G Mm / R²

400 = G Mm / (6400 x 10³ )²

G Mm = 400 x (6400 x 10³ )²

At height h from centre

mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²

= 400 / 4

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A skydiver reaches terminal velocity. Then he opens his parachute.

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Answer:

it is 0.9999.10896

Explanation:

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What happens when two positive charges are near each other?

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If two positive charges are near each other they will repel each other.

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2 years ago

A spotlight on the ground shines on a wall 12 meters away. If a man 2 meters high walks away from the spotlight towards the wall

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Answer: The length of the shadow on the wall is decreasing by 0.6m/s

Explanation:

the specified moment in the problem, the man is standing at point D with his head at point E.

At that moment, his shadow on the wall is y=BC.

The two right triangles ΔABC and ΔADE are similar triangles. As such, their corresponding sides have equal ratios:

ADAB=DEBC

8/12=2/y,∴y=3 meters

If we consider the distance of the man from the building as x then the distance from the spotlight to the man is 12−x.

(12−x) /12=2/y

1− (1 /12x )=2 × 1/y

Let's take derivatives of both sides:

−1 / 12dx = −2 × 1 / y^2 dy

Let's divide both sides by dt:

−1/12⋅dx/dt=−2/y^2⋅dy/dt

At the specified moment:

dxdt=1.6 m/s

y=3

Let's plug them in:

−1/121.6) = - 2/9 × dy/dt

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dy/dt = 1.6/12 × 9/2

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2 years ago

Titanium metal requires a photon with a minimum energy of 6.94×10−19J to emit electrons. If titanium is irradiated with light of

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Answer:

a) $1.59(10)^{-19} J$

b) $2.34(10)^{12} electrons$

Explanation:

The photoelectric effect consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

If the light is a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.

<u>This is what Einstein proposed: </u>

Light behaves like a stream of particles called photons with an energy $E$ :

$E=\frac{hc}{\lambda}$ (1)

So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the kinetic energy $K$ of the photoelectron:

$E=\Phi+K$ (2)

Where $\Phi=6.94(10)^{-19} J$ is the minimum amount of energy required to induce the photoemission of electrons from the surface of Titanium metal.

Knowing this, let's begin with the answers:

<h3 /><h3>a) Maximum possible kinetic energy of the emitted electrons ( $K$ )</h3>

From (1) we can know the energy of one photon of 233 nm light:

$E=\frac{hc}{\lambda}$

Where:

$h=6.63(10)^{-34}J.s$ is the Planck constant

$\lambda=233 (10)^{-9} m$ is the wavelength

$c=3 (10)^{8} m/s$ is the speed of light

$E=\frac{(6.63(10)^{-34}J.s)(3 (10)^{8} m/s)}{3 (10)^{8} m/s}$ (3)

$E=8.53(10)^{-19} J$ (4) This is the energy of one 233 nm photon

Substituting (4) in (2):

$8.53(10)^{-19} J=6.94(10)^{-19} J+K$ (5)

Finding $K$ :

$K=1.59(10)^{-19} J$ (5) This is the maximum possible kinetic energy of the emitted electrons

<h3>b) Maximum number of electrons that can be freed by a burst of light whose total energy is $2 \mu J=2(10)^{-6} J$ </h3>

Since one photon of 233 nm is able to free at most one electron from the Titanium metal, we can calculate the following relation:

$\frac{E_{burst}}{E}$

Where $E_{burst}=2(10)^{-6} J$ is the energy of the burst of light

Hence:

$\frac{E_{burst}}{E}=\frac{2(10)^{-6} J}{8.53(10)^{-19} J}=2.34(10)^{12} electrons$ This is the maximum number of electrons that can be freed by the burst of light.

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2 years ago