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3 years ago

You buy a 75-W lightbulb in Europe, where electricity is delivered at 240 V.

Physics

1 answer:

Lena [83]3 years ago

4 0

Answer:

a. 2.5 X 10^-1

Explanation:

First of all we shall calculate the resistance of the bulb of 75W , 240 V .

power = V² /R , V is potential and R is resistance

75 = 240²/ R

R = 240²/ 75

= 768 ohm .

Now this bulb is used with 120 V

current = 120 / 768

= .15625 A

power = V X I

potential x current

= 120 x .15625

= 18.75 W.

comparing it with 75 W

the ratio

18.75 / 75

= 2.5 x 10⁻¹.

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Answer the discussion question in complete sentences:

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Earth is divided into two hemispheres/coast. Which is south, north, east, and west.
California is west coast.

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3 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

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3 years ago

A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+138 determines the height of the rock abov

Likurg_2 [28]

Answer:

a) 138 ft

b) 4.62 s

c) 1.375 s

d) 168.25 ft

Explanation:

The height of a rock (thrown from the top of a bridge) above the level of water surface as it varies with time when thrown is given in the question as

h = f(t) = -16t² + 44t + 138

with t in seconds, and h in feet

a) The bridge's height above the water.

At t=0 s, the rock is at the level of the Bridge's height.

At t = 0,

h = 0 + 0 + 138 = 138 ft

b) How many seconds after being thrown does the rock hit the water?

The rock hits the water surface when h = 0 ft. Solving,

h = f(t) = -16t² + 44t + 138 = 0

-16t² + 44t + 138 = 0

Solving this quadratic equation,

t = 4.62 s or t = -1.87 s

Since time cannot be negative,

t = 4.62 s

c) How many seconds after being thrown does the rock reach its maximum height above the water?

At maximum height or at the maximum of any function, the derivative of that function with respect to the independent variable is equal to 0.

At maximum height,

(dh/dt) = f'(t) = (df/dt) = 0

h = f(t) = -16t² + 44t + 138

(dh/dt) = (df/dt) = -32t + 44 = 0

32t = 44

t = (44/32)

t = 1.375 s

d) What is the rock's maximum height above the water?

The maximum height occurs at t = 1.375 s,

Substituting this for t in the height equation,

h = f(t) = -16t² + 44t + 138

At t = 1.375 s, h = maximum height = H

H = f(1.375) = -16(1.375²) + 44(1.375) + 138

H = 168.25 ft

Hope this Helps!!!

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3 years ago

In a given reversible process, the temperature of an ideal gas is kept constant as the gas is compressed to a smaller volume. Wh

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Answer:

B. The gas must release heat to its surroundings.

Explanation:

Since the process is reversible , it must be slow . Since temperature is kept constant so the internal energy is constant , Since volume is decreased , pressure must be increased. Since work is done on the gas ( gas is compressed ) , so heat must be released by the gas so that its internal energy remains constant.

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4 years ago

A volcanic eruption throws a boulder that lands 1.00 km horizontally from the crater. If the volcanic rocks were launched at an

Juliette [100K]

Answer:

a) v₀ = 69.29 m / s , b) t = 18.84 s

Explanation:

a) For this exercise we will use the projectile launch equations

x = v₀ₓ t

y = y₀ + $v_{oy}$ t - ½ g t²

Let's fix our reference system on the volcano, so the horizontal distance x = 1 km = 1000 m and the vertical distance y = -900 m, the initial height of the crater is I = 0 m. Let's replace to find the speeds

v_{oy} = v₀ sin θ

v₀ₓ = v₀ cos θ

y = v₀ sin θ (x / v₀ cos θ) - ½ g (x / v₀ cos θ)2

y = x tan θ - ½ g x² / v₀² sec² θ

½ g x² sec² θ / v₀² = x tan θ - y

v₀² = ½ g x² sec² θ / (x tan θ –y)

Let's calculate

v₀² = ½ 9.8 1000² sec² 40 / (1000 tan 40 - (-900))

v₀ = √ (8.35 10⁶ / 1,739 10³)

v₀ = 69.29 m / s

b) Flight time

x = v₀ₓ t

t = x / v₀ cos θ

t = 1000 / 69.29 cos 40

t = 18.84 s

7 0

3 years ago