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2 years ago

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An object slides along a curved track of negligible friction, as shown in the figure. The potential energy U of the object as a

function of the position x is shown in the graph. The object starts at rest at x=0. Which of the following is a correct claim about the motion of the object?
(Would prefer an explanation, but it's not necessary)
A) The kinetic energy of the object is approximately the same at x=2m and x=5m.

B) At x=2m, the object has more potential energy than kinetic energy.

C) The object comes to rest when x=5.5m.

D) The rate of change of the potential energy is a minimum at x=0m.

E) The object achieves its maximum speed at approximately x=3m.

1 answer:

balandron [24]2 years ago

5 0

Answer:

E) The object achieves its maximum speed at approximately x=3m.

Explanation:

Since the object is moving in gravitational field

potential energy + kinetic energy = constant = 14 J

At x = 3 m , potential energy is minimum that is zero

kinetic energy will be maximum .

Kinetic energy = 14 - 0 = 14 J

Hence option ( E ) is the answer.

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Suppose that light from a laser with wavelength 633 nm is incident on a thin slit of width 0.500 mm. If the diffracted light pro

coldgirl [10]

Explanation:

It is given that,

Wavelength, $\lambda=633\ nm=633\times 10^{-9}\ m$

Slit width, $a=0.5\ mm=0.0005\ m$

Order, m = 2

If the diffracted light projects onto a screen at distance 1.50 m, L = 1.5 m

For the diffraction of light,

$y=\dfrac{m\lambda L}{a}$

$y=\dfrac{2\times 633\times 10^{-9}\times 1.5}{0.0005}$

y = 0.0037 m

So, the distance from the center of the diffraction pattern to the dark band is 0.0037 meters. Hence, this is the required solution.

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3 years ago

An initially uncharged 3.67 μF capacitor and a 8.01 k Ω resistor are connected in series to a 1.50 V battery that has negligible

inn [45]

Answer:

The initial current in the circuit is 0.1873 mA

Explanation:

Given;

capacitance is given as 3.67 x 10⁻⁶ F

resistance is given as 8010 Ω

voltage across the circuit is 1.5 V

Since the capacitor is initially uncharged, the capacitive reactance is zero.

From ohms law;

Voltage across the circuit is directly proportional to the opposition to the flow of current.

V = IR

I = V/R

I = 1.5/8010

I = 0.0001873 A = 0.1873 mA

Therefore, the initial current in the circuit is 0.1873 mA

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3 years ago

What can you find using newton's three laws of motion

topjm [15]

Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

Explanation:

hope this helps! :)

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2 years ago

Coffee is poured at a uniform rate at 20 cm3 / s into a cup whose inside is shaped like a truncated cone. if the upper and lower

Vladimir [108]

Let the cup is filled to height h after some time

now the total volume of coffee filled in the cup is given as

$\frac{2}{y} = \frac{4}{6+y}$

$2y = 6 + y$

$y = 6 cm$

now volume of the coffee will be

$V = \frac{1}{3}\pi r^2(y + 6) - \frac{1}{3}\pi 2^2 (6)$

here we know that

$\frac{r}{y+6} = \frac{2}{6}$

$r = \frac{y+6}{3}$

$V = \frac{1}{3}\pi (\frac{y+6}{3})^2(y+6) - \frac{1}{3}\pi 2^2(6)$

now we know that volume flow rate is given as

$Q = \frac{dV}{dt}$

$20 cm^3/s = \frac{1}{3}\pi (\frac{1}{9})(3(y+6)^2)\frac{dy}{dt}$

$20 \times 9 = \pi (y + 6)^2 v$

here y = 3 cm

$180 = \pi (9)^2 v$

$v = 0.71 cm/s$

so water will rise up with speed 0.71 cm/s

5 0

3 years ago

Two identical loudspeakers separated by distance dd emit 200 Hz sound waves along the x-axis. As you walk along the axis, away f

uysha [10]

Answer:

The first possible value of d is 0.85 m

The second possible value of d is 2.55 m

The third possible value of d is 4.25 m

Explanation:

Given that,

Distance =d

Frequency of sound wave= 200 Hz

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{v}{f}$

Put the value into the formula

$\lambda=\dfrac{340}{200}$

$\lambda=1.7\ m$

The separation between the speakers in the destructive interference is

$\Delta x= d$

The equation for destructive interference

$2\pi\times\dfrac{\Delta x}{\lambda}-\Delta\phi_{0}=(m+\dfrac{1}{2})2\pi$

The loudspeakers are in phase

So, $\Delta\phi_{0}=0$

The equation for destructive interference is

$2\pi\times\dfrac{d}{\lambda}=(m+\dfrac{1}{2})2\pi$ ....(I)

Here, m = 0,1,2,3.....

We need to calculate the first possible value of d

For, m = 0

Put the value in the equation (I)

$2\pi\times\dfrac{d_{1}}{1.7}=(0+\dfrac{1}{2})2\pi$

$d_{1}=\dfrac{1.7}{2}$

$d_{1}=0.85\ m$

We need to calculate the second possible value of d

For, m = 1

Put the value in the equation (I)

$2\pi\times\dfrac{d_{2}}{1.7}=(1+\dfrac{1}{2})2\pi$

$d_{2}=\dfrac{1.7\times3}{2}$

$d_{2}=2.55\ m$

We need to calculate the third possible value of d

For, m = 1

Put the value in the equation (I)

$2\pi\times\dfrac{d_{3}}{1.7}=(2+\dfrac{1}{2})2\pi$

$d_{3}=\dfrac{1.7\times5}{2}$

$d_{3}=4.25\ m$

Hence, The first possible value of d is 0.85 m

The second possible value of d is 2.55 m

The third possible value of d is 4.25 m

3 0

2 years ago