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Jobisdone [24]
3 years ago
10

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.

Physics
1 answer:
balandron [24]3 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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30. Light has been described as a ___ and a ___
andrezito [222]

Answer:

Explanation: light has been described as a particle and a wave

7 0
3 years ago
A particle moving uniformly along the x axis is located at 11.3m at 0.588s and at 3.38m at 4.67s. Find its displacement during t
Yakvenalex [24]
Displacement = (distance between start and end points) in the direction of (direction from start to end point). Distance = (11.3-3.38)= 7.92 m. Direction = the negative 'x' direction.
5 0
3 years ago
Read 2 more answers
In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spo
Marysya12 [62]

Answer:

14.9 mm

Explanation:

We know dsinθ = mλ where d = separating of slit, m = order of maximum = 3 and λ = wavelength = 510nm = 510 × 10⁻⁹ m

Also tanθ = L/D where L = distance of m order fringe from central bright spot = 17 mm = 0.017 m and D = distance of screen from slit = 1.6 m

So, sinθ = mλ/d

Since θ is small, sinθ ≅ tanθ

So,

mλ/d = L/D

d = mλD/L

Substituting the values of the variables into the equation, we have

d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m

d = 2448 × 10⁻⁹ m²/0.017 m

d = 144000 × 10⁻⁹ m

d = 1.44 × 10⁻⁴ m

d = 0.144 × 10⁻³ m

d = 0.144 mm

Now, for the second-order maximum, m' of the 670 nm wavelength of light,

m'λ'/d = L'/D where m' = order of maximum = 2, λ' = wavelength of light = 670 nm = 670 × 10⁻⁹ m, d = slit separation = 0.144 mm = 0.144 × 10⁻³ m, L' = distance of second order maximum from central bright spot and D = distance of screen from slit = 1.6m

So, L' = m'λ'D/d

So, substituting the values of the variables into the equation, we have

L' = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m

L' = 2144  × 10⁻⁹ m²/0.144 × 10⁻³ m

L' = 14888.89 × 10⁻⁶ m

L' = 0.01488 m

L' ≅ 0.0149 m

L' = 14.9 mm

4 0
3 years ago
Three persons wants to push a wheel cart in the direction marked x in Fig. The two person push with horizontal forces F1 and F2
Svetllana [295]

Answer:

<u>I had to search the Figure on Google to solve this question.</u>

a) The magnitude of the force F₃ is:

F_{3} = 87.47 N

And the direction of F₃:

\alpha = 79.04 ^{\circ}  (with respect to the y-direction, in the third quadrant)

b) P = 4.22 N  

Explanation:

<u>I had to search the Figure on Google to solve this question.</u>

a) We can find the force of the third person as follows:

\Sigma F_{x} = F_{1x} + F_{2x} + F_{3x} = 0

\Sigma F_{y} = F_{1y} - F_{2y} + F_{3y} = 0

So, in x-direction we have:

\Sigma F_{x} = 45 N*cos(70) + 75 N*cos(20) + F_{3x} = 0

F_{3x} = -85.87 N

In y-direction we have:

\Sigma F_{y} = 45 N*sin(70) - 75 N*sin(20) + F_{3y} = 0

F_{3y} = -16.63 N

The magnitude of the force F₃ is:

F_{3} = \sqrt{F_{3x}^{2} + F_{3y}^{2}} = \sqrt{(-85.87 N)^{2} + (-16.63 N)^{2}} = 87.47 N

To find the direction of F₃ we need to calculate its angle with respect to the y-direction (in the third quadrant):

tan(\alpha) = \frac{|F_{3x}|}{|F_{3y}|} = \frac{85.87 N}{16.63 N}

\alpha = 79.04 ^{\circ}

<em>b) If the third person exerts the force found in part (a) the car will stop, so the only way for the cart to accelerate at 200 m/s² is that the third person does not exert the force found in a. </em>      

<u>To find the weight of the cart​ when it accelerates at 200 m/s², we need to consider: F₃ = 0</u>.  

First, we need to find the cart's mass. Since the car is moving in the x-direction we have:

\Sigma F_{x} = F_{1x} + F_{2x} = ma

45 N*cos(70) + 75 N*cos(20) = m*200 m/s^{2}

m = \frac{45 N*cos(70) + 75 N*cos(20)}{200 m/s^{2}} = 0.43 kg

Now, the weight of the cart​ is:

P = mg = 0.43 kg*9.81 m/s^{2} = 4.22 N

I hope it helps you!                                                                                    

3 0
2 years ago
What is the velocity of a quarter dropped from a tower after 10 seconds?
Goryan [66]
U = 0, the initial vertical velocity

Assume g = 9.8 m/s² and ignore air resistance.

Use the formula
v = u + gt
where v =  vertical velocity after t seconds.
That is,
v = 0 + (9.8 m/s²)*(10 s) = 98.0 m/s

Answer: 98.0 m/s
3 0
2 years ago
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