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stepan [7]
3 years ago
12

An object of mass m is oscillating back and forth in simple harmonic motion. The maximum origin, x = 0, and moving in the x dire

ction. Find the following in terms of m, T, and A:(a)The equation of motion of the object, (b)The maximum speed of the object (c)The maximum distance from equilibrium is A, and the period of oscillation is T. At t = 0 the object is at the acceleration of the object.(d)The total energy of the object.​
Physics
1 answer:
cluponka [151]3 years ago
5 0

Answer:

x = A sin ω  t      equation of motion

max speed = A ω

max distance (max value of x) = A

max KE = 1/2 m V^2 = 1/2 m A^2 ω^2

f (frequency) =  ω / 2 π = 1 / T     or T =  2 π /  ω

ω = 2 π / T     substitute for ω to put in terms of T

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Some car manufacturers claim that their vehicles could climb a slope of 42 ∘. For this to be possible, what must be the minimum
Paladinen [302]

Answer:

D. 0.9

Explanation:

Calculating minimum coefficient of static friction, we first resolve the forces (normal and frictional) acting on the vehicle at an angle to the horizontal into their x and y components. After this, we can now substitute the values of x and y components into equation of static friction. Diagrammatic illustration is attached.

Resolving into x component:

                        ∑F_{x} = F_{s} - mgsin\alpha =0

                          F_{s} = mgsin\alpha     ------(1)

Resolving into y component:

                        ∑F_{y} = F_{n} - mgcos\alpha =0

                          F_{n} = mgcos\alpha      ------(2)

Static frictional force, F_{s} \leq μ F_{n}       ------(3)

substituting F_{s} from equation (1) and F_{n} from equation (2) into equation (3)

                         mgsin\alpha \leq μ mgcos\alpha

                         sin\alpha \leq μ cos\alpha

                         μ \geq \frac {sin\alpha}{cos\alpha}

                         μ \geq tan\alpha

The angle the vehicles make with the horizontal α = 42°

                         μ ≥ tan 42°

                         μ ≥ 0.9

6 0
3 years ago
A tourist stands at the top of the Grand Canyon, holding a rock, overlooking the valley below. Find the final velocity and displ
Anestetic [448]

Answer:

a. -39.2 m/s; -78.4 m

b. -31.2 m/s; -46.4 m

c. -47.2 m/s; -110.4 m

Explanation:

<h2>Part (a)</h2>

We are given/can infer these variables:

  • t = 4.0 s
  • a = -9.8 m/s²
  • v_0 = 0 m/s

We want to find the displacement and the final velocity of the rock.

  • Δx = ?
  • v = ?

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = 0 + (-9.8)(4.0)
  • v = -9.8 * 4.0
  • v = -39.2 m/s

The final velocity of the rock is -39.2 m/s.

Now we can use this equation to find the displacement of the rock:

  • Δx = v_0 t + 1/2at²

Plug in the known variables into this equation.

  • Δx = 0 * 4.0 + 1/2(-9.8)(4.0)²
  • Δx = 1/2(-9.8)(4.0)²
  • Δx = -4.9 * 16
  • Δx = -78.4 m

The displacement of the rock is -78.4 m.

<h2>Part (b)</h2>

We are given/can infer these variables:

  • v_0 = 8.0 m/s
  • a = -9.8 m/s²
  • t = 4.0 s

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = 8.0 + (-9.8)(4.0)
  • v = 8.0 + -39.2
  • v = -31.2 m/s

The final velocity of the rock is -31.2 m/s.

We can use this equation to find the displacement:

  • Δx = v_0 t + 1/2at²

Plug in known variables:

  • Δx = 8.0(4.0) + 1/2(-9.8)(4.0)²
  • Δx = 32 - 4.9(16)
  • Δx = -46.4 m

The displacement of the rock is -46.4 m.

<h2>Part (c)</h2>

We are given/can infer these variables:

  • v_0 = -8.0 m/s
  • a = -9.8 m/s²
  • t = 4.0 s

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = -8.0 + (-9.8)(4.0)
  • v = -8.0 - 39.2
  • v = -47.2 m/s

The final velocity of the rock is -47.2 m/s.

We can use this equation to find the displacement:

  • Δx = v_0 t + 1/2at²

Plug in known variables:

  • Δx = -8.0(4.0) + 1/2(-9.8)(4.0)²
  • Δx = -32 - 4.9(16)
  • Δx = -110.4 m

The displacement of the rock is -110.4 m.

8 0
4 years ago
Which of the following is the best example of kinetic energy being transformed into potential energy?
devlian [24]

Answer:

coasting down hill on a bicycle

Explanation:

Coasting down the hill on a bicycle is a typical example of how kinetic energy is being transformed to potential energy in a system.

Kinetic energy is the energy due to the motion of a body, it can be derived using the expression below;

      K.E  = \frac{1}{2} m v²

Potential energy is the energy due to the position of a body. It can be derived using;

     P.E  = mgh

m is the mass

v is the velocity

g is the acceleration due to gravity

h is the height

Now, at the top of the hill, the potential energy is at the maximum. As the bicycle coasts down the potential energy is converted to kinetic energy.  

6 0
3 years ago
Read 2 more answers
The red giant Betelgeuse has a surface temperature of 3000 K and is 600 times the diameter of our sun. (If our sun were that lar
Nat2105 [25]

Answer:

Explanation:

a )

Radius of the sun = .69645 x 10⁹ m .

600 times = 600 x .69645 x 10⁹ m

= 4.1787 x 10¹¹ m .

surface area A = 4π (4.1787 x 10¹¹)²

= 219.317 x 10²²

energy radiated E = σ A Τ⁴

= 5.67 x 10⁻⁸ x 219.317 x 10²² x (3000)⁴

= 100695 x 10²⁶ J

To know the wavelength of photon emitted

\lambda_mT= b

\lambda_m= \frac{b}{T}

= 2.89777 x 10⁻³ / 3000

= 966 nm

= 1275 /966 eV

1.32 x 1.6 x 10⁻¹⁹ J

= 2.112 x 10⁻¹⁹ J

No of photons radiated = 100695 x 10²⁶ / 2.112 x 10⁻¹⁹

= 47677.5 x 10⁴⁵

= .476 x 10⁵⁰ .

b )

energy radiated by our sun per second

E₂ = σ A 5800⁴

energy radiated by Betelgeuse per second

E₁ = σ  x 600²A x  3000⁴

E₁ / E₂  = σ  x 600²A x  3000⁴ / σ A 5800⁴

= 36 X 10⁴ x 3⁴ x 10¹² / 58⁴ x 10⁸

= 25.76 x 10⁸ x 10⁻⁵

= 25760 times .

4 0
4 years ago
Catherine likes the name she was given, but wishes it were shorter. She decides to start going by the new name of Cathy. It's cu
Blababa [14]

The answer would be conformity. This is a kind of social influence concerning a change in belief or conduct so one can fit in with a group. This alteration is in reply to real (physical occurrence of others) or fictional (pressure of social norms / prospects) group pressure.

3 0
3 years ago
Read 2 more answers
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