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Free_Kalibri [48]
3 years ago
14

An object in simple harmonic motion has an amplitude of 0.360 m and an oscillation period of 0.620 s. Determine the maximum spee

d of the motion.
Physics
1 answer:
steposvetlana [31]3 years ago
5 0

Answer:

3.648 m/s

Explanation:

Data provided in the question:

Amplitude, A = 0.360 m

oscillation period, T = 0.620 s

The maximum speed of the motion, V = A × ω

where,

ω = Angular speed

also,

ω = (2π) / T

on substituting the respective values, we get

ω = (2π) / 0.620

or

ω = 10.134 rad / s

therefore,

V = 0.360 × 10.134 = 3.648 m/s

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A wall has a negative charge distribution producing a uniformhorizontal electric field. A small plastic ball of mass .01kg carry
Sauron [17]

Answer:

a)  E = -4 10² N / C , b) x = 0.093 m, c)     a = 10.31 m / s², θ=-71.9⁰

Explanation:

For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball

X axis

             F_{e} - T_{x} = m a

Axis y

            T_{y} - W = 0

Initially the system is in equilibrium, so zero acceleration

            Fe = T_{x}  

            T_{y} = W

Let us search with trigonometry the components of the tendency

            cos θ = T_{y} / T

            sin θ = T_{x}  / T

           T_{y} = cos θ

           T_{x}  = T sin θ

We replace

            q E = T sin θ

            mg = T cosθ

             

a) the electric force is

                F_{e} = q E

                E = F_{e} / q

                E = -0.032 / 80 10⁻⁶

                E = -4 10² N / C

b) the distance to this point can be found by dividing the two equations

                q E / mg = tan θ

                θ = tan⁻¹ qE / mg

Let's calculate

              θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)

              θ = tan⁻¹ 0.3265

               θ = 18 ⁰

               sin 18 = x/0.30

               x =0.30 sin 18

               x = 0.093 m

c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations

X axis

           F_{e} = m aₓ

            aₓ = q E / m

           aₓ = 80 10⁻⁶ 4 10² / 0.01

           aₓ = 3.2 m / s²

Axis y

           W = m a_{y}

           a_{y} = g

           a_{y} = 9.8 m/s²

The total acceleration is can be found using Pythagoras' theorem

             a = √ aₓ² + a_{y}²

             a = √ 3.2² + 9.8²

             a = 10.31 m / s²

The Angle meet him with trigonometry

               tan θ = a_{y} / aₓ

               θ = tan⁻¹ a_{y} / aₓ

               θ = tan⁻¹ (-9.8) / 3.2

               θ = -71.9⁰

Movement is two-dimensional type with acceleration in both axes

8 0
3 years ago
A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A
Y_Kistochka [10]

Answer:

The tensions in T_{BC} is approximately 4,934.2 lb and the tension in T_{BD} is approximately  6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = T_{BC}

The tension in rope segment BD = T_{BD}

The tension in rope segment AB = T_{AB} = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = T_{AB} = 1200 lb

T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

T_{BC} × sin(26.0°) + T_{BD} × sin(21.0°) = 0...........................(2)

Which gives;

T_{BC} × sin(26.0°) = - T_{BD} × sin(21.0°)

T_{BC} = - T_{BD} × sin(21.0°)/(sin(26.0°))  ≈ - T_{BD} × 0.8175

Substituting the value of, T_{BC}, in equation (1), gives;

- T_{BD} × 0.8175 × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb

- T_{BD} × 0.7348  + T_{BD} ×0.9336 = 1200 lb

T_{BD} ×0.1988 = 1200 lb

T_{BD} ≈ 1200 lb/0.1988 = 6,035.6938 lb

T_{BD} ≈ 6,035.6938 lb

T_{BC} ≈ - T_{BD} × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

T_{BC} ≈ -4934.1733 lb

From which we have;

The tensions in T_{BC} ≈ -4934.2 lb and  T_{BD} ≈ 6,035.7 lb.

8 0
3 years ago
After observing the electric field in your trials above, where was the electric field the strongest? what was the direction of t
earnstyle [38]
<span>Answer: Answer is The direction of the electric field is always directed in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding the source charge.</span>
8 0
3 years ago
What is the amount of displacement of a runner who runs exactly 2 laps around a 400 meter track?
max2010maxim [7]

At the end of the laps, the runner's displacement is zero.

8 0
3 years ago
HELP ME PLEASE, I'M BEING TIMED FAST
Tom [10]

Answer:

A) Energy is tranferred from Joey to the water. The temperature of the water increases.

Explanation:

At first Joey jumps and gains a height above the water level of the pool, this way has an energy potential initial, as Joey falls into the water his speed is increased that is to say its energy potential is transformed into kinetic energy, and at the moment of impact with the water, this energy kinetic is transformed into heat which is transferred to the water. Therefore the temperature increment.

Note: This is one of the reasons why space agencies are studying  spatial asteroids that are directed toward the earth, as these come with great kinetic energy, and great potential energy, if these are of a considerable size can cause catastrophic damage, even if they fall into the ocean, due to the large amount of energy which can cause the instantaneous evaporation of large amounts of water and collateral damage in other areas.

8 0
3 years ago
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