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Irina18 [472]
3 years ago
9

A spring with a spring constant of 100 N/m is relaxed at the beginning. The spring is then compressed by 0.1 m. An object of 0.0

1 kg is attached to the compressed spring. If the compressed spring is released, the object will experience simple harmonic motion. What is the maximum possible speed of the object during the oscillation
Physics
1 answer:
tangare [24]3 years ago
6 0

Answer:

Velocity, v = 10 m/s

Explanation:

Spring constant of the spring, k = 100 N/m

Compression in the string, x = 0.1 m

Mass of the spring, m = 0.01 kg

Let v is the maximum possible speed of the object during the oscillation. It can be calculated by equating potential energy stored in the spring and the kinetic energy imparted to the object as :

\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{kx^2}{m}}

v=\sqrt{\dfrac{100\times (0.1)^2}{0.01}}

v = 10 m/s

So, the maximum possible speed of the object during the oscillation is 10 m/s. Hence, this is the required solution.

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A submarine is 2.84 102 m horizontally from shore and 1.00 102 m beneath the surface of the water. A laser beam is sent from the
xxMikexx [17]

Answer:

468 m

Explanation:

So the building and the point where the laser hit the water surface make a right triangle. Let's call this triangle ABC where A is at the base of the building, B is at the top of the building, and C is where the laser hits the water surface. Similarly, the submarine, the projected submarine on the surface and the point where the laser hit the surface makes a another right triangle CDE. Let D be the submarine and E is the other point.

The length CE is length AE - length AC = 284 - 234 = 50 m

We can calculate the angle ECD:

tan(\hat{ECD}) = \frac{ED}{EC} = \frac{100}{50} = 2

\hat{ECD} = tan^{-1} 2 = 63.43^o

This is also the angle ACB, so we can find the length AB:

tan(\hat{ACB}) = \frac{AB}{AC} = \frac{AB}{234}

2 = \frac{AB}{234}

AB = 2*234 = 468 m

So the height of the building is 468m

5 0
4 years ago
A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an
Dominik [7]

Answer:

position as a function of time is y = 0.05 × cos(9.9)t

Explanation:

given data

mass = 5 kg

length = 10 cm = 0.1 m

displaced = 5 cm

to find out

position as a function of time

solution

we will apply here equilibrium that is

mass × g = k × length

put here value and find k

k = \frac{5*9.8}{.01}

k = 490 N/m

and ω is

ω = \sqrt{\frac{k}{m} }

ω = \sqrt{\frac{490}{5} }

ω = 9.9

so here position w.r.t  time is

y = 0.05 × cosωt

y = 0.05 × cos(9.9)t

so position as a function of time is y = 0.05 × cos(9.9)t

8 0
3 years ago
si un disco fue lanzado con una fuerza de 1000 N, y el disc recorrió una distancia e o.6 m ¿Qué trabajo efectuó el lanzador ?
Alexeev081 [22]

Answer:

Trabajo realizado = 600 Nm

Explanation:

Dados los siguientes datos;

Fuerza = 1000 Newton

Distancia = 0.6 metros

Para encontrar el trabajo realizado;

Trabajo \; realizado = fuerza * distancia

Sustituyendo en la ecuación, tenemos;

Trabajo \; realizado = 1000 * 0.6

Trabajo realizado = 600 Nm

4 0
3 years ago
Help awnser physics​
kirill115 [55]

Answer:

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3 0
3 years ago
The change in the momentum of an object is represented by the following formula:
Ede4ka [16]

Hi there!

Recall that:

Change in momentum = mass × change in velocity

Or:

Δp = mΔv = m(vf - vi)

Plug in the given values. We can assign east to be positive and west to be negative in this instance (Velocity is a vector with direction).

Thus:

Δp = (1)(-21 - 10) = -31 kgm/s OR 31 kgm/s WEST.

The correct answer is B.

Change in momentum is EQUIVALENT to the quantity of IMPULSE.

The correct answer is H.

6 0
2 years ago
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