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Ber [7]
3 years ago
12

An object of mass 0.67 kg is attached to a spring with spring constant 15 N/m. If the object is pulled 14 cm from the equilibriu

m position and released.
What is the maximum speed of the object?
Physics
1 answer:
crimeas [40]3 years ago
5 0

Answer:

The maximum speed of the object is 0.662 m/s.

Explanation:

Given that,

Mass of the object, m = 0.67 kg

Spring constant of the spring, k = 15 N/m

The spring is pulled 14 cm or 0.14 m from the equilibrium position and released.

To find,

The maximum speed of the object.

Solution,

The maximum speed of the object is given by :

v=A\omega........(1)

Where

\omega=\sqrt{\dfrac{k}{m}}

\omega=\sqrt{\dfrac{15}{0.67}}

\omega=4.73\ rad/s

So,

v=0.14\times 4.73

v = 0.662 m/s

So, the maximum speed of the object is 0.662 m/s.

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Explanation: The Earth pushes you down at 100lbs, so you push down the earth by 100lbs which is enough to keep you firmly attached to the ground and not allow you to jump more than a couple of feet.

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3 years ago
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In an arcade game a 0.099 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and releas
Sonja [21]

Answer:

The speed of disk is 1.98 \frac{m}{s}

Explanation:

Given:

Mass of m = 0.099 kg

Spring constant k = 244 \frac{N}{m}

Compression of spring x = 4 \times 10^{-2} m

From energy conservation theorem,

Spring potential energy converted into kinetic energy,

   \frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}

  v = \sqrt{\frac{k x^{2} }{m} }

  v = \sqrt{\frac{244 \times 16 \times 10^{-4} }{0.099} }

  v = 1.98 \frac{m}{s}

Therefore, the speed of disk is 1.98 \frac{m}{s}

8 0
3 years ago
A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W
weqwewe [10]

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, P_{gauge \ pressure} = 2.95 atm

Atmospheric pressure, P_{atm} = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2

where;

P₁ = P_{gauge \ pressure} + P_{atm \ pressure}

P₂ = P_{atm}

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 =  P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 =  P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 =  P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} +  \rho gz_1 =  \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} +  \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} +  \rho gz_1)}{\rho} }

where;

\rho is the density of seawater = 1030 kg/m³

v_2 = \sqrt{ \frac{2(2.95*101325 \ + \  1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s

Therefore, the water is flowing at the rate of 28.04 m/s.

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