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stepladder [879]
4 years ago
6

A vertical spring (ignore its mass), whose spring constant is 594-N/m, is attached to a table and is compressed down by 0.196-m.

What upward speed (in m/s) can it give to a 0.477-kg ball when released?
Physics
1 answer:
Ray Of Light [21]4 years ago
6 0

Answer:

Speed, v = 6.91 m/s

Explanation:

Given that,

Spring constant, k = 594 N/m

It is attached to a table and is compressed down by 0.196 m, x = 0.196 m

We need to find the speed of the spring when it is released. Here, the elastic potential energy is balanced by the kinetic energy of the spring such that,

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

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

v=\sqrt{\dfrac{594\ N/m\times (0.196\ m)^2}{0.477\ kg}}

v = 6.91 m/s

So, the speed of the ball is 6.91 m/s. Hence, this is the required solution.

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(a) A force F = (3xî + 4yĵ), where F is in newtons and x and y are in meters, acts on an object as the object moves in the x-dir
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W = 53.6648 J

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Now replacing numeric

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The solar constant is a. the same for a sphere or for a circle. b. only about a quarter of all energy emitted by the sun. c. a c
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If f3=0 and f1=12n, what does the magnitude of f⃗ 2 have to be for there to be rotational equilibrium?
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4 0
4 years ago
Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele
nekit [7.7K]

Answer:

0.536\sqrt{\frac{GM}{R}}

Explanation:

We are given that

Mass of one  asteroid 1,m_1=M

Mass of asteroid 2,m_2=1.97 M

Initial distance between their centers,d=13.63 R

Radius of each asteroid=R

d'=R+R=2R

Initial velocity of both asteroids

u=0

We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

(m_1+m_2)u=m_1v_1+m_2v_2

(M+1.97 M)\times 0=Mv_1+1.97Mv_2

Mv_1=-1.97 Mv_2

v_1=-1.97v_2

According to law of conservation of energy

Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2

GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2

1.97M^2G(\frac{13.63-2}{27.26R})=\frac{1}{2}Mv^2_2(3.8809+1.97)

1.97MG(\frac{11.63}{27.26 R})=\frac{1}{2}(5.8509)v^2_2

v^2_2=\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}

v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}

v_2=0.536\sqrt{\frac{GM}{R}}

Hence, the speed of second asteroid =0.536\sqrt{\frac{GM}{R}}

8 0
3 years ago
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