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mr Goodwill [35]
3 years ago
11

a 1.00 kg mass is placed at the free end of a compressed spring. The force constant of the spring is 115 N/m. The spring has bee

n compressed 0.200 m from its neutral position. it is now released. neglecting the mass of the spring and assuming that the mass is sliding on a frictionless surface, how fast will the mass move as it passes the neutral position of the spring?
Physics
1 answer:
frosja888 [35]3 years ago
8 0

Here 1 kg mass is hold against compressed spring

when it is released and reached to natural position it will start moving with certain speed

As per energy concept we will now say that potential energy stored in the spring will convert into kinetic energy of the block

so here we will say

U_{spring} = KE_{block}

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

115(0.20)^2 = 1.00(v^2}

v = 2.145 m/s

so block will released with speed 2.145 m/s

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Fiesta28 [93]

Answer:

a) 1.75s b) 17.2 m/s (down)

Explanation:

d1= 15m d2= 0m (because it hits ground)

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Equation

the triangle means change in so d2-d1

Δd= v1 * t + 1/2 * a * t^2

0m-15m= v1*t + 1/2 a t^2

-15 m= 0m/s*t (goes away) + 1/2* a *t^2

-15mx2= t^2

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Square root (-30/-9.81m/s^2)

t=1.75 s

b) now v2!!

Im going to use v2= v1 + a*t

v2= 0m/s + -9.81 x 1.75s

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3 years ago
A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2,
Stella [2.4K]

Answer:

the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   P =  104.04 \hat{i} -314.432 \hat{j}

Explanation:

The free-body  diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;

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Now;

V_x = \frac{d}{dt}(12t^3+2) = 36 t^2

V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s

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-V_y = R* \omega

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-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   P =  104.04 \hat{i} -314.432 \hat{j}

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3 years ago
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Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving
Iteru [2.4K]

Explanation:

Let m_p is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.

The magnetic force is balanced by the centripetal force acting on the proton as :

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r is the radius of path,

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Time period is given by :

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Frequency of proton is given by :

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The wavelength of radiation is given by :

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So, the wavelength of radiation produced by a proton is \dfrac{2\pi m_pc}{qB}. Hence, this is the required solution.

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3 years ago
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