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2 years ago

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If spring has a spring constant of 500 N/m and is stretched .50 meters,how much energy is stored in the spring ((show work for f

ull credit, show equation))

1 answer:

Free_Kalibri [48]2 years ago

7 0

Answer:

The energy stored is: 62.5 Joules

Explanation:

Given

$k = 500N/m$ --- spring constant

$x = 0.5m$ --- stretch

Required

The amount of energy

This is calculated as:

$U = \frac{1}{2} kx^2$

$U = \frac{1}{2} * 500N/m * (0.5m)^2$

$U = 250N/m * (0.5m)^2$

$U = 62.5\ J$

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During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of

olasank [31]

Answer:

3.416 m/s

Explanation:

Given that:

mass of cannonball $m_A$ = 72.0 kg

mass of performer $m_B$ = 65.0 kg

The horizontal component of the ball initially $\mu_{xA}$ = 6.50 m/s

the final velocity of the combined system v = ????

By applying the linear momentum of conservation:

$m_A \mu_{xA}+m_B \mu_{xB} = (m_A+m_B) v$

$72.0 \ kg \times 6.50 \ m/s+65.0 \ kg \times 0 = (72.0 \ kg+65.0 \ kg) v$

$468 kg m/s + 0 = (137 kg)v$

$v = \dfrac{468\ kg m/s }{137 \ kg}$

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2 years ago

Coherent light with wavelength 599 nm passes through two very narrow slits with separation of 20 μm, and the interference patter

goblinko [34]

Answer:

134.77 mm

Explanation:

Wave length of light λ = 599 x 10⁻⁹ m

Slit separation d = 20 x 10⁻⁶ m

Screen distance D = 3 m

Distance of second dark fringe from centre

= 1.5 x λ D / d

Putting the values given above

distance = $\frac{1.5\times599\times10^{-9}\times 3}{20\times10^{-6}}$

= 134.77 x 10⁻³ m

= 134.77 mm.

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3 years ago

A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that

saveliy_v [14]

Answer:

The speed it reaches the bottom is

$v=6.51m/s$

Explanation:

Given: $m=5.0kg$ , $r=47cm\frac{1m}{100cm}=0.47m$

Using the conservation of energy theorem

$U_i=K_E+K_{ER}$

$m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2$

$v=r*w$ , $I=\frac{1}{2}*m*r^2$

$m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2$

$m*g*h=\frac{3}{4}*m*r^2*w^2$

$g*h=\frac{3}{4}*r^2*w^2$

Solve to w'

$w^2=\frac{4*g*h}{3*r^2}$

$h=x*sin(30)=6.5m*sin(30)=3.25m$

$w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}$

$w=27.74rad/s$

$v=27.74rad/s*0.235m=6.51m/s$

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2 years ago

When jogging outside you accidentaly bumb into a curb. Your feet stop but your body continues to movr foward and you end up on t

sattari [20]

B.) Newtons first law of motion

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2 years ago

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The drawing shows a large cube (mass = 28.6 kg) being accelerated across a horizontal frictionless surface by a horizontal force

MrRissso [65]

Answer:

P= 454.11 N

Explanation:

Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).

$P= (M+m)*a\\a = \frac{P}{28.6 +4.3}\\a = \frac{P}{32.9}$

The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:

$F = m * a*\mu \\F= 4.3*0.710*\frac{P}{32.9}\\F=3.053*\frac{P}{32.9}$

In order for the small cube to not slide down, the friction force must equal the weight of the small cube:

$3.053*\frac{P}{32.9} = 4.3 * g\\\\P = \frac{4.3*9.8*32.9}{3.053} \\P= 454.11 N$

The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N

8 0

2 years ago