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A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end

Triss [41]

Answer: m= 35.6 kg

Explanation:

For finding the mass of the stone we have the formula

v= $\sqrt{\frac{Tension}{Linear. Mass. density} }$

Here, Tension= m*g = m*9.81

and linear mass density= $\frac{8.25 g}{65 cm}$

Linear mass density= $\frac{8.25*10^-3}{65*10^-2}$

Linear mass density= 0.0127 kg/m

Velocity= $2*\frac{l}{t}$

Velocity= 2 * $\frac{65*10^-2}{7.84}$

Velocity= 165.8 m/s

So putting all these values in equation we get

v= $\sqrt{\frac{Tension}{Linear. Mass. density} }$

165.8= $\sqrt{\frac{m*9.81}{0.0127} }$

Solving we get

m= 35.58 kg

or m= 35.6 kg

3 0

4 years ago

For variables control, a circuit voltage will be measured using a sample of five circuits. The past average voltage for samples

astra-53 [7]

Answer:

Average :

UCL = 4.15

LCL = 2.65

Range :

UCL = 2.75

LCL = 0

Explanation:

Given :

Sample size, n = 5

Average, X = 3.4

Range, R = 1.3

A2 for n = 5 ; equals 0.577 ( X chart table)

For the average :

Upper Control Limit (UCL) :

X + A2*R

3.4 + 0.577(1.3) = 4.1501

Lower Control Limit (LCL) :

X - A2*R

3.4 - 0.577(1.3) = 2.6499

FOR the range :

Upper Control Limit (UCL) :

UCL = D4*R

D4 for n = 5 ; equals = 2.114

UCL = 2.114*1.3 = 2.7482

Lower Control Limit (LCL) :

LCL = D3*R

D3 for n = 5 ; equals = 0

LCL = 0 * 1.3 = 0

8 0

3 years ago

What is the difference between a vigorous sport and vigorous recreation ?

NNADVOKAT [17]

Answer:

Explanation:El ejercicio vigoroso previene en mayor medida el síndrome metabólico (un conjunto de enfermedades que aumentan el riesgo cardiovascular )

mientras que una reacción vigorosa se produce entre el aluminio y el gas cloro. Como consecuencia de la gran cantidad de energía liberada se producen luz y calor

7 0

3 years ago

g A lighter ball A with momentum PA=5.0 kg∙m/s in positive x-direction approaches a heavier ball B at rest before the collision.

Svet_ta [14]

Answer:

7kgm/s

Explanation:

Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Let P1A and P1B be the initial momentum of the bodies A and B respectively

Let P2A and P2B be the final momentum of the bodies A and B respectively after collision.

Based on the law:

P1A+P2A = P1B + P2B

Given P1A = 5kgm/s

P2A = 0kgm/s(ball B at rest before collision)

P2A = -2.0kgm/s (negative because it moves in the negative x direction)

P2B = ?

Substituting the values in the equation gives;

5+0 = -2+P2B

5+2 = P2B

P2B = 7kgm/s

3 0

3 years ago

A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

The formula for the acceleration, given in $\frac{m}{s^{2}}$ , is:

$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

Velocity can be derived from the following definition of acceleration:

$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

$\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}$

$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

3 0

3 years ago