05.06.2022

This is about bounds and density

. 4

Faq

Physics
Step-by-step answer
P Answered by PhD

I answered Number 4 (Solids and Elasticity)

Explanation:

Part a)

If a weight of 10 N stretches a certain spring 3 cm, a) Determine the spring constant k.

Solution:

To determine the spring constant we use the formula F = k x (X)  

where ,

F is the force applied in this case 10 N

x is the displacement of the spring’s end from its equilibrium position in this case 3 cm

F=kx\\10=k(3)\\10/3=k\\k=3.33\ N/cm

Part b)

b) Determine how much stretch will occur if you double the weight with a force of 20 N?

Solution:

If we double the weight to 20 N we need to find the need to find the new stretch of the spring which is x. But we know the spring constant which is k so it becomes,

F=kx\\20=3.33x\\20/3.33=x\\x=6.00\ cm

So we can see that if we double the Force the stretch is also doubled thus showing us a direct variation as seen in the formula F = k x (X)

Part c)

c) Using g=9.8 m/s^2 determine the mass of the 10 N weight and round your answer to 3 significant digits.

Solution:

The formula of weight is,

W=mg

where,

w is the weight of the object in Newtons (N) which is 10 in this case

m is the mass of the object in kilograms (kg) we to need to find in this case and,

g is acceleration due to gravity which is constant 9.8 m/s^2

so now,

W=mg\\10=m(9.8)\\10/9.8=m\\m=1.02\  kg

Physics
Step-by-step answer
P Answered by PhD

I answered Number 4 (Solids and Elasticity)

Explanation:

Part a)

If a weight of 10 N stretches a certain spring 3 cm, a) Determine the spring constant k.

Solution:

To determine the spring constant we use the formula F = k x (X)  

where ,

F is the force applied in this case 10 N

x is the displacement of the spring’s end from its equilibrium position in this case 3 cm

F=kx\\10=k(3)\\10/3=k\\k=3.33\ N/cm

Part b)

b) Determine how much stretch will occur if you double the weight with a force of 20 N?

Solution:

If we double the weight to 20 N we need to find the need to find the new stretch of the spring which is x. But we know the spring constant which is k so it becomes,

F=kx\\20=3.33x\\20/3.33=x\\x=6.00\ cm

So we can see that if we double the Force the stretch is also doubled thus showing us a direct variation as seen in the formula F = k x (X)

Part c)

c) Using g=9.8 m/s^2 determine the mass of the 10 N weight and round your answer to 3 significant digits.

Solution:

The formula of weight is,

W=mg

where,

w is the weight of the object in Newtons (N) which is 10 in this case

m is the mass of the object in kilograms (kg) we to need to find in this case and,

g is acceleration due to gravity which is constant 9.8 m/s^2

so now,

W=mg\\10=m(9.8)\\10/9.8=m\\m=1.02\  kg

Mathematics
Step-by-step answer
P Answered by Specialist

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  


Aplane lamina with constant density p(x, y) = p occupies a region in the xy-plane bounded by a simpl
Aplane lamina with constant density p(x, y) = p occupies a region in the xy-plane bounded by a simpl
Physics
Step-by-step answer
P Answered by Master

The question is incomplete. The complete question is :

A plate of uniform areal density $\rho = 2 \ kg/m^2$ is bounded by the four curves:

$y = -x^2+4x-5m$

$y = x^2+4x+6m$

$x=1 \ m$

$x=2 \ m$

where x and y are in meters. Point $P$ has coordinates $P_x=1 \ m$ and $P_y=-2 \ m$. What is the moment of inertia $I_P$ of the plate about the point $P$ ?

Solution :

Given :

$y = -x^2+4x-5$

$y = x^2+4x+6$

$x=1 $

$x=2 $

and $\rho = 2 \ kg/m^2$ , $P_x=1 \ $ , $P_y=-2 \ $.

So,

$dI = dmr^2$

$dI = \rho \ dA  \ r^2$  ,           $r=\sqrt{(x-1)^2+(y+2)^2}$

$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \  dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

$I=\frac{32027}{21} \times 2$

  $= 3050.19 \ kg \ m^2$

So the moment of inertia is  $3050.19 \ kg \ m^2$.


A plate of uniform areal density is bounded by the four curves: where and are in meters. Point has c
Physics
Step-by-step answer
P Answered by Master

The question is incomplete. The complete question is :

A plate of uniform areal density $\rho = 2 \ kg/m^2$ is bounded by the four curves:

$y = -x^2+4x-5m$

$y = x^2+4x+6m$

$x=1 \ m$

$x=2 \ m$

where x and y are in meters. Point $P$ has coordinates $P_x=1 \ m$ and $P_y=-2 \ m$. What is the moment of inertia $I_P$ of the plate about the point $P$ ?

Solution :

Given :

$y = -x^2+4x-5$

$y = x^2+4x+6$

$x=1 $

$x=2 $

and $\rho = 2 \ kg/m^2$ , $P_x=1 \ $ , $P_y=-2 \ $.

So,

$dI = dmr^2$

$dI = \rho \ dA  \ r^2$  ,           $r=\sqrt{(x-1)^2+(y+2)^2}$

$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \  dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

$I=\frac{32027}{21} \times 2$

  $= 3050.19 \ kg \ m^2$

So the moment of inertia is  $3050.19 \ kg \ m^2$.


A plate of uniform areal density is bounded by the four curves: where and are in meters. Point has c
Mathematics
Step-by-step answer
P Answered by Specialist

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  


Aplane lamina with constant density p(x, y) = p occupies a region in the xy-plane bounded by a simpl
Aplane lamina with constant density p(x, y) = p occupies a region in the xy-plane bounded by a simpl
Mathematics
Step-by-step answer
P Answered by PhD
Answer: 440 grams for 1.54 is the better value
Explanation:
Take the price and divide by the number of grams
1.54 / 440 =0.0035 per gram
1.26 / 340 =0.003705882 per gram
0.0035 per gram < 0.003705882 per gram
Mathematics
Step-by-step answer
P Answered by PhD

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