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3 years ago

Why is chemical energy a part of all chemical reactions?

2 answers:

5 0

Energy is always needed to preform anything, and chemical energy is one form of energy. Chemical energy is a part of chemical reactions because it involves teh chemicals.

vivado [14]3 years ago

4 0

For a rection to take place energy must be involved for it to stage

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A uniform drawbridge must be held at a 37 ? angle above the horizontal to allow ships to pass underneath. The drawbridge weighs

iVinArrow [24]

Answer:

Explanation:

a ) Let T be tension in the horizontal cable .

Taking torque about hinge of weight and tension

45000 x 7 x cos37 = 3.5 sin 37 x T

T = 119431 N

b ) If R be the reaction force on the hinge

R sin 37 = 45000 ( vertical force balancing each other )

R = 74775 N

c ) The direction of R will be along the bridge

d ) Moment of inertia of bridge

I = Ml² / 3 = (45000 / 9.8 ) x 14² / 3

= 300000.

When cable breaks , the weight creates a torque about the hinge will creates angular acceleration

Torque by weight about the hinge

= 45000 x 7 x cos 37

= 251570

angular acceleration = torque / moment of inertia

= 251570 / 300000

= .84 radian / s²

4 0

3 years ago

The integral with respect to time of a force applied to an object is a measure called impulse, and the impulse applied to an obj

aliina [53]

Answer:

Follows are the solution to this question:

Explanation:

By checking the value in which we have calculated by performing its differentiation of $\frac{a}{3}t^3+bt$ , the correct form of its integer value is calculating with regard to t, that also provides as expected $at^2+b = F(t)$ .

4 0

3 years ago

Can you die if you bust a hemorrhoid?

SVEN [57.7K]

I think that you can die

8 0

3 years ago

Read 2 more answers

Determine the moment of inertia Ixx of the mallet about the x-axis. The density of the wooden handle is 860 kg/m3 and that of th

Yuki888 [10]

Complete Question

Diagram for this shown on the first uploaded image

Answer:

The moment of inertia Ixx of the mallet about the x-axis is $I{xx}= 0.119 kg \cdot m^2$

Explanation:

From the question we are told that

The density `of wooden handle is $\rho_w = 860 kg/m^3$

The density `of soft-metal head is $\rho_s =8000kg/m^3$

Generally the mass of the wooden can be mathematically obtained with this formula

$m_w = \rho_w A_w l_w$

Where $A_w$ is mass of wooden handle which is mathematically obtain with the formula

$A_w = \frac{\pi}{4} d^2_w$

Where $d_w$ is the diameter of the wooden handle which from the diagram is

$27mm = \frac{27}{1000} = 0.027m$

So $A_w = \frac{\pi}{4} * 0.027^2$

$l_w$ is the length of the the wooden handle which is given in the diagram as $l_w = 315mm = \frac{315}{1000} = 0.315m$

Substituting these value into the formula for mass

$m_w = 860 * (\frac{\pi}{4} * 0.027^2 ) *0.315$

$= 0.155kg$

Generally the mass of the soft-metal head can be mathematically obtained with this formula

$m_s = \rho_s A_s l_s$

Where $A_s$ is mass of soft-metal head which is mathematically obtain with the formula

$A_s = \frac{\pi}{4} d^2_s$

Where $d_s$ is the diameter of the soft-metal head which from the diagram is

$36mm = \frac{36}{1000} = 0.036m$

So $A_s = \frac{\pi}{4} * 0.036^2$

$l_s$ is the length of the the soft-metal head which is given in the diagram

as $l_s = 90mm = \frac{90}{1000} = 0.090m$

Substituting these value into the formula for mass

$m_s = 8000 * (\frac{\pi}{4} * 0.036^2 ) *0.090$

$=0.733kg$

Generally the mass moment of inertia about x-axis for the wooden handle is

$(I_{xx})_w = [\frac{1}{3}m_w + l_w^2 ]$

Substituting values

$(I_{xx})_w = [\frac{1}{3}*0.155 + 0.315^2 ]$

$=5.12*10^{-3}kg \cdot m^2$

Generally the mass moment of inertia about x-axis for the soft-metal head is

$(I_{xx})_s = [\frac{1}{12}m_s l_s ^2 + b^2]$

Where b is the distance from the centroid to the axis of the head which is mathematically given as

$b=l_w +\frac{d_s}{2}$

Substituting values

$b = 0.315 + \frac{0.036}{2}$

$= 0.336m$

Now substituting values into the formula for mass moment of inertia about x-axis for soft-metal head

$(I_{xx})_s = [\frac{1}{12} *0.733* 0.090^2 + 0.336^2]$

$=0.113 kg \cdot m^2$

Generally the mass moment of inertia about x-axis is mathematically represented as

$I_{xx} = (I_{xx})_w + (I_{xx})_s$

$= [\frac{1}{3}m_w + l_w^2 ] + [\frac{1}{12}m_s l_s ^2 + b^2]$

Substituting values

$I_{xx} = 5.12*10^{-3} +0.113$

$I{xx}= 0.119 kg \cdot m^2$

8 0

3 years ago

What concept does the image above show?

kati45 [8]

There is no image??? i think you forgot it attach it

8 0

3 years ago