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stich3 [128]
2 years ago
9

Why is pangaea important?

Physics
1 answer:
Trava [24]2 years ago
4 0
Pangaea is important because it was all the continents in one giant land mass surrounded by water. When the plates shifted and broke apart we got the seven continents we know today. <span />
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A helium weather balloon is filled to a volume of 219 m3 on the ground, where the pressure is 754 torr and the temperature is 29
Novosadov [1.4K]

Answer:

V_2 = 606.9 m^3

Explanation:

By ideal gas equation law we know that

PV = nRT

now we know that when balloon rises to certain level then the number of moles will remains same

so we can say

n_1 = n_2

\frac{P_1V_1}{RT_1} = \frac{P_2V_2}{RT_2}

now plug in all data to find the final volume of the balloon

\frac{754\times 219}{R(298)} = \frac{210 \times V_2}{R(230)}

V_2 = \frac{230\times 754 \times 219}{210 \times 298}

V_2 = 606.9 m^3

8 0
2 years ago
A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elas
Karo-lina-s [1.5K]

Answer:

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

Explanation:

Given that

Yield strength ,Sy= 240 MPa

Tensile strength = 310 MPa

Elastic modulus ,E= 110 GPa

L=380 mm

ΔL = 1.9 mm

Lets find strain:

Case 1 :

Strain due to elongation (testing)

ε = ΔL/L

ε = 1.9/380

ε = 0.005

Case 2 :

Strain due to yielding

\varepsilon' =\dfrac{S_y}{E}

\varepsilon' =\dfrac{240}{110\times 1000}

ε '=0.0021

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

For computation of load strain due to testing should be less than the strain due to yielding.

4 0
3 years ago
Acceleration occurs whenever an ___________________ force acts on an object.
leonid [27]
Acceleration occurs whenever the forces on an object are unbalanced.

It's the group of forces on the object that's either balanced or unbalanced.
There's no such thing as "an unbalanced force".

8 0
3 years ago
Read 2 more answers
. Friction is a rubbing force that ___________ a spinning yo-yo.
Advocard [28]
The yo-yo speeds up when you rub it
3 0
2 years ago
Read 2 more answers
Illustrates an Atwood's machine. Let the masses of blocks A and B be 7.00 kg and 3.00 kg , respectively, the moment of inertia o
Harman [31]

Answer:  

A) 1.55  

B) 1.55

C) 12.92

D) 34.08

E)  57.82

Explanation:  

The free body diagram attached, R is the radius of the wheel  

Block B is lighter than block A so block A will move upward while A downward with the same acceleration. Since no snipping will occur, the wheel rotates in clockwise direction.  

At the centre of the whee, torque due to B is given by  

{\tau _2} = - {T_{\rm{B}}}R  

Similarly, torque due to A is given by  

{\tau _1} = {T_{\rm{A}}}R  

The sum of torque at the pivot is given by  

\tau = {\tau _1} + {\tau _2}  

Replacing {\tau _1} and {\tau _2} by {T_{\rm{A}}}R and - {T_{\rm{B}}}R respectively yields  

\begin{array}{c}\\\tau = {T_{\rm{A}}}R - {T_{\rm{B}}}R\\\\ = \left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R\\\end{array}  

Substituting I\alpha for \tau in the equation \tau = \left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R  

I\alpha=\left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R  

\frac{I\alpha}{R} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right  

The angular acceleration of the wheel is given by \alpha = \frac{a}{R}  

where a is the linear acceleration  

Substituting \frac{a}{R} for \alpha into equation  

\frac{I\alpha}{R} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right we obtain  

\frac{Ia}{R^2} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right  

Net force on block A is  

{F_{\rm{A}}} = {m_{\rm{A}}}g - {T_{\rm{A}}}  

Net force on block B is  

{F_{\rm{B}}} = {T_{\rm{B}}} - {m_{\rm{B}}}g  

Where g is acceleration due to gravity  

Substituting {m_{\rm{B}}}a and {m_{\rm{A}}}a for {F_{\rm{B}}} and {F_{\rm{A}}} respectively into equation \frac{Ia}{R^2} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right and making a the subject we obtain  

\begin{array}{c}\\{m_{\rm{A}}}g - {m_{\rm{A}}}a - \left( {{m_{\rm{B}}}g + {m_{\rm{B}}}a} \right) = \frac{{Ia}}{{{R^2}}}\\\\\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g - \left( {{m_{\rm{A}}} + {m_{\rm{B}}}} \right)a = \frac{{Ia}}{{{R^2}}}\\\\\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)a = \left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g\\\\a = \frac{{\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)}}\\\end{array}  

Since {m_{\rm{B}}} = 3kg and {m_{\rm{B}}} = 7kg  

g=9.81 and R=0.12m, I=0.22{\rm{ kg}} \cdot {{\rm{m}}^2}  

Substituting these we obtain  

a = \frac{{\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)}}  

\begin{array}{c}\\a = \frac{{\left( {7{\rm{ kg}} - 3{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{\left( {7{\rm{ kg}} + 3{\rm{ kg}} + \frac{{0.22{\rm{ kg/}}{{\rm{m}}^2}}}{{{{\left( {0.120{\rm{ m}}} \right)}^2}}}} \right)}}\\\\ = 1.55235{\rm{ m/}}{{\rm{s}}^2}\\\end{array}

Therefore, the linear acceleration of block A is 1.55 {\rm{ m/}}{{\rm{s}}^2}

(B)

For block B

{a_{\rm{B}}} = {a_{\rm{A}}}

Therefore, the acceleration of both blocks A and B are same

1.55 {\rm{ m/}}{{\rm{s}}^2}

(C)

The angular acceleration is \alpha = \frac{a}{R}

\begin{array}{c}\\\alpha = \frac{{1.55{\rm{ m/}}{{\rm{s}}^2}}}{{0.120{\rm{ m}}}}\\\\ = 12.92{\rm{ rad/}}{{\rm{s}}^2}\\\end{array}

(D)

Tension on left side of cord is calculated using

\begin{array}{c}\\{T_{\rm{B}}} = {m_{\rm{B}}}g + {m_{\rm{B}}}a\\\\ = {m_{\rm{B}}}\left( {g + a} \right)\\\end{array}

\begin{array}{c}\\{T_{\rm{B}}} = \left( {3{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2} + 1.55{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 34.08{\rm{ N}}\\\end{array}

(E)

Tension on right side of cord is calculated using

\begin{array}{c}\\{T_{\rm{A}}} = {m_{\rm{A}}}g - {m_{\rm{A}}}a\\\\ = {m_{\rm{A}}}\left( {g - a} \right)\\\end{array}

\begin{array}{c}\\{T_{\rm{A}}} = \left( {7{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2} – 1.55{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 57.82{\rm{ N}}\\\end{array}

6 0
2 years ago
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