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padilas [110]
3 years ago
11

As a liquid is added to a beaker, the pressure exerted by the liquid on the bottom

Physics
1 answer:
abruzzese [7]3 years ago
4 0

Answer: c) increases

Explanation:

Pressure increases with decreasing height

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Fill in the blank: An object is most likely to sink in water if _________________.
drek231 [11]
High density
random words to fill up 20 character minimum for answering questions :P
4 0
3 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
3 years ago
Why is it important that radioisotopes used in diagnostic tests have short half-lives? 20 A) These radioisotopes have a greater
LiRa [457]

Answer:

Option B) This minimizes the harmful side effects of the radiations

Explanation:

Half-life is the time taken for the decay of an radio-active atom in which it disintegrates such that it becomes half of its value at the beginning.... The nuclei should be in active mode for a longer duration sufficient for the treatment of the condition but these nuclei should have a sufficient  shorter half life so that they don't get enough time to cause any  damage to the health of the person other than treating the cause.

A shorter half life gives the assurance that the radiation after the treatment will leave the body without getting accumulated and cause harm to the body cells and other organs.

8 0
3 years ago
Which of the following is the smallest conceivable amount of time that could pass between a lunar eclipse and a solar eclipse .
pickupchik [31]
The time required for a moon to orbit around the earth is about 27-28 days

In order for lunar eclipse to occur the line that should be formed is:
Sun-Earth-Moon
because earth is making shade on moon

in order for solar eclipse to occur the line is now:
Sun-Moon-Earth
because moon is making a shade on earth (blocking sun = solar eclipse)

Therefore moon needs to make half of its orbit to go from behind the earth to in front of the earth.

28/2 = 14

Answer is 14
5 0
3 years ago
Question 2.
boyakko [2]

The tensile stress of the wire supporting 2 kg mass is determined as 6.1 x 10⁷ N/m².

<h3>Tensile stress of the wire</h3>

The tensile stress of the wire is calculated as follows;

σ = F/A

where;

  • A is area of the wire

A = πr² = πD²/4

where;

  • D is diameter = 0.64 mm

A = π x (0.64 x 10⁻³)²/4

A = 3.22 x 10⁻⁷ m²

σ = F/A = (mg)/A = (2 x 9.8)/( 3.22 x 10⁻⁷)

σ = 6.1 x 10⁷ N/m²

Learn  more about tensile stress here: brainly.com/question/25748369

#SPJ1

3 0
2 years ago
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