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3 years ago

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Two objects, labeled A and B, are the same size. Object A has a density of 1.21 g/cm3. Object B has a density of 1.37 g/cm3. Bot

h are placed in a beaker of water. Which will float higher in the water?

2 answers:

Sedaia [141]3 years ago

4 0

Neither object will float in water.

They both have densities greater than 1.0 g/cm³,
so both will sink to the bottom of the beaker.

Like rocks.

deff fn [24]3 years ago

4 0

Answer:

None of the object float in water

Explanation:

When an object float in water than we can say that buoyancy force due to water will counter balance the weight of the object

As we can say that

$W_g = F_b$

so here if object is just floating or we can say that if object is in the state of sinking then we have

$\rho_{object} V g = \rho_{water} Vg$

so we have

$\rho_{object} = \rho_{water}$

so from above relation we can say that

i) $\rho_{object} > \rho_{water}$ then it will sink inside water

ii) $\rho_{object} < \rho_{water}$ then it will float in water

since we know density of water is 1 gm/cm^3 so here given both objects are having density more than this

so both the objects will sink inside water

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An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the

alexandr402 [8]

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

5 0

3 years ago

The "Biltmore Agreement" stipulated that a. remote broadcasts could not emanate from hotels. b. NBC could not give away tickets

Anna35 [415]

Answer:

The "Biltmore Agreement" stipulated that:

Radio stations agreed to broadcast no longer than five minutes of news, twice per day, while using information supplied by the newspapers.

e. radio stations could only air five-minutes newscasts a day.

Explanation:

The Biltmore Agreement tried to reconcile within the press war between newspapers and radio, as during its golden age the newspapers´ revenues decreased. Radio´s brand new technology was more attractive and creative for advertising and could report breaking news faster than the newspapers, which through the press associations including the Associated Press and the United Press, pressured to stop providing news to radio stations beginning a war in 1933, which partially ended with the Biltmore Agreement, which restricted the radio´s broadcasting of news if the newspapers continued publishing radio listings, radio stations were to broadcast no longer than five minutes of news, twice per day, if information supplied by the newspapers was used, no sponsors were allowed, and no more that 30 words in a single story were allowed either; radio stations had to include: "See your daily newspaper for further details" in their announcements and, could only broadcast news after 9:30 AM for morning news, and after 9:00 PM for evening news, so people would have already received their newspapers.

7 0

3 years ago

Ilia_Sergeevich [38]

Answer:

Potential

Explanation:

The water in the bathtub has more potential energy than that in the tea cup because it has a greater number of water molecules.

Mass is a parameter that is very instrumental in determining potential energy.

The potential energy of a body is the energy due to the position of that body.

Potential energy = mass x acceleration due gravity x height

Mass is the amount of matter in a substance. Water in the tub will have more mass and contain a greater number of water molecules there in.

Since potential energy is directly proportional to mass, then, it will have a greater amount of potential energy.

3 0

2 years ago

Allen and Jason are chucking a speaker around. On one particular throw, Allen throws the speaker, which is playing a pure tone o

Sholpan [36]

Answer:

Explanation:

We shall apply Doppler's effect of sound .

speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at 6 m/s .

apparent frequency = $f_o\times\frac{V+v_o}{ V-v_s}$

V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .

Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f

apparent frequency = $f\times \frac{340+6}{340-10}$

= $f\times \frac{346}{330}$

So m = 346 , n = 330 .

8 0

3 years ago

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