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hichkok12 [17]
2 years ago
15

Three balls are in water. Ball 1 floats, with half of it exposed above the water level. Ball 2, with a density less than the den

sity of water is held below the surface by a cord anchored to the bottom of the container, so that it is fully submerged. Ball 3, of same radius as ball 2, but of greater mass, is suspended from a rope so that it is fully submerged.
Which is true for Ball 1? Choose the best answer.

1) The magnitude of the buoyancy force is exactly one half of that of the ball's weight.
2) The magnitude of the buoyancy force is more than that of the ball's weight.
3) The magnitude of the buoyancy force is less than that of the ball's weight.
4) The magnitude of the buoyancy force is equal to that of the ball's weight.
Physics
2 answers:
goblinko [34]2 years ago
7 0

Answer:

4) The magnitude of the buoyancy force is equal to that of the ball's weight

Explanation:

Since ball 1 floats without sinking then it is in equilibrium. This means that the net downward force (weight) is balanced by the net upward force (buoyancy forces).

Ymorist [56]2 years ago
4 0

Answer:

The magnitude of buoyancy force is equal to that of ball's weight.

Explanation:

Ball 1 is floating on water. Weight of ball 1 is Fg=m1g  is acting vertically downward

Force of buoyancy FB = ρVdisg is acting vertically upward.

Net force acting on the ball is zero, FB=Fg

Answer

The magnitude of buoyancy force is equal to that of ball's weight.

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Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

8 0
2 years ago
a closed tank is partially filled with glycerin. if the air pressure in the tank is 6 lb/in.2 and the depth of glycerin is 10 ft
KATRIN_1 [288]

Answer:

<u><em>note:</em></u>

<u><em>solution is attached due to error in mathematical equation. please find the attachment</em></u>

4 0
1 year ago
Umar has two copper pans, each containing 500cm3 of water. Pan A has a mass of 750g and pan B has a mass of 1.5kg. Which pan wil
Olin [163]

Answer:

heat required in pan B is more than pan A

Explanation:

Heat required to raise the temperature of the substance is given by the formula

Q = ms\Delta T

now we know that both pan contains same volume of water while the mass of pan is different

So here heat required to raise the temperature of water in Pan A is given as

Q_1 = (m_w s_w + m_ps_p)\delta T

Q_1 = (0.5(4186) + 0.750(s))\Delta T

Now similarly for other pan we have

Q_2 = (m_w s_w + m_ps_p)\delta T

Q_2 = (0.5(4186) + 1.50(s))\Delta T

So here by comparing the two equations we can say that heat required in pan B is more than pan A

3 0
2 years ago
The acceleration due to gravity for any object, including 1 washer on the string, is always assumed to be m/s2. The mass of 3 wa
Pachacha [2.7K]

Answer:

The force will increase in proportion to the mass of the objects

Explanation:

The acceleration due to gravity is always the same. It is expressed in meters per second squared or m/s². The figure of 9.81 m/s² is an average value that was taken after calculating the acceleration under different surfaces. In fact, the acceleration differs depending on the shape of the part of the earth in relation to the earth's magnetic field and force.

Thus, if one washer was 20 kg, the acceleration being 9.81 m/s² the weight will be:

F = ma

  = 20 * 9.81\\= 196.2 N

If there are there washers, the weight will be:

F = 3 * 20 * 9.81

  = 588.6 N

5 0
2 years ago
Read 2 more answers
A very long conducting tube (hollow cylinder) has inner radius a and outer radius b. It carries charge per unit length +α, where
patriot [66]

Answer:

A) i) E =α/ [2πrL(εo)]

ii) E=0

iii) E = α/(πrεo)

The graph between E and r for the 3 cases is attached to this answer ;

B) i) charge on the inner surface per unit length = - α

ii) charge per unit length on the outer surface = 2α

Explanation:

A) i) For r < a, the charge is in the cavity and takes a shape of the cylinder. Thus, applying gauss law;

EA = Q(cavity) / εo

Now, Qcavity = αL

So, E(2πrL) = αL/εo

Making E the subject of the formula, we have;

E =α/ [2πrL(εo)]

ii) For a < r < b; since the distance will be in the bulk of the conductor, therefore, inside the conductor, the electric field will be zero.

So, E=0

iii) For r > b; the total enclosed charge in the system is the difference between the net charge and the charge in the inner surface of the cylinder.

Thus, Qencl = Qnet - Qinner

Qinner will be the negative of Qnet because it should be in the opposite charge of the cavity in order for the electric field to be zero. Thus;

Qencl = αL - (-αL) = 2αL

Thus, applying gauss law;

EA = Qencl / εo

Thus, E = Qencl / Aεo

E = 2αL/Aεo

Since A = 2πrL,

E = 2αL/2πrLεo = α/(πrεo)

B) i) The charge on the cavity wall must be the opposite of the point charge. Therefore, the charge per unit length in the inner surface of the tube will be - α

ii)Net charge per length for tube is +α and there is a charge of - α on the inner surface. Thus charge per unit length on the outer surface will be = +α - (- α) = 2α

7 0
1 year ago
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