Identify the following items as scalar or vector.

Cylindrical beaker of height 0.100 mm and negligible weight is filled to the brim with a fluid of density rhorhorho = 890 kg/m3k

Nesterboy [21]

Incomplete part of the question

A ball of density ρb = 5000 kg/m3 and volume V = 60.0 cm3 is then submerged in the fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81 m/s2 . What is the weight Wb of the ball? Express your answer numerically in newtons.

What is the reading W2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale. Calculate your answer from the quantities given in the problem and express it numerically in newtons.

What is the force Fr applied to the ball by the rod? Take upward forces to be positive (e.g., if the force on the ball is downward, your answer should be negative). Express your answer numerically in newtons.

The rod is now shortened and attached to the bottom of the beaker. The beaker is again filled with fluid, the ball is submerged and attached to the rod, and the beaker with fluid and submerged ball is placed on the scale.

What weight W3 does the scale now show?

Answer:

(a) 2.94 N

(b) 1 N

(c) 2.42 N

(d) 3.42 N

Explanation:

(a)

From the definition of density, it's mass per unit volume hence mass is a product of density and volume. To get weight, we multiply mass by acceleration due to gravity

The weight of the ball is $W=\rho g V$

Where $\rho$ is the density, V is volume and g is acceleration due to gravity

Substituting density for 5000 Kg/m3 and g for 9.8 m/s2 and v for 0.00006 m3 then

$W= 5000 kg/m^{3} * 9.8 m/s^{2} * 0.00006 m^{3}=2.94 N$

(b)

Because the ball is being held up mostly by the rod, the fluid pressure on the bottom of the cylinder is just the same as before.

The scale does not "know" the ball is there at all.

That's why it still reads 1 N.

Therefore, the reading is 1 N

(c)

The buoyant force of the fluid on the ball is equal to the weight of the displaced fluid, namely,

$890 kg/m^{3} * 9.8 m/s^{2} * 0.00006 m^{3} = 0.52 N$

so the force needed for the rod to hold up the ball is 2.94 N - 0.52 N = 2.42 N.

(d)

Now the scale "feels" the weight of the ball,

so the scale reads the weight of the ball

PLUS the weight of the original fluid

MINUS the weight of the fluid that was displaced

= 2.94 N + 1.00 N - 0.52 N = 3.42 N

6 0

3 years ago

A rectangular block weighs 240 N. the area of the block in contact with the floor is 20 cm2.calculate the pressure on the floor(

maks197457 [2]

Answer:

12 N/cm²

Explanation:

From the question given above, the following data were obtained:

Weight (W) of block = 240 N

Area (A) = 20 cm²

Pressure (P) =?

Next, we shall determine the force exerted by the block. This can be obtained as follow:

Weight (W) of block = 240 N

Force (F) =.?

Weight and force has the same unit of measurement. Thus, we force applied is equivalent to the weight of the block. Thus,

Force (F) = Weight (W) of block = 240 N

Force (F) = 240 N

Finally, we shall determine the pressure on the floor as follow:

Force (F) = 240 N

Area (A) = 20 cm²

Pressure (P) =?

P = F/A

P = 240 / 20

P = 12 N/cm²

Therefore, the pressure on the floor is 12 N/cm².

7 0

3 years ago

A 87 kgkg skydiver can be modeled as a rectangular "box" with dimensions 18 cmcm ×× 47 cmcm ×× 180 cmcm . If he falls feet first

garik1379 [7]

Answer:

Therefore the terminal velocity = 1.45 m/s

Explanation:

Terminal velocity: Terminal velocity is the highest velocity of an object when it falls from rest trough a media.

$V_t=\sqrt{\frac{2w}{c_d\rho A}}$

$V_t$ = terminal velocity

w = weight of the object = mg

$c_d$ = drag coefficient=0.80

A= frontal area

$\rho$ = media density = 1.2 kg/m³

m = mass = 8 kg

g= acceleration due to gravity = 9.8 m/s²

Front area = length× breadth

= (18×47)cm²

=846 cm²

Therefore the terminal velocity

$v_t= \sqrt{\frac{2\times 87\times 9.8}{0.80\times 846 \times 1.2}$

=1.45 m/s

Therefore the terminal velocity = 1.45 m/s

8 0

4 years ago

Suppose the electric field in some region is found to be E = kr3 ˆr, in spherical coordinates (k is some constant). (a) Find the

Assoli18 [71]

Answer:

Part a)

$\rho = 3\epsilon_0 k r^2$

Part b)

$Q = 4\pi \epsilon_0kR^5$

Explanation:

Part a)

As we know that electric field intensity due to some given charge distribution is given as

$E = kr^3 \hat r$

now electric flux through a spherical surface of radius r is given as

$\phi = E. A$

$\phi = kr^3(4\pi r^2)$

now by Guass law we know that

$E.A = \frac{q}{\epsilon_0}$

$q = 4\pi \epsilon_0kr^5$

now volume charge density is given as

$\rho = \frac{q}{\frac{4}{3}\pi r^3}$

$\rho = 3\epsilon_0 k r^2$

Part b)

Total charge inside the radius R is given as

$Q = 4\pi \epsilon_0kR^5$

7 0

3 years ago

21. An AC generator has an rms output voltage of 274 V at a frequency of 57.0 Hz.

maks197457 [2]

The answers are as follows;

a) the inductive reactance is 322 ohm

b) The maximum voltage is 387.5 V

c) The rms and maximum currents in the inductor are 1.2 A and 0.85 A.

<h3>What is the reactance?</h3>

The reactance is obtained from;

XL = 2πfL

XL = 2 * 3.14 * 57.0 * 0.900

XL = 322 ohm

The maximum voltage is obtained as;

Vo = Vrms * √2

Vo = 274 V * √2

Vo = 387.5 V

Io = Vo/XL

Io = 387.5 V/ 322 ohm

Io = 1.2 A

Irms = 274 V/322 ohm

Irms = 0.85 A

Learn more about inductive reactance:brainly.com/question/17129912

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4 0

2 years ago