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3 years ago

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An object in a fluid experiences a buoyant force from the fluid. If the object is completely immersed, on which does the magnitu

de of this force depend on?

1 answer:

loris [4]3 years ago

4 0

Answer:

Weight of the fluid that the object displaces.

Explanation:

When the fluid is completely immersed in a fluid, it experiences pressure from all the direction. While the object is immersed in the fluid a force acts on it in the opposite direction, i.e., upwards. This force is termed as buoyant force.

Also, as per the Archimedes' Principle, the force experience by the object is the same as the weight of the fluid that gets displaced by the object.

Thus on complete immersion of the object in the fluid, it experiences the force same as the weight of the fluid that gets displaced

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An oscillating dipole antenna 1.73 m long with a maximum 36.0 mV potential creates a 500 Hz electromagnetic wave. (a) What is th

sergejj [24]

Answer:

$E_0=0.021v/m$

Explanation:

From the question we are told that:

Length $l=1.73m$

Voltage $V=36.0mV$

Frequency $F=500Hz$

Generally the equation for maximum Electric Field is mathematically given by

$E_0=\frac{\triangle}{r}$

$E_0=\frac{36*10^{-3}}{1.73}$

$E_0=0.021v/m$

5 0

3 years ago

A heavy rope 6.00 m long and weighing 29.4 N is attached at one end to a ceiling and hangs vertically. A 0.500-kg mass is suspen

aivan3 [116]

Answer:

a) $v=3.1252\ m.s^{-1}$

b) $v=39.0672\ m.s^{-1}$

c) $v=8.2685\ m.s^{-1}$

d) No,

No.

Explanation:

Given:

length of rope, $l=6\ m$

weight of the rope, $w=29.4\ N$

mass suspended at the lower end of the rope, $M=0.5\ kg$

<u>Now the mass of the rope:</u>

$m=\frac{w}{g}$

$m=\frac{29.4}{9.8}$

$m=3.01\ kg$

<u>So the linear mass density of rope:</u>

$\mu=\frac{m}{l}$

$\mu=\frac{3.01}{6}$

$\mu=0.5017\ kg.m^{-1}$

We know that the speed of wave in a tensed rope is given as:

$v=\sqrt{\frac{F_T}{\mu} }$

where:

$F_T=$ tension force in the rope

a)

At the bottom of the hanging rope we have an extra mass suspended. So the tension at the bottom of the rope:

$F_T=M\times g$

$F_T=0.5\times 9.8$

$F_T=4.9\ N$

Therefore the speed of the wave at the bottom point of the rope:

$v=\sqrt{\frac{4.9}{0.5017} }$

$v=3.1252\ m.s^{-1}$

b)

Tension at a point in the middle of the rope:

$F_T=M\times g+\frac{w}{2}$

$F_T=0.5\times 9.8+\frac{29.4}{2}$

$F_T=19.6\ N$

Now wave speed at this point:

$v=\sqrt{\frac{19.6}{0.5017} }$

$v=39.0672\ m.s^{-1}$

c)

Tension at a point in the top of the rope:

$F_T=M\times g+w$

$F_T=0.5\times 9.8+29.4$

$F_T=34.3\ N$

Now wave speed at this point:

$v=\sqrt{\frac{34.3}{0.5017} }$

$v=8.2685\ m.s^{-1}$

d)

Tension at the middle of the rope is not the average tension of tension at the top and bottom of the rope because we have an extra mass attached at the bottom end of the rope.

Also the wave speed at the mid of the rope is not the average f the speeds at the top and the bottom of the ropes because it depends upon the tension of the rope at the concerned points.

7 0

3 years ago

A soft-drink bottler purchases glass bottles from a vendor. The bottles are required to have internal pressure strength of at le

maria [59]

Answer: K =24 psi

Explanation:

Given: Standard deviation =3psi

Internal pressure strength =157psi

Number of random bottle =n=64

K= 3 × square root of 64

K= 3×8=24 psi

If mean internal pressure K fall below K,

157-1.3=155.7psi

At 2%:

0.16×64 = 10.24

5 0

3 years ago

A rod of propoer length l0 is at rest in frame S'. it ilies in the x',y' plane and amkes an angle of sin^-1(3/5) What must be th

Blizzard [7]

Answer:

v = 1.98*10^8 m/s

Explanation:

Given:

- Rod at rest in S' frame

- makes an angle Q = sin^-1 (3/5) in reference frame S'

- makes an angle of 45 degree in frame S

Find:

What must be the value of v if as measured in S the rod is at a 45 degree)

Solution:

- In reference frame S'

x' component = L*cos(Q)

y' component = L*sin(Q)

- Apply length contraction to convert projected S' frame lengths to S frame:

x component = L*cos(Q) / γ (Length contraction)

y component = L*sin(Q) (No motion)

- If the rod is at angle 45° to the x axis, as measured in F, then the x and y components must be equal:

L*sin(Q) = L*cos(Q) / γ

Given: γ = c / sqrt(c^2 - v^2)

c / sqrt(c^2 - v^2) = cot(Q)

1 - (v/c)^2 = tan(Q)

v = c*sqrt( 1 - tan^2 (Q))

For the case when Q = sin^-1 (3/5)::

tan(Q) = 3/4

v = c*sqrt( 1 - (3/4)^2)

v = c*sqrt(7) / 4 = 1.98*10^8 m/s

5 0

3 years ago

(c) If η = 60% and TC = 40°F, what is TH, in °F?

arlik [135]

2b2t hope that helps

5 0

3 years ago