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alexandr402 [8]
2 years ago
6

You have two identical pure silver ingots. You place one of them in a glass of water and observe it to sink to the bottom. You p

lace the other in a container full of mercury and observe that it floats. Comparing the buoyant forces in the two cases you conclude that
a.) the buoyant force in water is smaller than in mercury

b.) the buoyant force in the water is larger than that in mercury

c.) the buoyant force in the water is zero and that in mercury is non - zero

d.) the buoyant force in the water is equal to that in mercury

e.) no conclusion can be made about the respective values of the buoyant forces
Physics
1 answer:
Elis [28]2 years ago
3 0

Answer: a)

Explanation:

The buoyant force, as stated by Archimedes’ principle, is equal to the weight of the liquid that occupies the same volumen as the submerged object, as follows:

Fb = δ.V.g

If this force is larger than the weight of the object (that means that the fluid is denser than the solid), the object floats, which is the case for silver and mercury.

Instead, silver density is larger than water density, which explains why the pure silver ingot sinks.

Finally, as mercury is denser than water, we conclude that for a same object, the buoyant force in mercury is larger than in water (exactly 13.6 times greater).  

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A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.
Airida [17]

Answer:

Part A :  E =   \frac{1}{4\pi}ε₀ Q₁/R₁² Volt/meter

Part B :  V =  \frac{1}{4\pi}ε₀ Q₁/R₁ Volt

Explanation:

Given that,

Charge distributed on the sphere is Q₁

The radius of sphere is R

₁

The electric potential at infinity is 0

<em>Part A</em>

The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.  

If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

                          F = \frac{1}{4\pi}ε₀ Q₁/R₁²

Then the electric field at that point is

                                   E =  F/1

                            E =  \frac{1}{4\pi}ε₀ Q₁/R₁²  Volt/meter

Part B

The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.

Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

                           V =  \frac{1}{4\pi}ε₀ Q₁/R₁  Volt

4 0
1 year ago
A Wooden block has a mass of 0.200kg, a specific heat of 710 J (kg times degrees Celsius and is at a temperature of 20.0 degrees
olchik [2.2K]

Answer:

35°C

Explanation:

q = mCΔT

2130 J = (0.200 kg) (710 J/kg/°C) (T − 20.0°C)

T = 35°C

8 0
2 years ago
A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end tex
love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

\Delta x = v_{i}.t + \frac{a}{2}.t^{2}

v_{i} is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

\Delta x = 5(6) - \frac{4}{2}.6^{2}

\Delta x = 30 - 2.36

\Delta x = - 42

Displacement of the boat for t=6.0s interval is \Delta x = - 42m, i.e., 42 m to the left.

8 0
2 years ago
Jake uses a fire extinguisher to put out a small fire. When he squeezes the handle, the flame rettardant is released from the ex
Tpy6a [65]
Can you attach a picture of the actual problem?
7 0
2 years ago
Read 2 more answers
An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in ser
s2008m [1.1K]

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

V_L=V_oe^{-\frac{Rt}{L}}

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s

hence, after 1.6s the inductor will have a potential difference of 24V

3 0
2 years ago
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