Consider the following statements. A. Heat flows from an object at higher temperature to an object at lower temperature; B. Heat flows from an object in liquid state to an object in solid state; C. Heat flows from an object with higher thermal energy to one with lower thermal energy. Which statements are true? 1. A only 2. A and C only 3. B and C only 4. None is true.

Answers

Answer 1

Answer:

Only ' A ' is always true. (choice-1)

' B ' is not true when you drop a red hot spoon into cold soup.

' C ' is not true when you drop a red hot marble into a cool swimming pool.

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Each plate of a parallel‑plate capacitor is a square of side 0.0479 m, and the plates are separated by 0.479 × 10 − 3 m. The capacitor is charged and stores 8.11 × 10 − 9 J of energy. Find the electric field strength E inside the capacitor.
A trumpet player hears 5 beats per second when she plays a note and simultaneously sounds a 440 Hz tuning fork. After pulling her tuning valve out to slightly increase the length of her trumpet, she hears 3 beats per second against the tuning fork. Was her initial frequency 435 Hz or 445 Hz? Explain.
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A typical sugar cube has an edge length of 1 cm. If you had a cubical box that contained a mole of sugar cubes, what would its edge length be? (One mole = 6.02 ✕ 1023 units.)

A particle (q = 5.0 nC, m = 3.0 μg) moves in a region where the magnetic field has components Bx = 2.0 mT, By = 3.0 mT, and Bz = −4.0 mT. At an instant when the speed of the particle is 5.0 km/s and the direction of its velocity is 120° relative to the magnetic field, what is the magnitude of the acceleration of the particle in m/s2?

Answers

The acceleration of the particle is 38.87 kg.

Net magnetic field

The net magnetic field is calculated as follows;

$B_(net) = √(B_x^2 + B_y^2 + B_z^2) \n\nB_(net) = √(2^2 + 3^2 + 4^2) = 5.385 \ mT$

Magnetic force on the charge

The magnetic force on the charge is calculated as follows;

$F = qvB * sin(\theta)\n\nF = 5* 10^(-9) * 5* 10^3 * 5.385 * 10^(-3) * sin(120)\n\nF = 1.166 * 10^(-7) \ N$

Acceleration of the particle

The acceleration of the particle is calculated as follows;

$a = (F)/(m) \n\na = (1.166 * 10^(-7))/(3 * 10^(-9)) \n\na = 38.87 \ kg$

Learn more about magnetic force here: brainly.com/question/13277365

Explanation:

It is given that,

Charge on the particle, $q=5\ nC=5* 10^(-9)\ C$

Mass of the particle, $m=3\ \mu g=3* 10^(-6)\ g=3* 10^(-9)\ kg$

Magnetic field component, $B_x=2\ mT,B_y=3\ mT,B_z=-4\ mT$

Net magnetic field, $B=√(2^2+3^2+4^2)=5.38\ mT=0.00538\ T$

Speed of the particle, v = 5 km/s = 5000 m/s

Angle between velocity and magnetic field, $\theta=120$

Magnetic force is given by :

$F=qvB\ sin\theta$

$F=5* 10^(-9)* 5000\ m/s* 0.00538* sin(120)$

$F=1.16* 10^(-7)\ N$

Acceleration of the particle is given by, $a=(F)/(m)$

$a=(1.16* 10^(-7)\ N)/(3* 10^(-9)\ kg)$

$a=38.6\ m/s^2$

So, the acceleration of the particle is 38.6 m/s². Hence, this is the required solution.

Water on Earth was (a) transported here by comets; (b) accreted from the solar nebula; (c) produced by volcanoes in the form of steam; (d) created by chemical reactions involving hydrogen and oxygen shortly after Earth formed.

Answers

Answer: Water on Earth was transported here by comets. The correct option is A.

Explanation:

Comets are made up of water with ice, rock and minerals.

Alot of research and hypotheses has been made to prove the origin of water on planet earth. Extraplanetary source such as comets, trans-Neptunian objects, and water-rich meteoroids (protoplanets) are believed to have delivered water to Earth.

The type of function that describes the amplitude of damped oscillatory motion is _______. The type of function that describes the amplitude of damped oscillatory motion is _______. quadratic sinusoidal inverse exponential linear

Answers

Answer:

exponential

Explanation:

type of function that describes the amplitude of damped oscillatory motion is exponential because as we know that here function is

y = A × $e^{(-bt)/(2m)}$ × cos(ωt + ∅ ) ..................................... ( 1 )

here function A × $e^{(-bt)/(2m)}$ is amplitude

as per equation ( 1 )it is exponential

so that we can say that amplitude of damped oscillatory motion is exponential

1. The resistance of an electric device is 40,000 microhms. Convert that measurement to ohms2. When an electric soldering iron is used in a 110 V circuit, the current flowing through the iron is
2 A. What is the resistance of the iron?
3. A current of 0.2 A flows through an electric bell having a resistance of 65 ohms. What must be
the applied voltage in the circuit?

Answers

Answer:

(1) 0.04 ohms (2) 55 ohms (3) 13 volt

Explanation:

(1) The resistance of an electric device is 40,000 microhms.

We need to convert it into ohms.

$1\ \mu \Omega =10^(-6)\ \Omega$

To covert 40,000 microhms to ohms, multiply 40,000 and 10⁻⁶ as follows :

$40000 \ \mu \Omega =40000 * 10^(-6)\ \Omega\n\n=0.04\ \Omega$

(2) Voltage used, V = 110 V

Current, I = 2 A

We need to find the resistance of the iron. Using Ohms law to find it as follows :

V = IR, where R is resistance

$R=(V)/(I)\n\nR=(110)/(2)\n\nR=55\ \Omega$

(3) Current, I = 0.2 A

Resistance, R = 65 ohms

We need to find the applied voltage in the circuit. Using Ohms law to find it as follows :

V=IR

V = 0.2 × 65

V = 13 volt

Answer:

1. 0.04 Ohms

2. 55 Ohms

3. 13 Volts

Explanation:

Penn Foster

Points A, B, and C are at the corners of an equilateral triangle of side 8 m. Equal positive charges of 4 mu or micro CC are at A and B. (a) What is the potential at point C? 8.990 kV * [2.5 points] 2 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 8.990 OK (b) How much work is required to bring a positive charge of 5 mu or micro CC from infinity to point C if the other charges are held fixed? .04495 J * [2.5 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] .04495 OK (c) Answer parts (a) and (b) if the charge at B is replaced by a charge of -4 mu or micro CC. Vc= kV [2.5 points] 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] W =

Answers

Answer:

a) 8.99*10³ V b) 4.5*10⁻² J c) 0 d) 0

Explanation:

The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
For a point charge, it can be expressed as follows:

$V =(k*q)/(d)$

As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at any other corner, as follows:

$V = (2*q*k)/(d) = (2*8.99e9N*m2/C2*4e-6C)/(8m) =\n \n V= 8.99e3 V$

The potential at point C is 8.99*10³ V

The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:

$W = V * q = 8.99e3 V* 5e-6C = 4.5e-2 J$

The work needed is 0.045 J.

If we replace one of the charges creating the potential at the point C, by one of the same magnitude, but opposite sign, we will have the following equation:

$V = (8.99e9N*m2/C2*(4e-6C))/(8m) + ((8.99e9N*m2/C2*(-4e-6C))/(8m)) = 0$

This means that the potential due to both charges is 0, at point C.

If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.

The erg is a unit of work in units of centimeters (cm), grams (g), and seconds (s), and 1 erg=1 g⋅cm^2/s^2 . Recall that the SI unit of force is the newton (N) and is equal to kg⋅m/s^2 . You push an object 0.032 m by exerting 0.010 N of force. The work is the force times the distance.How much work have you done, expressed in ergs?

A: 3,200 ergs

B: 32 ergs

C: 0.32 ergs

D: 0.00032 ergs

Answers

For the given problem, the amount of work done expressed in ergs is 3200 ergs.

Answer: Option A

Explanation:

The work done on an objects are the force acting on it to move the object to a particular distance. So, work done on the object will be directly proportional to the force acting on it and the displacement.

Here, the force acting on the object is given as 0.010 N and the displacement of the object is 0.032 m. So, the work done on the object is

$\text { Work done }=\text { Force } * \text { displacement }$

$\text { Work done }=0.010 \mathrm{N} * 0.032 \mathrm{m}=0.00032 \mathrm{Nm}$

It is known that $1 N=1 \mathrm{kg} \mathrm{ms}^(-2)$

So, the work done can be expressed in $k g m s^(-2)$ as,

$\text { Work done }=0.00032 \mathrm{kgm}^(2) \mathrm{s}^(-2)$

It is known that $1 \mathrm{erg}=1 \mathrm{g} \mathrm{cm}^(2) / \mathrm{s}^(2)$ , so the conversion of units from Nm to erg will be done as follows:

$\text { Work done }=0.00032 \mathrm{kgm}^(2) \mathrm{s}^(-2) * \frac{1000 \mathrm{g}}{1 \mathrm{kg}} * \frac{100 * 100 \mathrm{cm}^(2)}{m^(2)}=3200 \mathrm{g} \mathrm{cm}^(2) \mathrm{s}^(-2)$

Thus, work done in ergs is 3200 ergs.

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If you have two substances, one with a density of 2.0 g/cm3 and one with a density of 1.3 g/cm3 and you combined them, which one would float on topother and why?

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.777 m and 2.67 kg, respectively. When the propellor rotates at 573 rpm (revolutions per minute), what is its rotational kinetic energy?