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Explanation:

a long straight wire __. a. increases with distance in a linear relationship b. increases with distance in a non-linear relationship c. decreases with distance in an inverse (1/r) relationship d. decreases with distance in an inverse-square (1/r2) relationship

In part A of the lab, we see the magnetic field of a long straight wire decreases with distance in an inverse (1/r) relationship, therefore the correct option is C.

A magnetic field could be understood as an area around a magnet, magnetic material, or an electric charge in which magnetic force is exerted. The SI unit of the magnetic field is tesla.

For a  long straight wire carrying the current, the relation with the distance as given below

B = μI/(2πr)

where B is the magnetic field

μ is the permeability of the free space

r is the distance from the wire

As we can see from the above relation

the magnetic field of a long straight wire decreases with distance in an inverse (1/r) relationship, therefore the correct answer is option C.

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Answer:

Alpha particles are more massive than beta particles.

Explanation:

The alpha particles are also called double-positive Heilum Nuclei because they have a charge of "+2" and a mass of 4 a.m.u. The properties of the alpha particles are as follows:

1. It possesses high energy due to high velocity. It is 7.7 MeV for most energetic from Rac (i.e: Bismuth-214)

2. It has a very high ionizing power. A 7.7 MeV particle produces about 0.2 x 10⁶ ions.

3. The range of alpha particles is very small. It is about 7 x 10⁻² m and only 4 x 10⁻⁵ m in aluminum for 7.7 MeV alpha-particle.

4. Alpha particles produce fluorescence on striking certain substances, such as zinc sulphide and bariumplatinocynide.

The beta particles are fast-moving electrons, which have a negligible mass.  

Hence, the correct option is:

Alpha particles are more massive than beta particles.

Answer:

what,     100.999m

Explanation:

convert 999 mm into meters, which is 0.999m and add that to a 100 m and that will make the total 100.999 m

The result of the addition of the two values is equal to 100.999 meters.

Given the following data:

To determine the result of the addition of the two values:

First of all, we would convert the value in millimeter (mm) to meter (m) as follows:

Conversion:

1 millimeter = 0.001 meter

999 millimeter = X meter

Cross-multiplying, we have:

[tex]X = 0.001 \times 999[/tex]

X = 0.999 meter.

For the result:

[tex]Result = 0.999 +100[/tex]

Result = 100.999 meters.

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Assuming ideal conditions, Boyle's law says that

P₁ V₁ = P₂ V₂

where P₁ and V₁ are the initial pressure and temperature, respectively, and P₂ and V₂ are the final pressure and temperature.

So you have

(455 mm Hg) (56.5 m³) = (632 mm Hg) V₂

==>   V₂ = (455 mm Hg) (56.5 m³) / (632 mm Hg) ≈ 40.7 m³

Answer:

1.24

Explanation:

The resistivity of copper[tex]\rho_1=2.65\times 10^{-8}\ \Omega-m[/tex]

The resistivity of Aluminum,[tex]\rho_2=1.72\times 10^{-8}\ \Omega-m[/tex]

The wires have same resistance per unit length.

The resistance of a wire is given by :

[tex]R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}[/tex]

According to given condition,

[tex]\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24[/tex]

So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.

Answer:

"the minimum frequency of radiation that will produce a photoelectric effect."

Explanation:

That answer was derived from gogle cuz my explanations was harder to explain but good luck

Solution :

Let [tex]$d_1=\frac{5.5}{2}[/tex]

          = 2.75 m

[tex]d_2 = 0.21 \ m[/tex]

And [tex]$d=|d_1-d_2|$[/tex]

       [tex]$d=(d_1+d_2) - (d_1-d_2)$[/tex]

       [tex]$d=(2.75+0.21) - (2.75-0.21)$[/tex]

       [tex]$d = 2.96-2.54$[/tex]

       [tex]d = 0.42 \ m[/tex]

a). At minimum,

[tex]$d=\frac{\lambda}{2}$[/tex]

[tex]$\lambda = 2d$[/tex]

  = 2 x 0.42

  = 0.84 m

Frequency, [tex]$\nu = \frac{v}{\lambda}$[/tex]

                      [tex]$=\frac{340}{0.84}$[/tex]

                      = 404.76 Hz

Therefore, the frequency of he sound, [tex]$\nu$[/tex] = 404.76 Hz

b). At maximum, λ = d = 0.42 m

Therefore, the frequency, [tex]$\nu = \frac{v}{\lambda}[/tex]

                                             [tex]$=\frac{350}{0.42}$[/tex]

                                             = 809.52 Hz

Answer:

0

Explanation:

The speed of the ball when it reaches the floor is 0 because when an object is at rest or in uniform motion, it has no speed/velocity

The final speed of the ball when it reaches the floor is 7.10 m/s.

The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but only converted from one form to another or transferred from one system to another. In other words, the total amount of energy in a closed system remains constant over time, even though it may be converted from one form to another.

This principle is based on the first law of thermodynamics, which states that the total energy of a closed system is always conserved, and can only be changed by the transfer of heat, work, or matter into or out of the system. The conservation of energy has important applications in various fields of physics, including mechanics, thermodynamics, and electromagnetism, and is a fundamental principle in the understanding of the natural world.

Here in the Question,

We can use the conservation of energy to solve this problem. Initially, the ball has kinetic energy due to its motion on the tabletop, but no potential energy since it is at a constant height. When the ball rolls off the edge of the table, it loses some kinetic energy due to friction but gains potential energy as it moves upward. When it reaches the floor, it has gained potential energy but lost kinetic energy due to friction. We can assume that the energy lost due to friction is converted to thermal energy, so the total energy of the system is conserved.

Let's start by calculating the potential energy gained by the ball as it moves from the edge of the table to the floor:

ΔPE = mgh

where ΔPE is the change in potential energy, m is the mass of the ball, g is the acceleration due to gravity, and h is the vertical distance traveled by the ball.

ΔPE = (0.50 kg)(9.81 m/s^2)(1.0 m) = 4.905 J

Now we can use the conservation of energy to find the final kinetic energy of the ball, which will allow us to calculate its final speed:

KEi + ΔPEi = KEf + ΔPEf

where KEi and ΔPEi are the initial kinetic and potential energies of the ball, respectively, and KEf and ΔPEf are the final kinetic and potential energies of the ball, respectively.

Since the ball is not bouncing, we can assume that its initial and final potential energies are zero. Therefore:

KEi = KEf + ΔKE

where ΔKE is the change in kinetic energy due to friction.

We can assume that the coefficient of kinetic friction between the ball and the incline is constant, and use the work-energy principle to find ΔKE:

Wfric = ΔKE

where Wfric is the work done by friction.

The work done by friction can be expressed as:

Wfric = ffricd

where ffric is the force of friction and d is the distance traveled by the ball on the incline.

The force of friction can be expressed as:

ffric = μmg

where μ is the coefficient of kinetic friction, and m and g have their usual meanings.

Putting it all together, we get:

KEi = KEf + ffricd

KEi = KEf + μmgd

(1/2)mv^2 = (1/2)mu^2 + μmgd

v^2 = u^2 + 2gd

where u is the initial speed of the ball on the tabletop, and v is the final speed of the ball on the floor.

Plugging in the given values, we get:

v^2 = (5.0 m/s)^2 + 2(9.81 m/s^2)(1.0 m)

v^2 = 50.405

v = 7.10 m/s

Therefore, the final speed of the ball when it reaches the floor is 7.10 m/s.

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Answer:

A freely falling object has weight W=mg, where W-weight, m-mass of the object and g-acceleration produced due to the earth's gravity. ... This happens because the normal reaction force exerted on the object in the lift is equal to zero, and normal force equals to mg, which in turn equals the weight of the object

Explanation:

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Answer:

Gradually increases until the maximum weight reaches the surface of the earth

Explanation:

Answer:

24m

Explanation:

you can use the formula

s=ut+1/2at²

s=0+1/2(3)(4)²

=1/2(3)(8)

=24m

I hope this helps

Answer:

A circuit with two 10 ohm resistors connected in series.

Explanation:

The formula for the equivalent resistance for resistors in parallel is

[tex]\frac{1}{Rt} = \frac{1}{R1} + \frac{1}{R2}[/tex]   So if R1=R2= 10  [tex]\frac{1}{Rt} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} <=> Rt =\frac{10}{2} =5 ohm[/tex]

The formula for the equivalent resistance for resistors in series is

Rt = R1 + R2  So Rt= 10 + 10 = 20

Answer:

Following are the solution to the given question:

Explanation:

For charging plates that are connected in a similar manner:

Calculating the total charge:

[tex]\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C[/tex]

Calculating the common potential:

[tex]\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\[/tex]

Calculating the charge after redistribution:

[tex]When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2[/tex]        

[tex]\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\ \\ \to q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C[/tex]

Answer:

A. True

Explanation:

Answer:

1.74×10⁻³ m

Explanation:

Applying,

ε = Stress/strain............. Equation 1

Where ε = Young's modulus

But,

Stress = F/A.............. Equation 2

Where F = Force, A = Area

Strain = e/L.............. Equation 3

e = extension, L = Length.

Substitute equation 2 and 3 into equation 1

ε = (F/A)/(e/L) = FL/eA............. Equation 4

From the question,

Given: F = 285 N, L = 29 cm = 0.29 m, ε = 5.00×10⁹ N/m²,

A = πd²/4 = 3.14(0.0011²)/4 = 9.4985×10⁻⁶ m²

Substitute these values into equation 4

5.00×10⁹ = (285×0.29)/(9.4985×10⁻⁶×e)

Solve for e

e = (285×0.29)/(5.00×10⁹×9.4985×10⁻⁶)

e = 82.65/4.74925×10⁴

e = 1.74×10⁻³ m

Answer:

Before the parachute opens: Immediately on leaving the aircraft, the skydiver accelerates downwards due to the force of gravity. There is no air resistance acting in the upwards direction, and there is a resultant force acting downwards. The skydiver accelerates towards the ground.

Once the parachute is opened, the air resistance overwhelms the downward force of gravity. The net force and the acceleration on the falling skydiver is upward. An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down.

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Answer:

the distance moved by the puck after 2.5 s is 7.8 m

Explanation:

Given;

mass of the puck, m = 2 kg

initial velocity of the puck, u = 0

applied force, F = 5 N

time of motion, t = 1.5 s

Acceleration of the puck is calculated from Newton's second law of motion;

F = ma

a = F/m

a = 5/2

a = 2.5 m/s²

The distance moved by the puck after 2.5 s is calculated as;

s = ut + ¹/₂at

s = 0 + ¹/₂at²

s = ¹/₂at²

s = 0.5 x 2.5 x (2.5)²

s = 7.8 m

Therefore, the distance moved by the puck after 2.5 s is 7.8 m

Answer:

The magnitude of the magnetic field is 1.83 x [tex]10^{-5}[/tex] T.

Explanation:

The flow of an electric current in a straight wire induces magnetic field around the wire. When current is flowing through two wires in the same direction, a force of attraction exists between the wires. But if the current flows in opposite directions, the force of repulsion is felt by the wires.

In the given question, the direction of flow of current through the wires is opposite, thus both wires applies the same field on each other. The result to repulsion between them.

The magnetic field (B) between the given wires can be determined by:

B = [tex]\frac{U_{o}I }{2\pi r}[/tex]

where: I is the current, r is the distance between the wires and [tex]U_{0}[/tex] is the magnetic field constant.

But, I = 11 A, r = 0.12 m and [tex]U_{0}[/tex] = 4[tex]\pi[/tex] x [tex]10^{-7}[/tex] Tm/A

So that;

B = [tex]\frac{4\pi *10^{-7}*11 }{2\pi *0.12}[/tex]

   = 1.8333 x [tex]10^{-5}[/tex]

B = 1.83 x [tex]10^{-5}[/tex] T

Answer:

253.85°C

Explanation:

Here is the formula for converting K to °C

527K − 273.15 = 253.85°C

Check attached photo

Check attached photo

Answer:

Explanation:

1. If C = A + B then the lines A and B may have the same magnitude or they may not. The direction of A for example may be northwest ↖️ and the direction of B must be south ⬇️ because the arrow of A and the point of B must connect. Then C’s direction is west ⬅️ because it shouldn’t be as equilibrium.

2. If C = 0 t means the force is at equilibrium. That means all forces add up to zero. A’s direction for example may be northeast ↗️ and the direction of B may be south ⬇️ and the direction of C must be west if it has to be at equilibrium.

The magnitude of A and B must be equal

First, balance the reaction:

_ KClO₃   ==>   _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃   ==>   2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃   ==>   2/3 KCl + O₂

Next, look up the molar masses of each element involved:

• K: 39.0983 g/mol

• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

Answer:

Explanation:

The formula for determining the Emf induced in a loop is:

[tex]\varepsilon = \dfrac{d \phi}{dt}[/tex]

[tex]\varepsilon = \dfrac{d (B*A)}{dt}[/tex]

[tex]\varepsilon = A \times \dfrac{dB}{dt}[/tex]

[tex]\varepsilon = (side (l))^2 \times \dfrac{dB}{dt}[/tex]

where;

square area A = ( l²)

l² = 6.0 cm = 6.0 × 10⁻²

∴

[tex]\varepsilon = ( 6.0 \times 10^{-2})^2 \times 5.0 \times 10^{-3} \ T/S[/tex]

[tex]\varepsilon =18 \times 10^{6} \ V[/tex]

Recall that:

The resistivity of copper = [tex]1.68 \times 10^{-8}[/tex] ohm m

We can as well say that the length of the copper wire = perimeter of the square loop;

The perimeter of the square loop = 4L

Thus, the length of the copper wire  = 4 (6.0 × 10⁻² )m

= 24× 10⁻² m

Finally, the current in the loop is determined from the formula:

V = IR

where,

V = voltage

I = current and R = resistance of the wire

Making "I" the subject:

I = V/R

where;

[tex]R = \dfrac{\rho \times l}{A}[/tex]

[tex]R = \dfrac{\rho \times l}{\pi * r^2}[/tex]

[tex]R = \dfrac{1.68 *10^{-8} \times 24*10^{-2}}{\pi * (1*10^{-3})^2}[/tex]

[tex]R = 0.001283 \ ohms[/tex]

∴

[tex]I = \dfrac{18*10^{-6}}{0.001283}[/tex]

I = 14.029 mA

Answer:

[tex]t_2=10[/tex]

Explanation:

From the question we are told that:

Velocity of both spaceships [tex]V=0.8c[/tex]

Time [tex]t_1=6[/tex]

Generally the equation for time of spaceship 2 through earth is mathematically given by

[tex]t_2=\frac{t_1}{\sqrt{1-v^2}}[/tex]

[tex]t_2=\frac{6}{\sqrt{1-0.8^2}}[/tex]

[tex]t_2=10[/tex]

Answer:

Hazards during late childhood

Explanation:

[tex]qV = \frac{1}{2}mv^2[/tex]

Multiply both sides by 2 and then divide by m to get

[tex]\dfrac{2qV}{m} = v^2[/tex]

Take the square root of both sides to get

[tex]v = \sqrt{\dfrac{2qV}{m}}[/tex]

Answer:

a)  a = 5.3 km, b) sum fulfills the commutative property

Explanation:

This is a vector exercise, If you drive east from north, we can find the vector using the Pythagorean theorem

              R² = a² + b²

where R is the resultant vector R = 7.5 km and the others are the legs

If we assume that the two legs are equal to = be

             R² = 2 a²

             r = √2 a

             a = r /√2

we calculate

             a = 7.5 /√2

             a = 5.3 km

therefore, you must drive 5.3 km east and then 5.3 km north and you will reach the same point

b) As the sum fulfills the commutative property, the order of the elements does not alter the result

         a + b = b + a

therefore, it does not matter in what order the path is carried out, it always reaches the same end point

Answer:

I think the answer is A...I'm not sure

Explanation:

A=(6,4)

B=(-2,-1)

A-B=(6-(-2)),(4-(-1))

=(6+2),(4+1)

=(8,5)

Answer:

[tex]6-(-2)=[/tex]

[tex]6+2[/tex]

[tex]=8[/tex]

[tex]4-\left(-1\right)[/tex]

[tex]=4+1[/tex]

[tex]=5[/tex]

[tex](8,5)[/tex]

[tex]\textbf{OAmalOHopeO}[/tex]

Answer:

Standard units are commonly used units of measurement, which help us measure length, height, weight, temperature, mass and more. These units are standardised, which means that everyone gets the same understanding of the size, weight and other properties of objects and things.

Explanation:

Your friend would have a better experience of this event, than you .

An eclipse is produced when a planetary body moves in front of another planetary body and is visible from a third planetary body. Considering the sun, moon, and earth's locations in relation to one another during the time of the eclipse,

there are various types of eclipses in our solar system. For instance, a lunar eclipse occurs when the earth passes between the moon and the sun.

For the solar eclipse to happen the light from the sun is obstructed by the moon observing from the earth.

The buddies left Earth because they could view the whole eclipse, but you were on the moon and only saw parts of the eclipse turn black.

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Answer:

a)  [tex]P=0.80[/tex]

b)  [tex]1.25Hz[/tex]

c)  [tex]A=25cm[/tex]

Explanation:

From the question we are told that:

Travel Time [tex]T=0.40s[/tex]

Distance [tex]d=50cm[/tex]

a)

Period

Time taken to complete one oscillation

Therefore

[tex]P=2*T\\\\P=2*0.40[/tex]

[tex]P=0.80[/tex]

b)

Frequency is

[tex]F=\frac{1}{T}\\\\F=\frac{1}{0.80}[/tex]

[tex]1.25Hz[/tex]

c)

Amplitude:the distance between the mean and extreme position

[tex]A=\frac{50}{2}[/tex]

[tex]A=25cm[/tex]