A phonograph record has an initial angular speed of 37 rev/min. The record slows to 14 rev/min in 1.6 s. What is the record’s average angular acceleration during this time interval? Answer in units of rad/s 2

This is a physics problem related to Boyle's Law, which says that the volume of a gas decreases if its pressure increases, assuming constant temperature. We can't solve for the final volume specifically in this case since the initial pressure and volume aren't given, but if they were, we'd use the formula V₂ = (P₁V₁) / P₂, where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume.

This question is related to the concept of Boyle's Law in physics, which states that the pressure and volume of a gas have an inverse relationship, when temperature is held constant. If the pressure of the gas inside a cylinder is increased, the volume decreases, assuming the amount of gas and the temperature remain constant.

In this case without knowing the initial pressure and volume of the cylinder, it's impossible to calculate the exact final volume after the pressure has increased to 1.6 atm. However, the base formula you'd use to find out your final volume, given you had initial values for volume (V₁) and pressure (P₁), is: P₁V₁ = P₂V₂.

In this formula, P₁ and V₁ refer to the initial pressure and volume, while P₂ and V₂ refer to the final pressure and volume. To solve for the final volume (V₂), you would rearrange the formula to be V₂ = (P₁V₁) / P₂. So, if you were given the initial pressure and volume, you could substitute those values into the formula, along with the final pressure of 1.6 atm, to find the final volume.

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In this case, the final volume is 31.1 mL.

The final volume of the helium in milliliters can be calculated using Boyle's Law, which states that the product of pressure and volume is constant for a given amount of gas at constant temperature. Initially, the pressure is 447 torr, and the volume is 86.4 mL. When the pressure increases to 1,240 torr, the final volume can be found using the formula:

P1V1 = P2V2

Solving for the final volume, V2, we get:

V2 = (P1V1) / P2

Substitute the given values:

V2 = (447 torr× 86.4 mL) / 1,240 torr

= 31.1 mL

Explanation:

The given data is as follows.

       Mass of evaporating dish = 3.375 g

    Total mass = Mass of solid sample + evaporating dish

That is, Mass of solid sample + evaporating dish = 26.719 g

Therefore, we will calculate the mass of solid sample as follows.

    Mass of solid sample = (Mass of solid sample + evaporating dish) - mass of evaporating dish

                            = 26.719 g – 3.375 g

                            = 23.344 g

Thus, we can conclude that mass of his solid sample must be 23.344 g.

The mass of the solid sample is 23.344 g.

In order to find the mass of the solid sample, we need to subtract the mass of the evaporating dish from the combined mass of the dish and the sample. The mass of the solid sample can be calculated by subtracting 3.375 g (mass of the evaporating dish) from 26.719 g (combined mass of dish and sample).

Mass of solid sample = 26.719 g - 3.375 g = 23.344 g.

Therefore, the mass of the solid sample is 23.344 g.

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Answer:

The total charge on the rod is 47.8 nC.

Explanation:

Given that,

Charge = 5.0 nC

Length of glass rod= 10 cm

Force = 840 μN

Distance = 4.0 cm

The electric field intensity due to a uniformly charged rod of length L at a distance x on its perpendicular bisector

We need to calculate the electric field

Using formula of electric field intensity

[tex]E=\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}[/tex]

Where, Q = charge on the rod

The force is on the charged bead of charge q placed in the electric field of field strength E

Using formula of force

[tex]F=qE[/tex]

Put the value into the formula

[tex]F=q\times\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}[/tex]

We need to calculate the total charge on the rod

[tex]Q=\dfrac{Fx\sqrt{(\dfrac{L}{2})^2+x^2}}{kq}[/tex]

Put the value into the formula

[tex]Q=\dfrac{840\times10^{-6}\times4.0\times10^{-2}\sqrt{(\dfrac{10.0\times10^{-2}}{2})^2+(4.0\times10^{-2})^2}}{9\times10^{9}\times5.0\times10^{-9}}[/tex]

[tex]Q=47.8\times10^{-9}\ C[/tex]

[tex]Q=47.8\ nC[/tex]

Hence, The total charge on the rod is 47.8 nC.

Final answer:

Using Coulomb's Law, the charge on the rod is calculated to be 7.5 *10⁻⁶ C by rearranging the formula and solving for the rod's charge with the provided force and the charge on the bead.

Explanation:

To determine the total charge on the rod, we can use Coulomb's Law, which states that the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the glass bead is assumed to act like a point charge, and we can treat the charge distribution on the thin glass rod as uniform.

The formula for Coulomb's Law is:

F = k * |q₁ * q₂| / r²

where:

F is the magnitude of the force between the charges,

k is Coulomb's constant (8.988 * 10⁹ N m²/C²),

q₁ and q₂ are the amounts of the charges, and

r is the distance between the centers of the two charges.

Given that the force F is 840 * 10⁻⁶ N and charge on bead q₁ is 5.0 * 10⁻⁹ C, and the distance r is 4.0 * 10⁻² m, we can rearrange the formula to solve for the charge on the rod (q₂):

q₂ = F * r² / (k * q₁)

Plugging in the values, we can calculate as follows:

q₂ = (840 *10⁻⁶ N) * (4.0 *10^⁻² m)² / (8.988 *10⁹ N m²/C² * 5.0 *10⁻⁹ C)

Calculating the charge on the rod:

q₂ = 0.0075 C

or

q₂ = 7.5 * 10⁻⁶ C

This value represents the total charge on the rod.

Answer:

66.375 x 10⁻⁶ C/m

Explanation:

Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as ;

∅ = Q / ε₀        -----------------(i)

Where;

∅ = 7.5 x 10⁵ Nm²/C

ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²

Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;

7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)

Solve for Q;

Q = 7.5 x 10⁵ x 8.85 x 10⁻¹²

Q = 66.375 x 10⁻⁷ C

Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e

L = Q / l     ----------------------(ii)

Where;

Q = 66.375 x 10⁻⁷ C

l = length of the rod = 10.0cm = 0.1m

Substitute these values into equation (ii) as follows;

L = 66.375 x 10⁻⁷C / 0.1m

L = 66.375 x 10⁻⁶ C/m

Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.

The linear charge density of the rod is found using Gauss's law and is equal to 6.642 x 10^-5 C/m.

The concept in this question is Gauss's law, which states that the electric flux through any closed surface is equal to the total charge enclosed divided by the permittivity of free space, ε0. Mathematically, it is written as Φ = Q/ε0, where Φ is the electric flux, Q is the charge, and ε0 = 8.854 x 10^-12 C2/N·m2.

In this case, we have Φ = 7.50 x 105 N·m2/C. To find Q, we just rearrange the formula and multiply both sides by ε0. This gives Q = Φ * ε0 = 7.50 x 105 N·m2/C * 8.854 x 10^-12 C2/N·m2 = 6.642 x 10^-6 C.

The linear charge density λ is the total charge Q divided by the total length L of the rod. Here, L = 10.0 cm = 0.1 m, so λ = Q/L = 6.642 x 10^-6 C / 0.1 m = 6.642 x 10^-5 C/m.

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Answer:

Explanation:

Let the vertical height by which it descends be h . Let it acquire velocity of v .

1/2 mv² = mgh

v² = 2gh

As it leaves the surface of sphere , reaction force of surface  R = 0 , so

centripetal force = mg cosθ where θ is the angular displacement from the vertex .  

mv² / r = mg cosθ

(m/r )x 2gh = mg cosθ

2h / r = cosθ

cosθ = (r-h) / r

2h / r =  r-h / r

2h = r-h

3h = r

h = r / 3

Final answer:

Through conservation of energy and dynamics principles, the puck descends through a height of R/2 from the base of the sphere before losing contact, due to the gravitational force no longer providing sufficient centripetal force.

Explanation:

The question asks through what vertical height a small frictionless puck will descend before it leaves the surface of a fixed sphere of radius R, when given a tiny nudge down the sphere. Using the principles of energy conservation and dynamics, it can be determined that the puck will lose contact with the sphere when the centripetal force is no longer sufficient to provide the necessary force for circular motion, which happens at a height of R/2 from the base of the sphere. This happens because, at this point, the gravitational component acting towards the center of the sphere is exactly equal to the required centripetal force for circular motion. As a result, any further descent would mean this balance is disturbed, causing the puck to leave the surface of the sphere.

Answer:

a) Temperatura, b) Temperature, c)    Constant , d)  None of these , e) Gibbs enthalpy and free energy (G)

Explanation:

a) the expression for ideal gases is PV = nRT

     Temperature

b) The internal energy is E = K T

      Temperature

c)  S = ΔQ/T

In an isolated system ΔQ is zero, entropy  is constant

       Constant

d) all parameters change when changing status

        None of these

e) Gibbs enthalpy and free energy

Answer:

a)  y= 3.5 10³ m, b)   t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

          y₁ = y₀ + v₀ t + ½ a t²

          y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

         y₁ = ½ 16 10²

         y₁ = 800 m

At the end of this stage it has a speed

        v₁ = vo + a₁ t₁

        v₁ = 0 + 16 10

        v₁ = 160 m / s

Stage 2

        y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

        y₂ = 800 + 150 5 + ½ 11 5²

        y₂ = 1092.5 m

Speed ​​is

        v₂ = v₁ + a₂ t

        v₂ = 160 + 11 5

        v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

         v₃² = v₂² - 2 g y₃

         0 = v₂² - 2g y₃

         y₃ = v₂² / 2g

         y₃ = 215²/2 9.8

         y₃ = 2358.4 m

The total height is

          y = y₃ + y₂

          y = 2358.4 + 1092.5

          y = 3450.9 m

           y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

          v₃ = v₂ - g t₃

          v₃ = 0

          t₃ = v₂ / g

          t₃ = 215 / 9.8

          t₃ = 21.94 s

The time to climb is

          [tex]t_{s}[/tex] = t₁ + t₂ + t₃

          t_{s} = 10+ 5+ 21.94

          t_{s} = 36.94 s

The time to descend from the maximum height is

          y = v₀ t - ½ g t²

When it starts to slow down it's zero

         y = - ½ g t_{b}²

         t_{b}  = √-2y / g

        t_{b} = √(- 2 (-3450.9) /9.8)

        t_{b} = 26.54 s

Flight time is the rise time plus the descent date

        t = t_{s} + t_{b}

        t = 36.94 + 26.54

        t =63.84 s

        t = 64 s

Answer:

Explanation:

a ) Height to be cleared = 5 - 1.6 = 3.4 m

Horizontal distance to be cleared = 5 m .

angle of throw = 56°

here y = 3.4 , x = 5 , θ = 56

equation of trajectory

y = x tanθ - 1/2 g ( x/ucosθ)²

3.4 = 5 tan56 - 1/2 g ( 5/ucos56)²

3.4 = 7.4 - 122.5 / .3125u²

122.5 / .3125u² = 4

u² = 98

u = 9.9 m /s

Range = u² sin 2 x 56 / g

= 9.9 x 9.9 x .927 / 9.8

= 9.27 m

horizontal distance be-yond the fence will the rock land on the ground

= 9.27 - 5

= 4.27 m

Answer:

a) [tex]u=13.3032\ m.s^{-1}[/tex]

b) [tex]s=3.7597\ m[/tex]

Explanation:

Given:

horizontal distance between the fence and the point of throwing, [tex]r=14\ m[/tex]

height of the fence from the ground, [tex]h_f=5\ m[/tex]

height of projecting the throw above the ground, [tex]h'=1.6\ m[/tex]

angle of projection of throw from the horizontal, [tex]\theta=56^{\circ}[/tex]

The horizontal component of the velocity that remains constant throughout the motion:

[tex]u_x=u\cos\theta[/tex]

Now the time taken to reach the distance of the fence:

use equation of motion,

[tex]t_f=\frac{r}{u_x}[/tex]

[tex]t_f=\frac{14}{u.\cos56}[/tex] .................................(1)

Now the time taken to reach the fence height (this height must be attained on the event of descending motion of the rock for the velocity to be minimum).

Maximum Height of the projectile:

[tex]v_y^2=u_y^2-2\times g.h_m[/tex]

[tex]h_m=\frac{u_y^2}{19.6}[/tex] ........................(4)

Now the height descended form the maximum height to reach the top of the fence:

[tex]\Delta h=h_m-h'[/tex]

[tex]\Delta h=(\frac{u_y^2}{19.6} -3.4)\ m[/tex]

time taken to descent this height from the top height:

[tex]\Delta h=u_{yt}.t_d+\frac{1}{2} \times g.t_d^2[/tex]

where:

[tex]u_{yt}=[/tex]  initial vertical velocity at the top point

[tex]t_d=[/tex] time of descend

[tex](\frac{u_y^2}{19.6} -3.4)=0+0.5\times 9.8\times t_d^2[/tex]

[tex]t_d=\sqrt{(\frac{u_y^2}{96.04} -\frac{3.4}{4.9} )}[/tex]..............................(2)

So we find the time taken by the rock to reach the top of projectile where the vertical velocity is zero:

[tex]v_y=u_y-g.t_t[/tex]

where:

[tex]u_y=[/tex] initial vertical velocity

[tex]v_y=[/tex] final vertical velocity

[tex]t_t=[/tex] time taken to reach the top height of the projectile

[tex]0=u_y-g.t_t[/tex]

[tex]t_t=\frac{u_y}{9.8}\ seconds[/tex] .................................(3)

Now the combined events of vertical and horizontal direction must take at the same time as the projectile is thrown:

So,

[tex]t_f=t_t+t_d[/tex]

[tex]\frac{14}{u.\cos56}=\frac{u_y}{9.8} +\sqrt{(\frac{u_y^2}{96.04} -\frac{3.4}{4.9} )}[/tex]

[tex]\frac{14}{u.\cos56}=\frac{u\sin56}{9.8} +\sqrt{(\frac{(u.\sin56)^2}{96.04} -\frac{3.4}{4.9} )}[/tex]

[tex]\frac{196}{u^2.\cos^2 56} +\frac{u^2\sin^2 56}{96.04} -2.857\times \tan56=\frac{u^2\sin^2 56}{96.04} -0.694[/tex]

[tex]u=13.3032\ m.s^{-1}[/tex]

Max height:

[tex]h_m=\frac{(u.\sin 56)^2}{19.6}[/tex]

[tex]h_m=\frac{(13.3032\times \sin56)^2}{19.6}[/tex]

[tex]h_m=6.2059\ m[/tex]

Now the rock hits down the ground 1.6 meters below the level of throw.

Time taken by the rock to fall the gross height [tex]h_g=h_m+h'[/tex]:

[tex]h_g=u_{yt}.t_g+\frac{1}{2} g.t_g^2[/tex]

[tex]7.8059=0+0.5\times 9.8\times t_g^2[/tex]

[tex]t_g=1.2621\ s[/tex]

Time taken to reach the the top of the fence from the top, using eq. (2):

[tex]t_d=\sqrt{(\frac{u_y^2}{96.04} -\frac{3.4}{4.9} )}[/tex]

[tex]t_d=\sqrt{(\frac{(u.\sin56)^2}{96.04} -\frac{3.4}{4.9} )}[/tex]

[tex]t_d=0.7567\ s[/tex]

Time difference between falling from top height and the time taken to reach the top of fence:

[tex]\Delta t=t_g-t_d[/tex]

[tex]\Delta t=1.2621-0.7567[/tex]

[tex]\Delta t=0.5054\ s[/tex]

b)

Now the horizontal distance covered in this time:

[tex]s=u.\cos56\times\Delta t[/tex]

[tex]s=13.3032\times \cos56\times 0.5054[/tex]

[tex]s=3.7597\ m[/tex] is the horizontal distance covered after crossing the fence.

The charge enclosed by a concentric spherical surface depends on the size of the sphere.

To calculate the charge enclosed by a concentric spherical surface, we need to find the charge within that surface. The charge is uniformly distributed throughout the interior volume of the insulating sphere, so the charge within any smaller sphere that is completely enclosed by the larger sphere will be the same.

(a) For r = 1.00 cm, the smaller sphere is completely enclosed by the larger sphere. Therefore, the charge enclosed by the smaller sphere is 6.50 µC.

(b) For r = 6.50 cm, the smaller sphere is the same size as the larger sphere, so the charge enclosed by the smaller sphere will also be 6.50 µC.

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To answer this question it is necessary to use the Bohr Model as a theoretical reference. According to this model, the energy levels of an atom are divided by discrete and characteristic energies. When there is the emission or absorption of a photon with a characteristic energy there is a transition of another energy level, which is equal to the level of atomic energy. Since the energy of a photo is directly proportional to its frequency, the emitted or absorbed photons have characteristic frequencies to the difference in energy between atomic energy levels.

Answer:

a. 0.683secs and 2.99secs

b.(x,y)=(23.99,11.3)m/s, (23.99,-11.3)m/s\

c. V=30.0m/s at angle ∝=-36.9 Degree

Explanation:

Data given

Velocity, V=30m/s

Angle,∝=36.9 Degree

The motion describe by the baseball is a  projectile motion, the velocity at the x-axis and y-axis are given as

[tex]V_{x}=Vcos\alpha\\ V_{x}=30cos36.9\\ V_{x}=23.99m/s\\V_{y}=Vsin\alpha \\V_{y}=30sin36.9\\ V_{y}=18.01m/s[/tex]

a.To calculate the time at which the baseball was at a height of 10m, we use the equation describing the vertical distance traveled

[tex]y=V_{y}t-\frac{1}{2}gt^{2}\\ y=10m\\10=18.01t-\frac{1}{2}*9.81t^{2}\\10=18.01t-4.9t^{2}\\-4.9t^{2}+18.01t-10[/tex]

solving the quadratic equation using the formula method

[tex]t=\frac{-b±\sqrt{b^{2}-4ac }}{2a} \\a=-4.9, b=18.01,c=-10\\t=\frac{18.01±\sqrt{18.01^{2}-4*(-4.9)(-10) }}{2*(-4.9)} \\t=0.683s, t=2.99s[/tex]

Hence the two times required 0.683secs and 2.99secs

b. Note that no acceleration in the hotizontal component, so the velocity remain the same. at a time t=0.683secs,

[tex]V_{x}=23.99m/s, \\V_{y}=V_{yi}-gt\\V_{y}=18.01-9.8*0.683\\V_{y}=11.3m/s\\[/tex]

at a time t=2.99secs,

[tex]V_{x}=23.99m/s, \\V_{y}=V_{yi}-gt\\V_{y}=18.01-9.8*2.99\\V_{y}=-11.3m/s\\[/tex]

c.The landing velocity is the same as the initial projected velocity but in opposite direction i.e V=30.0m/s at angle ∝=-36.9 Degree

To find the times when the baseball is at a height of 10.0 m, solve the vertical motion equation. Calculate the horizontal and vertical components of velocity at each time. Finally, find the magnitude and direction of the velocity when it returns to the starting level.

To find the two times when the baseball is at a height of 10.0 m, we can use the equation for vertical motion:

y = yo + voyt - (1/2)gt^2

Setting y = 10 m and using the given initial vertical velocity, we can solve for t. Plugging in the values, we find two solutions: t = 2.46 s and t = 5.06 s.

Next, we can calculate the horizontal and vertical components of the baseball's velocity at each of the two times. The horizontal component remains constant throughout the motion and can be calculated using:

vx = vocosθ

Plugging in the given initial velocity and angle, we find that vx = 30.0 m/s * cos(36.9°) = 24.3 m/s.

The vertical component of the velocity changes due to acceleration from gravity. We can calculate it using:

vy = voy - gt

Plugging in the given initial vertical velocity and the acceleration due to gravity (-9.8 m/s^2), we find the vertical velocity at t = 2.46 s to be -6.16 m/s and at t = 5.06 s to be -14.56 m/s.

To find the magnitude of the baseball's velocity when it returns to the level at which it left the bat, we can use the Pythagorean theorem:

v = √(vx^2 + vy^2)

Plugging in the calculated values, we find the magnitude of the velocity to be 25.8 m/s.

The direction of the velocity can be found using the inverse tangent function:

θ = atan(vy/vx)

Plugging in the calculated values, we find the direction to be 180° - 36.9° = 143.1° below the horizontal.

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Answer:

[tex]T=94.54N[/tex]

Explanation:

The tension in a cable is given by:

[tex]T=\mu v^2(1)[/tex]

Where [tex]\mu[/tex] is the mass density of the cable and v is the speed of the cable's pulse. These values are defined as:

[tex]\mu=\frac{m}{L}(2)\\v=\frac{d}{t}[/tex]

The pulse makes four trips down and back along the cable, so [tex]d=4(2L)[/tex]

[tex]v=\frac{8L}{t}(3)[/tex]

Replacing (2) and (3) in (1), we calculate the tension in the cable:

[tex]T=\frac{m}{L}(\frac{8L}{t})^2\\T=\frac{64mL}{t^2}\\T=\frac{64(0.21kg(3.80m))}{(0.735s)^2}\\T=94.54N[/tex]

Answer:

A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is the surface charge density of the sheet"?

Explanation:

Surface charge density is a measure of how much electric charge is accumulated over a surface. It can be calculated as the charge per unit area.

We will convert all parameters in SI units.

Charge = Q = -5.02nC

Q  = -5.02×[tex]10^{-9}[/tex]C

As it is clear from question that Sheet is a square (All sides will be of equal length)

Area = A = (21.8×[tex]10^{-2}[/tex]m) (21.8×[tex]10^{-2}[/tex]m)  = 4.75×[tex]10^{-4}[/tex]m²

A  = 4.75×[tex]10^{-4}[/tex]m²

Surface charge density = Q/A

Surface charge density = (-5.02×[tex]10^{-9}[/tex]C)/(4.75×[tex]10^{-4}[/tex]m²)

Surface charge density = -1.057×[tex]10^{-5}[/tex] C[tex]m^{-2}[/tex]

The force exerted by each rope is approximately 230 N.

Let's denote the magnitude of the force exerted by each rope as F. Given that the resultant pull is 372 N directly upward, we can use vector decomposition to find the individual forces exerted by each rope.

The forces F acting at an angle of 72° between them can be decomposed into their horizontal and vertical components as follows:

[tex]\[ F_{\text{horizontal}} = F \cdot \cos(72^\circ) \]\\\ F_{\text{vertical}} = F \cdot \sin(72^\circ) \][/tex]

The vertical components of the forces add up to the resultant pull, so:

[tex]\[ 2 \cdot F_{\text{vertical}} = 372 \, \text{N} \]\\\\\ F_{\text{vertical}} = \frac{372 \, \text{N}}{2} \\\\= 186 \, \text{N} \][/tex]

Now, we can use the vertical component to find F:

[tex]\rm \[ F = \frac{F_{\text{vertical}}}{\sin(72^\circ)} \][/tex]

Substitute the known values:

[tex]\rm \[ F = \frac{186 \, \text{N}}{\sin(72^\circ)} \][/tex]

Calculate F:

[tex]\rm \[ F \approx 230 \, \text{N} \][/tex]

So, the force exerted by each rope is approximately 230 N.

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To find the pull exerted by each rope with an angle of 72°, we can use vector components. Each rope exerts a pull of approximately 436.2 N at an angle of 72° for both the vertical and horizontal components.

To find the pull exerted by each rope, we can use the concept of vector components. Let's call the magnitude of each pull T. Since the resultant pull is 372 N directly upward, we can calculate the vertical component of each rope's pull using the equation T*sin(72°) = 372 N.

Solving for T, we find that the magnitude of each rope's pull is approximately 436.2 N. To find the horizontal component of each rope's pull, we use the equation T*cos(72°).

Since the forces are equal-magnitude and have the same angle between them, each rope exerts the same pull of approximately 436.2 N at an angle of 72° for both the vertical and horizontal components of the pull.

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Answer:

B

Explanation:

The electroscope device is used to detect the presence of charge and it's relative amount. If a charged object is brought near the top of the electroscope(for a example positive charge) the leaves at the bottom spread apart. The leaves diverge further: The positive charge on the leaves has increased further. This happens when positive charge is produced on the leaves by the charged object. This is possible when the object is positively charged.The greater the charge, the farther apart they move.

Answer:

Explanation:

An electroscope is a device used to detect the presence of electric charges. However, to detect charge in an object, it requires hundreds of volts, that's why is only used with high voltage sources.

An important matter is that an electroscope follows the Coulomb electrostatic force, which is a law of physics that quantifies the amount of force between two charged particles, which are stationary.

Having said that, when two charged particles have the same nature, then they will spread apart. Same nature means both negative or both positive.

Therefore, the right answer is B.

Answer:

[tex]\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }[/tex]

Explanation:

Additional information:

The ball has charge [tex]-q_0[/tex], and the ring has  positive charge [tex]+Q[/tex] distributed uniformly along its circumference.

The electric field at distance [tex]z[/tex] along the z-axis due to the charged ring is

[tex]E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.[/tex]

Therefore, the force on the ball with charge [tex]-q_0[/tex] is

[tex]F=-q_oE_z[/tex]

[tex]F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}[/tex]

and according to Newton's second law

[tex]F=ma=m\dfrac{d^2z}{dz^2}[/tex]

substituting [tex]F[/tex] we get:

[tex]- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}[/tex]

rearranging we get:

[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0[/tex]

Now we use the approximation that

[tex]z^2+a^2\approx a^2[/tex] (we use this approximation instead of the original [tex]d^2+a^2\approx a^2[/tex] since [tex]z<d[/tex], our assumption still holds )

and get

[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0[/tex]

[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0[/tex]

Now the last equation looks like a Simple Harmonic Equation

[tex]m\dfrac{d^2z}{dz^2}+kz=0[/tex]

where

[tex]\omega=\sqrt{ \dfrac{k}{m} }[/tex]

is the frequency of oscillation. Applying this to our equation we get:

[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}[/tex]

[tex]\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}[/tex]

Answer:

Air Resistance

Explanation:

If you were to drop both items on a plant without an atmosphere, they would both hit the ground at the same time. Since a feather doesn't have much mass compared to the hammer, it takes more time for the feather to "push" itself through and overcome the opposite push from the air

Final answer:

On Earth, air resistance prevents a feather and a hammer from hitting the ground simultaneously when dropped from the same height. On the Moon, the absence of an atmosphere means no air resistance, allowing both objects to land at the same time in accordance with Galileo's principle of the universality of free fall.

Explanation:

If you drop a feather and a steel hammer at the same moment, they should hit the ground at the same time according to Galileo's principle of the universality of free fall. However, this does not occur on Earth due to the presence of air resistance. The feather experiences a significant amount of air resistance because of its shape and light weight, causing it to flutter and fall slower than the hammer.

On the Moon, where there is no atmosphere, there is no air resistance to act on the objects. When Apollo 15 astronaut David Scott conducted the experiment on the Moon, both the hammer and feather fell at the same acceleration and hit the lunar surface simultaneously. This specific demonstration was a perfect illustration of the universality of free fall in the absence of external forces besides gravity. On the Moon, the acceleration due to gravity is only 1.67 m/s², which is less than on Earth, but since it acts equally on all objects, both the feather and the hammer fell at the same rate.

Answer:

a) [tex]h=250\ m[/tex]

b) [tex]\Delta h=0.0835\ m[/tex]

Explanation:

Given:

a)

Maximum height reached by the helicopter:

using the equation of motion,

[tex]h=u.t+\frac{1}{2} a.t^2[/tex]

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

[tex]h=0+0.5\times 5\times 10^2[/tex]

[tex]h=250\ m[/tex]

b)

height fallen freely by Austin:

[tex]h_f=u.t_f+\frac{1}{2} g.t_f^2[/tex]

where:

[tex]u=[/tex] initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

[tex]t_f=[/tex] time of free fall

[tex]h_f=0+0.5\times 9.8\times 7^2[/tex]

[tex]h_f=240.1\ m[/tex]

Velocity just before opening the parachute:

[tex]v_f=u+g.t_f[/tex]

[tex]v_f=0+9.8\times 7[/tex]

[tex]v_f=68.6\ m.s^{-1}[/tex]

Time taken by the helicopter to fall:

[tex]h=u.t_h+\frac{1}{2} g.t_h^2[/tex]

where:

[tex]u=[/tex] initial velocity of the helicopter just before it begins falling freely = 0

[tex]t_h=[/tex] time taken by the helicopter to fall on ground

[tex]h=[/tex] height from where it falls = 250 m

now,

[tex]250=0+0.5\times 9.8\times t_h^2[/tex]

[tex]t_h=7.1429\ s[/tex]

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

remaining time,

[tex]t'=t_h-t_f[/tex]

[tex]t'=7.1428-7[/tex]

[tex]t'=0.1428\ s[/tex]

Now the height fallen in the remaining time using parachute:

[tex]h'=v_f.t'+\frac{1}{2} a_p.t'^2[/tex]

[tex]h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2[/tex]

[tex]h'=9.8165\ m[/tex]

Now the height of Austin above the ground when the helicopter crashed on the ground:

[tex]\Delta h=h-(h_f+h')[/tex]

[tex]\Delta h=250-(240.1+9.8165)[/tex]

[tex]\Delta h=0.0835\ m[/tex]

The questions deal with centripetal acceleration, speed, and the path of a supersonic airplane making a turn. The key is to use the formulae for centripetal acceleration, circular motion, and circle properties to calculate the desired quantities.

The questions pertain to the physics concept of centripetal acceleration. The centripetal acceleration of a body moving in a circular path is given by the equation a = v2/r. In the case of the supersonic airplane, we can plug in the given values of speed (v = 2570 km/h = 713.89 m/s) and radius (r = 80.5 km = 80500 m) into this formula to calculate the centripetal acceleration.

Next, to find the time it would take for the airplane to turn from North to East, we need to understand that the airplane is essentially making a 90-degree turn, or 1/4 of a full circular path. Therefore, the time would be 1/4 of the total time it would take to complete a full circle (T = 2πr/v). The distance covered during the turn would also equivalently be 1/4 of the total circumference of the path, which we can calculate using the formula for the circumference of a circle (C = 2πr).

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Answer:

The charge on the ball bearing 4.507 × 10^-8 C

Explanation:

From Coulomb's law

F = kq1q2/r²

make q2 the subject

q2 = Fr²/kq1

q2 = (1.8×10^-2 × 0.026²) ÷ (9×10^9 × 30×10^-9)

q2 = 4.507 × 10^-8 C

Using Coulomb's Law, the charge on the ball bearing is found to be approximately -1.92 x 10^-8 C. The negative sign indicates that the charges on the bead and the ball bearing are of opposite nature.

To determine the charge on the ball bearing, we will use Coulomb's Law, that implies that the product of the two charges divided by their distance from one another is what determines the force between two charges. The formula is: F = k * |q1*q2| / r^2 where F is the electric force, q1 and q2 are the charges, r is the distance between the charges, and k is Coulomb's Constant (8.988x10^9 N * m^2/C^2).

We can rearrange the equation to find the charge on the ball bearing (q2): q2 = F * r^2 / (k * q1)

Substituting the given values: q2 = 1.80x10^-2 N * (2.6x10^-2 m)^2 / (8.988x10^9 N * m^2/C^2 * 3.0x10^-8 C)

With some calculations, we get that the charge on the ball bearing, q2 is approximately  -1.92 x 10^-8 C. The negative sign represents the opposite nature of the charges, meaning the bead and ball bear opposite charges.

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Answer:

a) The force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another) is - 2.01 × 10⁻⁹ N

b) The force of repulsion at this same separation distance is 2.01 × 10⁻⁹ N

Explanation:

F = kq₁q₂/r²

r = 0.35 + 0.129 (since the ions are just touching each other)

r = 0.479 nm = 4.79 × 10⁻¹⁰ m

Since the first ion is a divalent cation, Z₁ = +2 and the monovalent anion, Z₂ = -1

q = Ze; e = 1.602 × 10⁻¹⁹ C

K = 8.99 × 10⁹ Nm²/C²

F = (8.99 × 10⁹)(1.602 × 10⁻¹⁹)²(2)(-1)/(4.79 × 10⁻¹⁰)² = - 2.01 × 10⁻⁹ N

b) At equilibrium,

Force of attraction + Force of repulsion = 0

Force of repulsion = -(Force of attraction) = 2.01 × 10⁻⁹ N

Answer:

a)Proportional limit =  27.756ksi

b) The modulus of elasticity = 15860ksi

c)Poisson ratio = 0.343

Explanation:

The detailed steps and calculation is as shown in the attached file.

Complete Question

The complete question is shown on the first uploaded image

Answer:

  The dopant density is ND   ≈  8.135*10¹²  cm⁻³

Explanation:

The explanation is shown on the second , third ,fourth and fifth uploaded image

To solve this problem it is necessary to apply the concepts related to density, such as the relationship between density and Volume.

The volume of a sphere can be expressed as

[tex]V = \frac{4}{3} \pi r^3[/tex]

Here r is the radius of the sphere and V is the volume of Sphere

Using the expression of the density we know that

[tex]\rho = \frac{m}{V} \rightarrow V = \frac{m}{\rho}[/tex]

The density is given as

[tex]\rho = (19.5g/cm^3)(\frac{10^3kg/m^3}{1g/cm^3})[/tex]

[tex]\rho = 19.5*10^3kg/m^3[/tex]

Now replacing the mass given and the actual density we have that the volume is

[tex]V = \frac{60kg}{19.5*10^3kg/m^3 }[/tex]

[tex]V = 3.0769*10^{-3} m ^3[/tex]

The radius then is,

[tex]V = \frac{4}{3} \pi r^3[/tex]

[tex]r = \sqrt[3]{\frac{3V}{4\pi}}[/tex]

Replacing,

[tex]r = \sqrt[3]{\frac{3(3.0769*10^{-3})}{4\pi}}[/tex]

The radius of a sphere made of this material that has a critical mass is 9.02 cm.

Answer:

This is the correct question.

Apply the junction rule to the junction labeled with the number1 (at the bottom of the resistor of resistance R_2).

Answer in terms of given quantities,together with the meter readings I_1 and I_2 and the current I_3.

b) Apply the loop rule to loop 2 (the smaller loop on the right).Sum the voltage changes across each circuit element around this loop going in the direction of the arrow. Remember that the current meter is ideal.

Express the voltage drops in terms ofV_b, I_2, I_3, the given resistances, and any other givenquantities.

c) Now apply the loop rule to loop 1(the larger loop spanning the entire circuit). Sum the voltagechanges across each circuit element around this loop going in thedirection of the arrow.

Express the voltage drops in terms ofV_b, I_1, I_3, the given resistances, and any other givenquantities.

Explanation:

a. Junction rule: Kirchhoff current law (KCL)

Then, total current entering a junction equals to current leaving the junction.

Therefore,

i1=i2+i3

b. Apply Kirchoff voltage law ( KVL) to loop 2

Then, sum of voltage in the loop equals to zero.

-i3R3+i2R2=0.

Then,

i2R2=i3R3

c. Apply KVL to the loop 1

-Vb+i1R1+i3R3

Therefore,

Vb=i1R1+i3R3.

The circuit diagram is in the attachment

Kirchhoff's first rule, or the junction rule, states that all currents entering a junction must equal all currents leaving that junction. For the network with a junction labeled 1 and currents I1, I2, and I3, the application of the rule yields the equation: I1 = I2 + I3.

To apply Kirchhoff's first rule, also known as the junction rule, we need to consider that all currents entering a junction must equal all currents leaving that junction. This principle is based on the conservation of charge where the total amount of electricity is maintained. Given the question, we are working with a junction labeled 1 at the bottom of the resistor of resistance R2, with meter readings I1 and I2, and current I3.

According to the junction rule for this scenario, we would express this as: I1 = I2 + I3, where I1 is the current flowing into the junction, and I2 and I3 are currents flowing out. This equation states that the current entering the junction (I1) must be equal to the sum of the currents leaving the junction ( I2 and I3).

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Answer:

B. It is an inelastic collision because the train cars stick together and move as one.

Explanation:

Momentum

When two or more objects collide in a closed system (no external forces are acting) the total momentum is conserved:

[tex]m_1v_1+m_2v_2+...+m_nv_n=m_1v_1'+m_2v_2'+...+m_nv_n'[/tex]

where m1...m2 are the masses of each object, v1...vn are their velocities before the collision takes place and v'1...v'n are their velocities after the collision.

If a collision is elastic, then the kinetic energy is also conserved. when the collision is inelastic, part of the initial kinetic energy is lost. A typical case of inelastic collision occurs when the objects join and remain together after the collision. The velocity is common to all of them and the mass is the sum of the individual masses.

This is exactly the case described in the question: serveral train cars are joined and continue moving together after the collision. It corresponds to a inelastic collision described in the option B.

Final answer:

The bulk specific gravity (SSD) of the sand is 0.221.

Explanation:

The bulk specific gravity (SSD) of the sand can be calculated using the given information. The combined weight of the sand and water is 1697 g, and the weight of the jar filled with water only is 1390 g. To find the weight of the sand in the SSD condition, we subtract the weight of the jar filled with water from the combined weight: 1697 g - 1390 g = 307 g. Therefore, the bulk specific gravity (SSD) of the sand is the weight of the sand divided by the weight of an equal volume of water, which is 307 g / 1390 g = 0.221.

Answer:

33.7410805301 s

22336.5953109 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = [tex]4\times 331[/tex]

s = Displacement

a = Acceleration = [tex]4g=4\times 9.81\ m/s^2[/tex]

g = Acceleration due to gravity = 9.81 m/s²

[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{4\times 331-0}{4\times 9.81}\\\Rightarrow t=33.7410805301\ s[/tex]

The time taken is 33.7410805301 s

[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 4\times 9.81\times 33.7410805301^2\\\Rightarrow s=22336.5953109\ m[/tex]

The plane would travel 22336.5953109 m

The shortest time is approximately 33.74 seconds, and the distance traveled is approximately 22.32 kilometers.

To solve this problem, we will first need to convert Mach 4 to a speed in meters per second, then use the kinematic equations for constant acceleration to find the time and distance.

(a) The speed of sound in cold air is given as 331 m/s. Therefore, Mach 4 is:

[tex]\[ v = 4 \times 331 \text{ m/s} = 1324 \text{ m/s} \][/tex]

The acceleration that the pilot can withstand without graying out is less than 4 g. Since 1 g is equal to 9.81 m/s², the maximum acceleration the pilot can withstand is:

[tex]\[ a_{\text{max}} = 4 \times 9.81 \text{ m/s}^2 = 39.24 \text{ m/s}^2 \][/tex]

Using the kinematic equation that relates initial velocity, final velocity, acceleration, and time:

[tex]\[ v = u + at \]\\ where \( v \) is the final velocity, \( u \) is the initial velocity (0 m/s since the pilot starts from rest), \( a \) is the acceleration, and \( t \) is the time. We can solve for \( t \): \[ 1324 \text{ m/s} = 0 \text{ m/s} + (39.24 \text{ m/s}^2)t \] \[ t = \frac{1324 \text{ m/s}}{39.24 \text{ m/s}^2} \] \[ t \approx 33.74 \text{ s} \][/tex]

(b) To find the distance traveled during this time, we use the kinematic equation:

[tex]\[ s = ut + \frac{1}{2}at^2 \]\\ where \( s \) is the distance, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. Since \( u = 0 \) m/s: \[ s = 0 \times t + \frac{1}{2}(39.24 \text{ m/s}^2)t^2 \] \[ s = \frac{1}{2}(39.24 \text{ m/s}^2)(33.74 \text{ s})^2 \] \[ s \approx \frac{1}{2}(39.24 \text{ m/s}^2)(1137.52 \text{ s}^2) \] \[ s \approx 22320.54 \text{ m} \] \[ s \approx 22.32 \text{ km} \][/tex]

Therefore, the shortest time the pilot can take to reach Mach 4 without graying out is approximately 33.74 seconds, and the distance traveled during this period of acceleration is approximately 22.32 kilometers.

Final answer:

The normal strain in the pipe subjected to a tension force is calculated using Hooke's Law, which relates stress to strain via the elastic modulus. The cross-sectional area is found by the area difference between outer and inner circles, considering the pipe's diameter and wall thickness. The final strain is the normal stress divided by the elastic modulus.

Explanation:

To determine the normal strain in the pipe due to the tension force, we can use Hooke's Law, which relates stress and strain via the elastic modulus. Normal strain (ε) is calculated by dividing the normal stress (σ) by the elastic modulus (E). The normal stress can be found by dividing the tension force (P) by the cross-sectional area (A) of the pipe.

The cross-sectional area for a pipe is the area of a ring, which can be calculated as the area of the outer circle minus the area of the inner circle. Given the outside diameter (do) is 45 mm and wall thickness (t) is 5 mm, the inside diameter (di) will be do - 2t = 35 mm. So the area is calculated with A = π/4 • (do2 - di2).

After the area is determined, the normal stress is P/A and therefore the strain is ε = σ/E. Keep in mind to convert E to the correct units to match those of stress (N/mm²).

For the provided tension force P = 90 kN and elastic modulus E = 150 GPa, the strain calculation will use these values and the cross-sectional area calculated earlier.

Answer:

[tex]x=9.78\times 10^0[/tex]

Explanation:

In this case, we need to find the value of expression :

[tex]x=\dfrac{(6.67\times 10^{-11})\times (5.97\times 10^{24})}{(6.38\times 10^6)^2}[/tex]

On solving, we get the value of given expression as :

x = 9.7827

In scientific notation, we get the value of x as :

[tex]x=a\times 10^k[/tex]

[tex]x=9.78\times 10^0[/tex]

a = 9.78

k = 0

Hence, this is the required solution.

To express the given expression in scientific notation, we find a = 9.78 and k = -1 after evaluating the expression using the laws of exponentiation and division.

The student's question involves evaluating an expression using scientific notation and expressing the result in proper scientific notation.

To solve (6.67×10−11)(5.97×1024) / (6.38×106)2, we need to use the laws of exponentiation and multiplication. Simplify the expression within the numerator and denominator separately before dividing them.

Numerator: (6.67×10−11)(5.97×1024) = 39.8309×1013

Denominator: (6.38×106)2 = 40.7044×1012

Divide the simplified numerator by the simplified denominator:
39.8309×1013 / 40.7044×1012 = 0.978×101

To express this in proper scientific notation, rewrite 0.978×101 as 9.78×10−0.

Therefore, the values of a and k when the value of this expression is written in scientific notation are a = 9.78 and k = −0.