An air-filled parallel plate capacitor has circular plates with radius r=20.0 cm, separated distance 4.00 mm. The capacitor is connected to 70.0 V battery. a) Find the capacitance, Co, and the charge on the plates, Qo. b) Find the electric field strength between the plates. c) While connected to the battery, a dielectric sheet with dielectric constant k is inserted between the plates (it fills the entire space). What are the capacitance, the charge, and the voltage now?

Answer:

(a): [tex]\rm -5.627\times 10^3\ N.[/tex]

(b):  [tex]\rm 7.626\times 10^2\ N.[/tex]

Explanation:

Given:

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by

[tex]\rm F=k\cdot\dfrac{q_1q_2}{r^2}[/tex]

where,

k is called the Coulomb's constant, whose value is [tex]\rm 9\times 10^9\ Nm^2/C^2.[/tex]

From Newton's third law of motion, both the spheres experience same force.

Therefore, the magnitude of the force that each sphere experiences is given by

[tex]\rm F=k\cdot\dfrac{q_1q_2}{r^2}\\=9\times 10^9\times \dfrac{(-19.8\times 10^{-6})\times (+40.7\times 10^{-6})}{(3.59\times 10^{-2})^2}\\=-5.627\times 10^3\ N.[/tex]

The negative sign shows that the force is attractive in nature.

Part (b):

The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.

Total charge on both the spheres, [tex]\rm Q=q_1+q_2=-19.8\ \mu C+40.7\ \mu C = 20.9\ \mu C.[/tex]

The new charges on both the spheres are equal and given by

[tex]\rm q_1'=q_2'=\dfrac Q2 = \dfrac{20.9}{2}\ \mu C=10.45\ \mu C = 10.45\times 10^{-6}\ C.[/tex]

The magnitude of the force that each sphere now experiences is given by

[tex]\rm F'=k\cdot \dfrac{q_1'q_2'}{r^2}'\\=9\times 10^9\times \dfrac{10.45\times 10^{-6}\times 10.45\times 10^{-6}}{(3.59\times 10^{-2})^2}\\=7.626\times 10^2\ N.[/tex]

Answer:

[tex]r_2 = 976.65 m[/tex]

Direction is 19 degree South of East

Explanation:

Let say initial position is our reference

so we will have campsite position given as

[tex]r = 1.50 km[/tex] at 20 degree E of N

now we will have

[tex]r = 1500 sin20\hat i + 1500 cos20\hat j[/tex]

[tex]r = 513 \hat i + 1409.5\hat j[/tex]

now our displacement to walk around is given as

[tex]d_1 = 600 \hat j[/tex]

then we move 20 degree W of N and move 1200 m

so we will have

[tex]d_2 = 1200 sin20(-\hat i) + 1200cos20\hat j[/tex]

so our final position is given as

[tex]r_1 = d_1 + d_2[/tex]

[tex]r_1 = 600\hat j - 410.4 \hat i + 1127.6\hat j[/tex]

[tex]r_1 = -410.4 \hat i + 1727.6\hat j[/tex]

now we know that

[tex]r_1 + r_2 = r[/tex]

so final leg of the displacement is given as

[tex]r_2 = r - r_1[/tex]

[tex]r_2 = (513 \hat i + 1409.5\hat j) - (-410.4 \hat i + 1727.6\hat j)[/tex]

[tex]r_2 = 923.4\hat i - 318.1 \hat i[/tex]

so magnitude is given as

[tex]r_2 = \sqrt{923.4^2 + 318.1^2}[/tex]

[tex]r_2 = 976.65 m[/tex]

direction is given as

[tex]\theta = tan^{-1}\frac{y}{x}[/tex]

[tex]\theta = tan^{-1}\frac{-318.1}{923.4}[/tex]

[tex]\theta = -19 degree[/tex]

so it is 19 degree South of East

Answer:

The second system must be set in motion [tex]t=0.70s[/tex] seconds later

Explanation:

The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:

[tex]T=2\pi\sqrt(\frac{m}{k} )[/tex]

One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.

If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion [tex]t=\frac{\pi}{2}\sqrt(\frac{m}{k})=\frac{\pi}{2}\sqrt(\frac{0.55}{2.8}= 0.70s)[/tex] seconds later

Answer:

Part a)

[tex]q = -21.2 \times 10^{-6} C[/tex]

Part b)

[tex]E = 1.02 \times 10^{-7} N/C[/tex]

Explanation:

Part a)

As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force

so here we have

[tex]mg = qE[/tex]

[tex]1.47 \times 10^{-3} (9.81) = q(680)[/tex]

[tex]q = 21.2 \times 10^{-6} C[/tex]

and its sign must be negative so that it will have upward electric force

so it is

[tex]q = -21.2 \times 10^{-6} C[/tex]

Part b)

as we know that weight of proton is balanced by electric force

so we will have

[tex]qE = mg[/tex]

[tex](1.6 \times 10^{-19})E = (1.67 \times 10^{-27})(9.81)[/tex]

[tex]E = 1.02 \times 10^{-7} N/C[/tex]

A. The charge (sign and magnitude) of the particle of mass 1.47 g is –2.12×10¯⁵ C

B. The magnitude of the electric field in which the electric force on the proton is equal in magnitude to its weight is 1.02×10¯⁷ N/C

This is simply defined as the force per unit charge. Mathematically, it is expressed as:

E = F / Q

Where

E = F/Q

E = mg / Q

680 = (1.47×10¯³ × 9.81) / Q

Cross multiply

680 × Q = (1.47×10¯³ × 9.81)

Divide both side by 680

Q = (1.47×10¯³ × 9.81) / 680

Q = 2.12×10¯⁵ C

Q = –2.12×10¯⁵ C (since it is stationary)

E = mg / Q

E = (1.67×10¯²⁷ × 9.81) / 1.60×10¯¹⁹

E = 1.02×10¯⁷ N/C

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Answer:

The answer is [tex]V = 1.785 \times 10^{10}\ Km^3[/tex].

Explanation:

You are asking only for the total volume, so the important data here is the surface area of the earth's crust,

[tex]A = 5.10 \times 10^8\ Km^2[/tex]

and the average thickness of the earth's crust,

[tex]h = 35\ Km[/tex].

Imagine that you can stretch the surface area as if it were a blanket, now if you want to calculate its volume, you just need to multiply its area by its thickness,

[tex]V = A*h = 1.785 \times 10^{10}\ Km^3[/tex].

Answer:

16.17 m/s

Explanation:

h = 3.2 m

u = 18.1 m/s

Angle of projection, θ = 49°

Let H be the maximum height reached by the ball.

The formula for the maximum height is given by

[tex]H=\frac{u^{2}Sin^{2}\theta }{2g}[/tex]

[tex]H=\frac{18.1^{2}\times Sin^{2}49 }{2\times 9.8}=9.52 m[/tex]

The vertical distance fall down by the ball, h'  H - h = 9.52 - 3.2 = 6.32 m

Let v be the velocity of ball with which it strikes the ground.

Use third equation of motion for vertical direction

[tex]v_{y}^{2}=u_{y}^{2}+2gh'[/tex]

here, uy = 0

So,

[tex]v_{y}^{2}=2\times 9.8 \times 6.32[/tex]

vy = 11.13 m/s

vx = u Cos 49 = 18.1 x 0.656 = 11.87 m/s

The resultant velocity is given by

[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]

[tex]v=\sqrt{11.87^{2}+11.13^{2}}[/tex]

v = 16.27 m/s

The maximum potential is [tex]6.45 \times 10^5\ V[/tex] and the maximum charge is [tex]71 \mu C[/tex].

The electric field is the gradient of the potential. So, the electric field in terms of the potential is given as:

[tex]V=E \times r[/tex]

Here, V is the potential, E is the electric field and r is the distance.

Given:

Diameter of the dome, [tex]d=43\ cm[/tex]

Maximum electric field, [tex]E=3.00 \times10^6\ V/m[/tex]

(a) The maximum potential is computed as:

[tex]V_{max}={E}{d/2}\\V_{max}= {3.00 \times 10^6} \times \frac{0.43}{2}\\V_{max}= 6.45 \times 10^5\ V[/tex]

(b) The maximum charge is computed as:

[tex]Q= \frac{Er}{k}\\Q= \frac{3.00 \times 10^6 \times 0.215}{9 \times 10^9}\\Q=71 \mu C[/tex]

Therefore, the maximum potential is [tex]6.45 \times 10^5\ V[/tex] and the maximum charge is [tex]71 \mu C[/tex].

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The maximum potential of the Van de Graaff generator with a diameter of 43.0 cm is 645 kV, and the maximum charge it can hold, before sparks occur due to air breakdown, is 5.11 μC.

The maximum potential of a Van de Graaff generator can be calculated using the formula V = (kQ)/r, where V is the potential, k is Coulomb's constant, Q is the charge, and r is the radius of the sphere. Given that the Van de Graaff generator has a diameter of 43.0 cm (which means the radius r is 21.5 cm or 0.215 m) and it is surrounded by air with a breakdown electric field of 3.00 x 106 V/m, the maximum surface potential (V) equals the breakdown electric field (E) times the radius (r), which gives V = E × r = 3.00 x 106 V/m × 0.215 m = 645,000 V or 645 kV.

The maximum charge on the dome can be calculated using the relationship between the electric field and charge on a sphere, which is E = (kQ)/(r2). Re-arranging this for Q gives Q = Er2/k. Inserting the known values, we get Q = (3.00 x 106 V/m) × (0.215 m)2 / (8.988 x 109 N·m2/C2) = 5.11 x 10−5 C or 5.11 μC, where k is Coulomb's constant.

Damage to the eardrum might start to occur at a depth of approximately [tex]\[ 2.25 \times 10^6 \, \text{m} \][/tex]  underwater when scuba diving in the ocean

To calculate the depth at which damage to the eardrum might occur while scuba diving, we need to consider the increase in pressure as you descend underwater. The pressure increases with depth due to the weight of the water above you. This pressure increase is given by the hydrostatic pressure formula:

[tex]\[ P = \rho \cdot g \cdot h \][/tex]

Where:

- [tex]\( P \)[/tex] is the pressure,

- [tex]\( \rho \)[/tex] is the density of the fluid (seawater in this case),

- [tex]\( g \)[/tex] is the acceleration due to gravity, and

- [tex]\( h \)[/tex]is the depth.

The force exerted on the eardrum is equal to the pressure difference between the external pressure and the pressure inside the ear canal, multiplied by the area of the eardrum. This can be represented as:

[tex]\[ F = A \cdot (P_{\text{internal}} - P_{\text{external}}) \][/tex]

Given that the force on the eardrum should not exceed 1.50 times the force from atmospheric pressure, we have:

[tex]\[ F_{\text{limit}} = 1.5 \cdot P_{\text{atm}} \cdot A \][/tex]

We can solve for the depth [tex](\( h \))[/tex] when this pressure difference exceeds the threshold.

First, let's find [tex]\( P_{\text{atm}} \)[/tex], the atmospheric pressure at sea level, which is approximately [tex]\( 101.3 \, \text{kPa} \)[/tex]. Then, we can calculate the pressure at depth and find the depth where the pressure difference reaches the limit.

[tex]\[ P_{\text{atm}} = 101.3 \, \text{kPa} \][/tex]

Now, let's calculate the depth at which damage to the eardrum might occur:

[tex]\[ F_{\text{limit}} = 1.5 \cdot P_{\text{atm}} \cdot A \]\[ P_{\text{internal}} = P_{\text{atm}} + \rho \cdot g \cdot h \]\[ F_{\text{limit}} = A \cdot (P_{\text{atm}} + \rho \cdot g \cdot h - P_{\text{external}}) \][/tex]

Given that [tex]\( P_{\text{external}} \)[/tex] is atmospheric pressure, we can rewrite this as:

[tex]\[ F_{\text{limit}} = A \cdot (\rho \cdot g \cdot h) \][/tex]

Now, we can rearrange to solve for [tex]\( h \):[/tex]

[tex]\[ h = \frac{F_{\text{limit}}}{A \cdot \rho \cdot g} \][/tex]

Substitute the values:

[tex]\[ h = \frac{1.5 \cdot P_{\text{atm}}}{A \cdot \rho \cdot g} \]\[ h = \frac{1.5 \cdot 101.3 \, \text{kPa}}{\pi \cdot (8.20 \, \text{mm})^2 \cdot (1.03 \times 10^3 \, \text{kg/m}^3) \cdot (9.81 \, \text{m/s}^2)} \]\[ h \approx \frac{1.5 \times 10^5 \, \text{Pa}}{\pi \times (8.20 \times 10^{-3} \, \text{m})^2 \times (1.03 \times 10^3 \, \text{kg/m}^3) \times (9.81 \, \text{m/s}^2)} \]\[ h \approx \frac{1.5 \times 10^5}{\pi \times 6.69 \times 10^{-5} \times 1.03 \times 10^3 \times 9.81} \][/tex]

[tex]\[ h \approx \frac{1.5 \times 10^5}{6.66 \times 10^{-2}} \]\[ h \approx 2.25 \times 10^6 \, \text{m} \][/tex]

So, damage to the eardrum might start to occur at a depth of approximately [tex]\[ 2.25 \times 10^6 \, \text{m} \][/tex]  underwater when scuba diving in the ocean.

Answer:

26.17  V

Explanation:

In a alternator , an ac voltage is produced whose magnitude of voltage varies sinusoidally.

Maximum voltage induced = nBAω

where n is no of turns , B is magnetic field , A is area of the coil and ω is angular velocity .

Rate of rotation   n  = 1000rpm = 1000/60 rps = 16.67 rps

angular velocity ω = 2π n = 2π x 16.67 = 104.68 rad / s

Putting all values in the given equation above

Max voltage = 250 x .1 x .01 x 104.68

= 26.17  V

Answer

The resistor has to be 100

Explanation:

We will have to use the Current Divider Rule, that rule states:

[tex]Ig=\frac{Req}{Rg}*(It)\\[/tex]

where:

Ig= Galvanometer current

It= Total current

Rg= Galvanometer Resistor

Req= Equivalent circuit resistor

For the case of two resistor in parallel:

[tex]Req=\frac{R1*Rg}{R1+Rg}[/tex]

now:

[tex]Req=\frac{2mA}{4mA}*100\\[/tex]

Req=50Ω

having the Equivalent resistor we can calculate R1 reformulating the Req formula:

[tex]R1=\frac{(Rg*Req)}{Rg-Req}\\[/tex]

R1=100 Ω

So now when a 4mA current flows into the new circuit, 2mA  will go through the Galvanometer deflecting the full scale.

To convert the galvanometer to a milliammeter capable of reading up to 4.00 mA, add a 100 Ω shunt resistor in parallel.

To convert a galvanometer with an internal resistance of 100 Ω and a full-scale deflection of 2.00 mA into a milliammeter capable of reading up to 4.00 mA, you need to add a shunt resistor parallel to the galvanometer.

The shunt resistor will divert the excess current, allowing the galvanometer to properly read the increased current range.

Calculate the voltage corresponding to the galvanometer's full-scale deflection:

Determine the current that will flow through the shunt resistor:

Calculate the resistance of the shunt resistor (Rsh) using Ohm's law:

Therefore, a resistor of 100 Ω should be connected in parallel to the galvanometer to enable it to read up to 4.00 mA.

Answer:

mass of bigger object M= 3 kg and mass of smaller object m= 2 kg

Explanation:

force between two objects F= 1.00×10^-8 N

distance between them R= 20 cm = 0.2 m

total mass M+m= 5 kg

We know that

[tex]F= G\frac{Mm}{R^2}[/tex]

putting the values we get

[tex]10^{-8}= 6.67\times10^{-11}\frac{M\times m}{0.2^2}[/tex]

M×m= 6

and M+m= 5

therefore we can calculate that M= 3 kg and m= 2 kg

mass of bigger object M= 3 kg and mass of smaller object m= 2 kg

Answer: Kilogram- newton is wrong unit for Torque

Idem pound-inch is also wrong unit for Torque

Explanation: As it is well known the torque is defined as :

Τorque= F x R

so its UNITS are Newton*meter (SI)

or in the Imperial System is often use  Pound-foot

Answer:

AD

Explanation:

A motion diagram is used to represent the motion of an object. Each point represents the location of the object after equal intervals of time, and the velocity of the object is represented using arrows.

The motion diagram for this problem has been attached here. The question is asking us to analyze the motion at point A. We can make the following observations for the motion diagram at point A:

- The direction of the arrows does not change --> this means that the object is moving on a straight line

- The length of the arrows remains constant at A --> this means that the object is moving at constant speed

Of all the situations listed in the question, only two of them are compatible with these observations:

A) A car is moving along a straight road at a constant speed.

D) A hockey puck slides along a horizontal icy (frictionless) surface. (Frictionless surface means there is no friction acting on the puck, so the acceleration is zero and therefore the speed is constant).

All the other options involves a non-zero acceleration, so that the velocity is not constant. Therefore, only A and D are the correct options.

Explanation:

Given

Refractive Index [tex](n_1)=1.65[/tex]

[tex]n_2=1.35[/tex]

angle of incident[tex]=35 ^{\circ}[/tex]

Light originates in the medium of higher refractive index

Let [tex]\theta _1[/tex] and [tex]\theta _2 [/tex]be the incident and refraction angles respectively  thus according to snell's law

[tex]n_1sin\theta _1=n_2sin\theta _2[/tex]

[tex]1.65\times sin35=1.35\times sin\theta _2[/tex]

[tex]sin\theta _2=0.701[/tex]

[tex]\theta _2=44.507^{\circ}[/tex]

(b)Ray originates from lighter medium

[tex]1.35\times sin35=1.65\times sin\theta _2'[/tex]

[tex]sin\theta _2'=0.469[/tex]

[tex]\theta _2'=27.96\approx 28^{\circ}[/tex]

Answer:

14 N

Explanation:

The tension in the second string is puling both the masses of 20 kg and 8 kg with acceleration of 0.5 m s⁻²

So tension in the second string = total mass x acceleration

= 28 x .5 = 14 N . Ans..

Answer:

The flux through one face of a cube is  [tex]- 5.2730\times 10^{4} Wb[/tex]

Given:

Charge, Q = [tex]- 2.80\micro C = -2.80\times 10^{- 6} C[/tex]

Solution:

The electric flux, [tex]\phi_{E} = \frac{Q}{\epsilon_{o}}[/tex]

Since, there are 6 faces in a cube, the flux through one face:

[tex]\phi_{E} = \frac{Q}{6\epsilon_{o}}[/tex]

[tex]\phi_{E} = \frac{-2.80\times 10^{- 6}}{6\epsilon_{o}}[/tex]

[tex]\phi_{E} = - 5.2730\times 10^{4} Wb[/tex]

Answer:

9.6 m

Explanation:

This is a  case of motion under variable acceleration . So no law of motion formula will be applicable here. We shall have to integrate the given equation .

a = 3.6 t + 5.6

d²x / dt² = 3.6 t + 5.6

Integrating on both sides

dx /dt = 3.6 t² / 2 + 5.6 t + c

where c is a constant.

dx /dt = 1.8  t²  + 5.6 t + c

when t = 0 , velocity dx /dt is zero

Putting these values in the equation above

0 = 0 +0 + c

c = 0

dx /dt = 1.8  t²  + 5.6 t

Again integrating on both sides

x = 1.8 t³ / 3 + 5.6 x t² /2 + c₁

x = 0.6 t³  + 2.8  t²  + c₁

when t =0,  x = 0

c₁ = 0

x = 0.6 t³  + 2.8  t²  

when t = 1.6

x = .6 x 1.6³ + 2.8 x 1.6²

= 2.4576 + 7.168

= 9.6256

9.6 m

The sled moves approximately 16.896 meters from t=0 s to t=1.6 s when it accelerates according to a = (3.6 m/s 3 )t + (5.6 m/s 2 )

The calculation for the distance moved by the sled involves using the equations of motion. For a sled that starts from rest and accelerates according to the function a = (3.6 m/s³) t + (5.6 m/s²), we first find the velocity function by integrating the acceleration function. That is, v(t) = ∫a dt = ∫[(3.6 m/s³)t + (5.6 m/s²)] dt = (1.8 m/s²) t² + (5.6 m/s) t + C. Given that the sled starts from rest, the constant C is zero. Therefore the velocity function is v(t) = (1.8 m/s²) t² + (5.6 m/s) t.

Next, we find the displacement function by integrating the velocity function. That is, x(t) = ∫v dt = ∫[(1.8 m/s²) t² + (5.6 m/s) t] dt = (0.6 m/s) t³ + (2.8 m/s) t² + D. Given the initial condition that the sled starts from rest, the constant D is also zero. So the displacement function is x(t) = (0.6 m/s) t³ + (2.8 m/s) t².

Finally, substituting t = 1.6 s into the displacement function will yield the distance moved by the sled in the time interval from t = 0 to t = 1.6 s, which comes out to be x(1.6 s) = (0.6 m/s) (1.6 s)³ + (2.8 m/s) (1.6 s)² = 9.728 m + 7.168 m = 16.896 m. So, the sled moves approximately 16.896 m.

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Answer:

F = 2 N

Explanation:

Let the rod is made up of large number of point charges

so here force due to one small part which is at distance "x" from the end of the rod on the given charge is

[tex]dF = \frac{kdq q}{(d + x)^2}[/tex]

here we know that

[tex]dq = \frac{Q}{L} dx[/tex]

so we have

[tex]F = \int \frac{k Q q dx}{L(d + x)^2}[/tex]

[tex]F = \int_0^L \frac{kQq dx}{L(d + x)^2}[/tex]

[tex]F = \frac{kQq}{L} ( \frac{1}{d} - \frac{1}{d + L})[/tex]

now plug in all data

[tex]F = \frac{(9\times 10^9)(2C)(10^{-9})}{2.5} (\frac{1}{2} - \frac{1}{(2 + 2.5)})[/tex]

[tex]F = 2N[/tex]

Answer:

1.708*10^6 tons.

Explanation:

Number of Aluminum cans used by 1 person in 1 year = 365/2=182.5 say it as 183 cans per year.

Total number of people in US= 280,000,000

Total number of cans used by americans.

[tex] = 5.12×10^10 cans[/tex]

Weight of 1 can =1/15 pounds

Weight of all cans used in 1 year

[tex]= \frac{5.12*10^10}{15} =3.41*10^9pounds.[/tex]

we know that

1ton=2000pounds.

[tex]\frac{3.41*10^9 pounds}{2000} = 1.708*10^6 tons.[/tex]

Answer:

ρ=0.0102lbm/ft^3

Explanation:

To solve this problem we must take into account the equation of continuity, this indicates that the sum of the mass flows that enter a system is equal to the sum of all those that leave.

Therefore, to find the mass flow of exhaust gases we must add the mass flows of air and fuel.

m=0.59+60=60.59lbm/s( mass flow of exhaust gases)

The equation that defines the mass flow (amount of mass that passes through a pipe per unit of time) is as follows

m=ρVA

Where

ρ=density

V=velocity

m=mass flow

A=cross-sectional area

solving for density

ρ=m/VA

ρ=60.59/{(1485)(4)}

ρ=0.0102lbm/ft^3

Answer:

The total charge Q of the sphere is [tex]2.094\times10^{-10}\ C[/tex].

Explanation:

Given that,

Radius = 5 cm

Charge density [tex]J= 400\ nC/m^3[/tex]

We need to calculate the total charge Q of the sphere

Using formula of charge

[tex]q=\rho V[/tex]

Where, [tex]\rho[/tex] = charge density

V = volume

Put the value into the formula

[tex]q=\rho\times(\dfrac{4}{3}\pi r^3)[/tex]

Put the value into the formula

[tex]q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3[/tex]

[tex]q=2.094\times10^{-10}\ C[/tex]

Hence, The total charge Q of the sphere is [tex]2.094\times10^{-10}\ C[/tex].

Answer and Explanation:

The gravitational acceleration 'g' depend directly on the mass of the object or body and inversely on the distance or radius squared:

[tex]g = \frac{Gm_{o}}{R^{2}}[/tex]

where

[tex]m_{o}[/tex] = mass of the object

G = Gravitational constt

Thus the plastic ball is lighter and have low mass as compared to the steel and golf balls.

This is the reason that a plastic ball have a low value of acceleration as compared to that of steel and golf balls with higher values of acceleration.

Final answer:

The acceleration due to gravity is a constant (g) for all objects in the same gravitational field, and any observed difference in falling rates is likely due to air resistance, not gravitational pull. Experiments have confirmed that objects fall at the same rate regardless of their mass or composition, assuming no air resistance.

Explanation:

The acceleration due to gravity should not vary with the material of the object if we ignore effects such as air resistance. This is because, according to Newton's second law, the force acting on an object is the product of its mass and the acceleration (F = ma). Here, the force due to gravity is mg, where m is the mass and g is the acceleration due to gravity. Since a = F/m, the mass cancels out, leaving a = g, which is constant for all objects in the same gravitational field.

The question assumes that the plastic ball has a lower acceleration due to gravity, which contradicts known physics principles. All objects, regardless of their mass or composition, feel the same acceleration due to gravity near the Earth's surface, assuming no other forces, like air resistance, play a significant role. Historical experiments by scientists such as Galileo and Eötvös have confirmed the equality of gravitational acceleration (g) for different substances within exceptionally high precision.

If you observe differing acceleration rates, this can be often attributed to air resistance, not a difference in gravitational pull. Objects with a larger surface area or less aerodynamic shape, like the plastic ball, may experience greater air resistance and thus appear to fall slower, even though their acceleration due to gravity is the same as denser objects like the steel or golf ball.

Final answer:

The force of attraction between the divalent cation and monovalent anion can be calculated using Coulomb's law. Plugging the values into the formula gives a force of -1.742 N.

Explanation:

The force of attraction between two ions can be calculated using Coulomb's law. The formula is given by F = k * (Q1 * Q2) / r^2, where F is the force, k is the Coulomb constant (9 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges of the ions, and r is the distance between their centers. In this case, since the anion is monovalent and the cation is divalent, the charges would be -1 and +2 respectively.

To calculate the force of attraction, we need to find the equilibrium interionic separation. Given that the atomic radii of the divalent cation and monovalent anion are 0.19 nm and 0.126 nm respectively, their total separation would be the sum of their radii, which is 0.316 nm.

Plugging these values into the formula, we get:

F = (9 x 10^9 Nm^2/C^2) * (-1 C) * (+2 C) / (0.316 x 10^-9)^2 = -1.742 N

Answer:[tex]\theta =57.99\approx 58 ^{\circ}[/tex]

Explanation:

Given

wall is 10 m from student

8 m high

Therefore student should launch at an angle \theta such that its maximum height is 8 m and its range is 20 m

For maximum height(h)[tex]=\frac{u^2sin^2\theta }{2g}[/tex]

Range (R)[tex]=\frac{u^2sin2\theta }{g}[/tex]

[tex]8=\frac{12^2sin^2\theta }{2g}-----1[/tex]

[tex]20=\frac{12^2sin2\theta }{g}----2[/tex]

divide 1 & 2

[tex]\frac{20}{8}=\frac{2sin2\theta }{sin^2\theta }[/tex]

[tex]\frac{5}{2}=\frac{2\times 2\times sin\theat \cdot cos\theta }{sin^2\theta }[/tex]

[tex]tan\theta =\frac{8}{5}[/tex]

[tex]\theta =57.99\approx 58 ^{\circ}[/tex]

Answer:

a) TB = m2 * w^2 * 2*d

b) TA = m1 * w^2 * d + m2 * w^2 * 2*d

Explanation:

The tension on the strings will be equal to the centripetal force acting on the boxes.

The centripetal force is related to the centripetal acceleration:

f = m * a

The centripetal acceleration is related to the radius of rotation and the tangential speed:

a = v^2 / d

f = m * v^2 / d

The tangential speed is:

v = w * d

Then

f = m * w^2 * d

For the string connecting boxes 1 and 2:

TB = m2 * w^2 * 2*d

For the string connecting box 1 to the shaft

TA = m1 * w^2 * d + m2 * w^2 * 2*d

Answer:

The height of the elevator at [tex]T_{i}[/tex] is [tex]\frac{gT_{2}^{2}}{2(1 + \frac{T_{2}}{T_{i}})}[/tex]

Solution:

As per the question:

Let us assume:

The velocity with which the elevator ascends be u'

The height attained by the elevator at time, [tex]T_{i}[/tex] be h

Thus

[tex]u' = \frac{h}{T_{i}}[/tex]     (1)_

Now, with the help of eqn (2) of motion, we can write:

[tex]h = - u'T_{2} + \frac{1}{2}gT_{2}^{2}[/tex]

Using eqn (1):

[tex]h = - \frac{h}{T_{i}}T_{2} + \frac{1}{2}gT_{2}^{2}[/tex]

[tex]h + \frac{h}{T_{i}}T_{2} = \frac{1}{2}gT_{2}^{2}[/tex]

[tex]h(1 + \frac{h}{T_{i}}T_{2}) = \frac{1}{2}gT_{2}^{2}[/tex]

[tex]h = \frac{gT_{2}^{2}}{2(1 + \frac{T_{2}}{T_{1}})}[/tex]

Answer:

acceleration is 1.59 m/s²

required % is 16.22 %

Explanation:

given data

radius = 63 m

speed = 10 m/s

to find out

acceleration of the car and acceleration due to gravity

solution

we know car is moving in circular path

so acceleration will be, a = [tex]\frac{v^2}{r}[/tex]   .........1

put here value

a =  [tex]\frac{10^2}{63}[/tex]

a = 1.59 m/s²

so acceleration is 1.59 m/s²

so acceleration due to gravity as a percentage is 9.8 m/s²

required % = [tex]\frac{1.59}{9.8}*100[/tex]

required %  = 16.22 %

Answer:

857.5 m

2.8583×10⁻⁶ seconds

Explanation:

Time taken by the sound of the thunder to reach the student = 2.5 s

Speed of sound in air is 343 m/s

Speed of light is 3×10⁸ m/s

Distance travelled by the sound = Time taken by the sound × Speed of sound in air

⇒Distance travelled by the sound = 2.5×343 = 857.5 m

⇒Distance travelled by the sound = 857.5 m

Time taken by light = Distance the light travelled / Speed of light

[tex]\text{Time taken by light}=\frac{857.5}{3\times 10^8}\\\Rightarrow  \text{Time taken by light}=2.8583\times 10^{-6}[/tex]

Time taken by light = 2.8583×10⁻⁶ seconds

We want to find the distance between the student and the lighting given that we know the time that passes between the moment he sees the lighting and the moment he hears it.

The solutions are:

Now, notice that there is an intrinsic problem with this question.

The student sees the "flash" when the light impacts on his eyes. Thus, the chain of events is:

So there is a little time between when the lighting happens and the student sees it, that will also have an impact on the distance to the lighting.

then we have:

(3*10^8 m/s)*t₁ = D

(343m/s)*t₂ = D

Where t₁ is the time that the light needs to reach the student, t₂ is the time the sound needs to reach the student, and D is the distance between the student and the ligthing.

We also know that:

t₂ - t₁ = 2.5s

Then we can write:

t₂ = t₁ + 2.5s and replace it on the second equation:

(343m/s)*(t₁ + 2.5s) = D

Then we have a system of equations:

(3*10^8 m/s)*t₁ = D

(343m/s)*(t₁ + 2.5s) = D

We can see that both of these left sides are equal to the same thing, then we must have:

(3*10^8 m/s)*t₁ = (343m/s)*(t₁ + 2.5s)

Then we solve it for t₁

t₁*(3*10^8 m/s - 343m/s) =  (343m/s)*2.5s = 857.5m

t₁ = (857.5m)/(3*10^8 m/s - 343m/s) = 2.858*10^(-6) s

Now that we know the value of t₁, we can find the value of D.

(3*10^8 m/s)*t₁ = D

(3*10^8 m/s)* 2.858*10^(-6) s = D = 857.501 m

Then the solutions are:

If you want to learn more, you can read:

https://brainly.com/question/4598472

Answer:

The equation which describes the simple harmonic motion is [tex]x=0.5\cos(15.81t)[/tex]

Explanation:

Given that,

Mass = 0.4 kg

Spring constant = 100 N/m

Maximum displacement = 0.5 m

We need to calculate the angular frequency

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

[tex]\omega=\sqrt{\dfrac{100}{0.4}}[/tex]

We need to find the equation which describes the simple harmonic motion

Using equation of simple harmonic motion

[tex]x = A\cos\omega t[/tex]

Where, A = amplitude

[tex]\omega [/tex] = angular frequency

Put the value of angular frequency

[tex]x=A\cos\sqrt{\dfrac{k}{m}t}[/tex]

Put the value in the equation

[tex]x=0.5\cos\sqrt{\dfrac{100}{0.4}t}[/tex]

[tex]x=0.5\cos(15.81t)[/tex]

Hence, The equation which describes the simple harmonic motion is [tex]x=0.5\cos(15.81t)[/tex]