A 0.350kg bead slides on a curved fritionless wire,
startingfrom rest at point A. At point B the bead
collideselastically with a 0.530kg ball at rest. find distance call
risesas it moves up the wire. Point A is 2.20 m from ground andfree
fall accel is 9.80 m/s. round answer to 3 significantfigures.

Answer:

The right answer is (D) 340m

Explanation:

If the package is dropped from a helicopter moving upward and the air resistance is negligible, the only acting force in the package is the gravity force. Therefore you can consider this as a Vertical Projectile Motion problem.

The initial speed of the package V₀ is 15m/s.

The acceleration of the package G is 9.8m/s².

The initial hight is Y₀. Therefore:

Y(t)=Y₀+V₀·t-0.5·G·t²

But we know than T(10s)=0m

0m=Y₀+150m-490m

Y₀=340m

The average velocity between the first two signs is 9.1 m/s, between the second and third signs is 10.4 m/s, and the car's acceleration is approximately 0.26 m/s².

To solve for the average velocity and acceleration of the car passing three equally spaced traffic signs, we can use the kinematic equations for uniformly accelerated motion.

Part a: average velocity between the first two signs

The average velocity is calculated by dividing the displacement (distance between the signs) by the time interval: vavg1,2 = d /
[tex](t_2 - t_1)[/tex]. Substituting the given values, we have vavg1,2 = 25 m / (3.9 s - 1.3 s), resulting in an average velocity of 9.1 m/s.

Part b: average velocity between the second and third signs

Similarly, for the average velocity between the second and third signs: vavg2,3 = d / ([tex]t_3 - t_2[/tex]). Using the given times, we calculate vavg2,3 = 25 m / (5.5 s - 3.9 s), giving an average velocity of 10.4 m/s.

Part c: magnitude of the acceleration

To find the acceleration, we can use the velocities we just calculated, and the time between the signs. We can set up an equation for motion for the time intervals. As initial velocity v0 for the second interval is the final velocity of the first interval, we get v0 = vavg1,2 and final velocity v = vavg2,3. Then we can use the equation v = v0 + a[tex](t_3 - t_2)[/tex] to find the acceleration. Rearranging for a, we have a = [tex](v - v_0) / (t_3 - t_2)[/tex]. After calculating, we find that the acceleration is approximately 0.26 m/s².

Answer:

5.4 m

Explanation:

mass, m = 34 kg

initial velocity, u = 9 m/s

final velocity, v = 0 m/s

coefficient of friction, μ = 0.6

Angle of inclination, θ = 10°

Let teh distance traveled before to come into rest is d.

According to the work energy theorem, the work done by all the forces is equal to the change in kinetic energy of the block.

Work done by the gravitational force = W1 = -  mg Sinθ x d  

Work done by the frictional force = W2 = - μ N = - μ mg Cosθ x d

negative sign shows that the direction of force and the direction of displacement is opposite to each other.

Total work done W = W1 + w2

W = - 34 x 9.8 x Sin 10 x d - 0.6 x 34 x 9.8 x cos 10 x d

W = - 254.86 d

Change in kinetic energy = 0.5 x m (v^2 - u^2)

                                         = 0.5 x 34 (0 - 81) = - 1377

So, W = change in KE

- 254.86 d = - 1377

d = 5.4 m

Answer:

31.174°

Explanation:

Bragg's condition is occur when the wavelength of radiation is comparable with the atomic spacing.

So, Bragg's reflection condition for n order is,

[tex]2dsin\theta=n\lambda[/tex]

Here, n is the order of maxima, [tex]\lambda[/tex] is the wavelength of incident radiation, d is the inter planar spacing.

Now according to the question, first order maxima occur at angle of 15°.

Therefore

[tex]2dsin(15^{\circ})=\lambda\\sin(15^{\circ})=\frac{\lambda}{2d}[/tex]

Now for second order maxima, n=2.

[tex]2dsin\theta=2\times \lambda\\sin\theta=\frac{2\lambda}{2d}[/tex]

Put the values from above conditions

[tex]sin\theta=2\times sin(15^{\circ})\\sin\theta=2\times 0.258819045103\\sin\theta=0.517638090206\\\theta=sin^{-1}0.517638090206\\ \theta=31.174^{\circ}[/tex]

Therefore, the second order maxima occurs at 31.174° angle.

Answer:

1.43 x 10¹⁷.

They will accelerate away from each other.

Explanation:

Force on each charged sphere F = mass x acceleration

= 8.55 x 10⁻³ x 25 x 9.8

= 2.095 N

Let Q be the charge on each sphere

F = [tex]\frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}[/tex]

2.095 =[tex]\frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}[/tex]

Q² =[tex]\frac{2.095\times(15)^2\times10^{-4}}{9\times10^9}[/tex]

Q = 2.289 X 10⁻⁶

No of electrons = Charge / charge on a single electron

= [tex]\frac{2.289\times10^{-6}}{1.6\times10^{-19}}[/tex]

=1.43 x 10¹³.

They will accelerate away from each other.

The number of electrons would have to add to each sphere to accelerate at 25.0g is 1.43×10¹³ and accelerate away from each other.

Electric force is the force of attraction of repulsion between two bodies.

According to the Coulombs law, the force of attraction of repulsion between charged two bodies is directly proportional to the product of charges of them and inversely proportional to the square of distance between them. It can be given as,

[tex]F=\dfrac{KQ_1Q_2}{r^2}[/tex]

Here, (k) is the coulombs constant,  (q1 and q2) is the charges of two bodies and (r) is the distance between the two charges.

Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.

Two spheres accelerate at 25.0g when released, and the mass of each sphere is 8.55 g. Thus, the force on the sphere can be given as,

[tex]F=8.55\times{10^{-3}}\times25\times9.8\\F=2.095\rm \;N[/tex]

Two very small spheres are 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.As the charge on both the sphere is same (say Q) Put these values in the above formula as,

[tex]2.095=\dfrac{(9\times10^{9})QQ}{(15\times10^{-2})^2}\\Q=2.289\times10^{-6}\rm \;C[/tex]

It is known that the charge on one electron is 1.6×10⁻¹⁹ C. Thus the number of electron in the above charge is,

[tex]n=\dfrac{2.289\times10^{-6}}{1.6\times10^{-19}}\\n=1.43\times10^{13}[/tex]

The direction of acceleration of both the sphere will be opposite to each other. Thus, they accelerate away from each other.

Hence, the number of electrons would have to add to each sphere to accelerate at 25.0g is 1.43×10¹³ and accelerate away from each other.

Learn more about the electric force here;

https://brainly.com/question/14372859

Answer:

The energy of the capacitors connected in parallel is 0.27 J

Given:

C = [tex]2.0\micro F = 2.0\times 10^{- 6} F[/tex]

C' = [tex]4.0\micro F = 4.0\times 10^{- 6} F[/tex]

Potential difference, V = 300 V

Solution:

Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:

[tex]C_{eq} = C + C' = 2.0 + 4.0 = 6.0\micro F = 6.0\times 10^{- 6} F[/tex]

The energy of the capacitor, E is given by;

[tex]E = \frac{1}{2}C_{eq}V^{2}[/tex]

[tex]E = \frac{1}{2}\times 6.0\times 10^{- 6}\times 300^{2} = 0.27 J[/tex]

Answer:

9.934 m/s²

Explanation:

Given:

Initial speed of the Bugatti Veyron Super Sport = 0 mi/h

Final speed of the Bugatti Veyron Super Sport = 60 mi/h

Now,

1 mi/h = 0.44704 m / s

thus,

60 mi/h = 0.44704 × 60 = 26.8224 m/s

Time = 2.70 m/s

Now,

The acceleration (a) is given as:

[tex]a=\frac{\textup{Change in speed}}{\textup{Time}}[/tex]

thus,

[tex]a=\frac{26.8224 - 0 }{2.70}[/tex]

or

a = 9.934 m/s²

Answer:

T₁ = 19.39N   with an angle of 52° with the horizontal.

T₂ = 15.64N  with an angle of 40° with the horizontal.

Explanation:

1)We calculate the weight of holiday decoration

W = m×g

W: holiday decoration weight in Newtons (N)

m: holiday decoration weight = 3 kg

g: acceleration due to gravity = 9.8 m/s²

W = m×g = 3 kg × 9.8 m/s² = 29.4 N

2)We apply Newton's first law to the system in equilibrium:

ΣFx = 0

T₁Cos52° - T₂Cos40°=0

T₁Cos52° = T₂Cos40°

T₁ = T₂(Cos40° / T₁Cos52°)

T₁ = 1,24×T₂  Equation (1)

ΣFy = 0

T₁Sin52° + T₂Sin40° - 29.4 = 0 Equation (2)

We replace T₁ of  Equation (1) in the Equation (2)

1.24×T₂×Sin52° + T₂Sin40° - 29.4 = 0

1.88*T₂ = 29.4

T₂ = 29.4/1.88

T₂ = 15.64N

We replace T₂ = 15.64N in the Equation (1)  to calculate T₁:

T₁ = 1.24×15.64N

T₁ = 19.39N

Answer:

1. 6.672 kPa

2. 49.05 mm of mercury

Explanation:

h = 6400 m

Absolute pressure, p = 46 kPa = 46000 Pa

density of air, d = 0.823 kg/m^3

density of mercury, D = 13600 kg/m^3

(a) Absolute pressure = Atmospheric pressure + pressure due to height

46000 = Atmospheric pressure + h x d x g

Atmospheric pressure = 46000 - 6400 x 0.823 x 10 = 6672 Pa = 6.672 kPa

(b) To convert the pressure into mercury pressure

Atmospheric pressure = H x D x g

Where, H is the height of mercury, D be the density of mercury, g be the acceleration due to gravity

6672 = H x 13600 x 10

H = 0.04905 m

H = 49.05 mm of mercury

To calculate the local atmospheric pressure in the city, use the formula P = P0 * e^(-Mgh/RT), where P is the pressure at the current location, P0 is the pressure at sea level, M is the molar mass of air, g is the acceleration due to gravity, h is the altitude, R is the ideal gas constant, and T is the temperature. Plug in the given values into the formula and solve for P0 to find the local atmospheric pressure in kPa. To convert this pressure to mmHg, use the conversion factor that 101.325 kPa is equal to 760 mmHg.

To calculate the local atmospheric pressure in the city, we can use the relationship between altitude, pressure, and density of air. Altitude affects atmospheric pressure because as we go higher in the atmosphere, the air becomes less dense and therefore exerts less pressure. The formula to calculate pressure is given by:

P = P0 * e^(-Mgh/RT)

Where P is the local atmospheric pressure, P0 is the pressure at sea level, M is the molar mass of air, g is the acceleration due to gravity, h is the altitude, R is the ideal gas constant, and T is the temperature. Since we know the altitude and pressure at the current location, we can rearrange the formula to solve for P0:

P0 = P / e^(-Mgh/RT)

Using the given values, we can substitute them into the formula. The density of air is 0.828 kg/m³, the altitude is 6400 m, the absolute pressure is 46 kPa, the molar mass of air is the average molar mass of air (28.97 g/mol), the acceleration due to gravity is 9.8 m/s², the ideal gas constant is 8.314 J/(mol·K), and the temperature can be assumed to be the average temperature at that altitude (around 0°C or 273 K).

Plugging in these values into the formula, we get:

Calculating this expression will give us the local atmospheric pressure in kPa.

To convert this pressure to mmHg, we can use the conversion factor that 101.325 kPa is equal to 760 mmHg. So, to find the pressure in mmHg, we can multiply the pressure in kPa by the conversion factor.

Answer:

6.93 m/s

Explanation:

mass of marble, m = 2 g

Diameter of the bowl = 4.9 m

radius of bowl, r = half of diameter of the bowl = 2.45 m

The length of the string is 2.45 m

Now it is executing simple harmonic motion, thus the potential energy at the bottom is equal to the kinetic energy at the highest point.

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

where, m be the mass of marble, v be the velocity of marble at heighest point, g be the acceleration due to gravity and r be the radius of the path which is equal to the length of the string

By substituting the values

9.8 x 2.45 = 0.5 x v^2

v = 6.93 m/s

Thus, the speed is 6.93 m/s.

Answer:

[tex]9.11\times 10^{-15}.[/tex]

Explanation:

The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of  -40 nC is removed from the water droplet.

The charge on one electron, [tex]\rm e=-1.6\times 10^{-19}\ C.[/tex]

Let the N number of electrons have charge -40 nC, such that,

[tex]\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.[/tex]  

Now, mass of one electron = [tex]\rm 9.11\times 10^{-31}\ kg.[/tex]

Therefore, mass of N electrons = [tex]\rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.[/tex]

It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.

Let it is m times the total mass of the droplet which is [tex]25\ \rm mg = 25\times 10^{-6}\ kg.[/tex]

Then,

[tex]\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.[/tex]

It is the required fraction of mass of the droplet.

Answer:

Final volumen first process [tex]V_{2} = 98,44 cm^{3}[/tex]

Final Pressure second process [tex]P_{3} = 1,317 * 10^{10} Pa[/tex]

Explanation:

Using the Ideal Gases Law yoy have for pressure:

[tex]P_{1} = \frac{n_{1} R T_{1} }{V_{1} }[/tex]

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

[tex]n_{1} = n_{2} = n [/tex]

In an isobaric process the pressure is constant so:

[tex]P_{1} = P_{2} [/tex]

[tex]\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }[/tex]

[tex]\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }[/tex]

[tex]V_{2} = \frac{T_{2} V_{1} }{T_{1} }[/tex]

Replacing : [tex]T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}[/tex]

[tex]V_{2} = 98,44 cm^{3}[/tex]

Replacing on the ideal gases formula the pressure at this piont is:

[tex]P_{2} = 3,92 * 10^{9} Pa[/tex]

For Temperature the ideal gases formula is:

[tex]T = \frac{P V }{n R }[/tex]

For the second process you have that [tex]T_{2} = T_{3} [/tex]  So:

[tex]\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }[/tex]

[tex]P_{2} V_{2}  = P_{3} V_{3} [/tex]

[tex]P_{3} = \frac{P_{2} V_{2}}{V_{3}} [/tex]

[tex]P_{3} = 1,317 * 10^{10} Pa[/tex]

Answer:

The acceleration of proton will be [tex]431.137\times 10^8m/sec^2[/tex]

Explanation:

We have given electric field E = 450 N/C

Charge on proton [tex]=1.6\times 10^{-19}C[/tex]

Force on electron due to electric field is given by [tex]F=qE=1.6\times 10^{-19}\times 450=720\times 10^{-19}N[/tex]

Mass of electron [tex]m=1.67\times 10^{-27}kg[/tex]

Now according to second law of motion [tex]F=ma[/tex]

So [tex]720\times10^{-19}=1.67\times 10^{-27}a[/tex]

[tex]a=431.137\times 10^8m/sec^2[/tex]

So the acceleration of proton will be [tex]431.137\times 10^8m/sec^2[/tex]

Answer:

El flujo del campo U a través de dicha superficie es [tex]\Phi_U=4\pi r^2K[/tex]

Explanation:

El flujo vectorial de un campo U a través de una superficie S está dado por

[tex]\Phi_U=\int_S \vec{U} \cdot \hat{n} \ dS[/tex]

donde [tex]\hat{n}[/tex] representa el vector unitario normal a la superficie. Teniendo en cuenta que el campo se encuentra en dirección radial, se tiene

[tex]\Phi_U=\int_S K\hat{r} \cdot \hat{r} \ dS = K\int_S dS = KS=K*4\pi r^2[/tex]

Answer:

The minimum time to get the car under max. speed limit of 79 km/h is 2.11 seconds.

Explanation:

[tex]a=\frac{V_f-V_0}{t}[/tex]

isolating "t" from this equation:

[tex]t=\frac{V_f-V_0}{a}[/tex]

Where:

a=[tex]-4.6m/s^2[/tex] (negative because is decelerating)

[tex]V_f= 79 km/h[/tex]

[tex]V_0= 114 km/h[/tex]

First we must convert velocity from km/h to m/s to be consistent with units.

[tex]79\frac{km}{h}*\frac{1000m}{1 km}*\frac{1h}{3600s}=\frac{79*1000}{3600}=21.94 m/s[/tex]

[tex]114\frac{km}{h}*\frac{1000m}{1 km}*\frac{1h}{3600s}=\frac{114*1000}{3600}=31.67 m/s[/tex]

So;

[tex]t=\frac{V_f-V_0}{a}=\frac{21.94 m/s-31.66m/s}{-4.6 m/s^2}=2.11 s[/tex]

Final answer:

The motorist's total displacement is 179.58 km north, and his average velocity is approximately 63.38 km/h north for the entire trip.

Explanation:

Displacement and Average Velocity Calculation

For the given problem, we can find the motorist's total displacement by adding the distances traveled in each segment of the journey, all in the same direction (north). First, the motorist drives north for 35.0 minutes at 85.0 km/h. The distance for this part of the trip is:

Distance = Speed × Time = 85.0 km/h × (35.0 min / 60 min/h) = 49.58 km (rounded to 2 decimal places)

The next segment involves a stop, so the displacement remains unchanged.

Finally, the motorist drives an additional 130 km north.

Therefore, the total displacement is the sum of these distances: Total Displacement = 49.58 km + 130 km = 179.58 km north.

To find the average velocity, we need to consider the total displacement and the total time, including the stop:

Total time = Driving time + Stopping time = (35.0 min + 2.00 h + 15.0 min) = 2.833 hours (2 hours and 50 minutes).

Average Velocity = Total Displacement / Total Time = 179.58 km / 2.833 h ≈ 63.38 km/h north.

The motorist's total displacement is 179.58 km north, and his average velocity over the entire trip is approximately 63.38 km/h north.

Answer:

Clockwise, converging towards

Explanation:

At the center of surface , system with low pressure in Northern hemisphere have air around the center that rotates in anti clockwise direction.

As a result of low pressure, the air is directed slightly inwards thus converges towards the center of the system.

In the high pressure systems, air rotates in clockwise direction and diverges away from the center of the system.

Answer:

Explanation:

Relative velocity = 100 - 15 = 85 m /s

Distance to be covered up with this relative speed = 1.3km

= 1300m

If t be the time taken

85 x t = 1300

t = 1300/ 85

= 15.29

Time of fall of the bomb t₁ by 1 km

1000 = 1/2 g t₁²

t₁² = 2000/9.8

t₁ = 14.28 s

Since t₁≠ t₂

It will not score a direct hit.

Answer:595.32 N

Explanation:

Given

Mechanic shop has a car of weight([tex]F_1[/tex]) 16,500 N

Area of lift([tex]A_1[/tex])=[tex]125 cm^2[/tex]

Area of small piston([tex]A_2[/tex])=[tex]4.51 cm^2[/tex]

According to pascal's law pressure transmit will be same in all direction in a closed container

[tex]P_1=P_2[/tex]

[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]

[tex]F_2=F_1\frac{A_2}{A_1}[/tex]

[tex]F_2=16,500\times \frac{4.51}{125}=595.32 N[/tex]

Answer:

The speed of the block when it reaches the bottom is 17 m/s.

(b) is correct option.

Explanation:

Given that,

Height = 15 m

Mass = 2 kg

We need to calculate the speed of the block when it reaches the bottom

Using conservation of energy

[tex]mgh=\dfrac{1}{2}mv^2[/tex]

[tex]v^2=2gh[/tex]

[tex]v=\sqrt{2gh}[/tex]

Where, g = acceleration due to gravity

h = height

Put the value into the formula

[tex]v=\sqrt{2\times9.8\times15}[/tex]

[tex]v=17\ m/s[/tex]

Hence, The speed of the block when it reaches the bottom is 17 m/s.

Answer:

Explanation:

given,

mass of neutron = 1.675 × 10⁻²⁷ kg

1 kg =  5.58 × 10³⁵ eV/c²                                        

a) mass of proton = 1.675 × 10⁻²⁷ ×  5.58 × 10³⁵ = 9.34 × 10⁸ eV/c²

b) mass of electron = 9.109 x 10⁻³¹ × 5.58 × 10³⁵ = 5.08 × 10⁵eV/c²

c) mass of hydrogen = 1.675 × 10⁻²⁷  × 5.58 × 10³⁵ = 9.34 × 10⁸ eV/c²

Answer:

After 0.19288 sec second object hit the ground after hitting first object    

Explanation:

We have given two objects are dropped from at the same time from height 0.49 m and 1.27 m

As the object is dropped so its initial velocity will be zero u = 0 m/sec

For object 1 height h = 0.49 m

From second equation of motion

[tex]h=ut+\frac{1}{2}gt^2[/tex]

[tex]0.49=0\times t+\frac{1}{2}\times 9.8\times t^2[/tex]

[tex]t^2=0.1[/tex]

t = 0.31622 sec

For second object

[tex]h=ut+\frac{1}{2}gt^2[/tex]

[tex]1.27=0\times t+\frac{1}{2}\times 9.8\times t^2[/tex]

[tex]t^2=0.1[/tex]

t = 0.5091 sec

So difference in time = 0.5091-0.31622 = 0.19288 sec

So after 0.19288 sec second object hit the ground after hitting first object

Answer: 20 m/s

Explanation: To solve this problem we have to consider the expression of the kinetic energy given by:

Ekinetic=(1/2)*(m*v^2)

then E=0.5*30Kg*(20 m/s)^2=15*400=6000J

Answer:

a) 94.176 m/s

b) 452.04 m

Explanation:

t = Time taken by the quarter to reach the ground = 9.6 s

u = Initial velocity

v = Final velocity

s = Displacement

a)

[tex]v=u+at\\\Rightarrow v=0+9.81\times 9.6\\\Rightarrow v=94.176 m/s[/tex]

Speed of the quarter just before it hits the ground is 94.176 m/s

b)

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 9.81\times 9.6^2\\\Rightarrow s=452.04\ m[/tex]

The building is 452.04 m high

If[tex]\( q_1 = 0 \), then \( q_2 = 11 \)[/tex] which is positive.

Thus, [tex]\( q_1 = 0 \, \text{C} \) and \( q_2 = 11 \, \text{C} \)[/tex] are the valid solutions.

To find the charges [tex]\( q_1 \) and \( q_2 \),[/tex] we can use Coulomb's law, which states that the magnitude of the force between two point charges is given by:

[tex]\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \][/tex]

Given that the force exerted by one charge on the other has a magnitude of 8 mN (milli-Newtons) when they are separated by 4.5 m, we can set up the equation as follows:

[tex]\[ 8 \times 10^{-3} \, \text{N} = \frac{(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \cdot |q_1 \cdot q_2|}{(4.5 \, \text{m})^2} \][/tex]

Solving for[tex]\( |q_1 \cdot q_2| \[/tex]), we get:

[tex]\[ |q_1 \cdot q_2| = \frac{8 \times 10^{-3} \, \text{N} \cdot (4.5 \, \text{m})^2}{8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2} \][/tex]

[tex]\[ |q_1 \cdot q_2| = \frac{8 \times 10^{-3} \times 20.25}{8.99 \times 10^9} \, \text{C}^2 \][/tex]

[tex]\[ |q_1 \cdot q_2| = \frac{0.162}{8.99 \times 10^9} \, \text{C}^2 \][/tex]

[tex]\[ |q_1 \cdot q_2| = 1.802 \times 10^{-11} \, \text{C}^2 \][/tex]

Now, since both charges are positive and repel each other, we know that [tex]\( q_1 \)[/tex]is smaller than [tex]\( q_2 \)[/tex]. So, let's assume [tex]\( q_1 = x \) and \( q_2 = 11 - x \)[/tex]. Substituting these values into the equation for [tex]\( |q_1 \cdot q_2| \)[/tex], we get:

[tex]\[ x \cdot (11 - x) = 1.802 \times 10^{-11} \, \text{C}^2 \][/tex]

[tex]\[ 11x - x^2 = 1.802 \times 10^{-11} \, \text{C}^2 \][/tex]

Now, solve this quadratic equation to find  x , which represents [tex]\( q_1 \).[/tex] Once we find x , we can find [tex]\( q_2 = 11 - x \).[/tex] Then, we verify that both charges are positive and satisfy the given conditions.

To solve the quadratic equation [tex]\( 11x - x^2 = 1.802 \times 10^{-11} \, \text{C}^2 \)[/tex], we first rearrange it to the standard form:

[tex]\[ x^2 - 11x + 1.802 \times 10^{-11} = 0 \][/tex]

We can use the quadratic formula to find[tex]\( x \):[/tex]

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

where [tex]\( a = 1 \), \( b = -11 \), and \( c = 1.802 \times 10^{-11} \).[/tex]

Plugging in these values:

[tex]\[ x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(1)(1.802 \times 10^{-11})}}{2(1)} \][/tex]

[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]

[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]

[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]

[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]

[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]

[tex]\[ x \approx \frac{11 \pm \sqrt{121}}{2} \][/tex]

[tex]\[ x \approx \frac{11 \pm 11}{2} \][/tex]

[tex]\[ x_1 \approx \frac{11 + 11}{2} \][/tex]

[tex]\[ x_1 \approx \frac{22}{2} \][/tex]

[tex]\[ x_1 \approx 11 \][/tex]

[tex]\[ x_2 \approx \frac{11 - 11}{2} \][/tex]

[tex]\[ x_2 \approx \frac{0}{2} \][/tex]

[tex]\[ x_2 \approx 0 \][/tex]

So, we have [tex]\( x_1 = 11 \) and \( x_2 = 0 \).[/tex]

Since [tex]\( q_1 \)[/tex]  must be smaller than [tex]\( q_2 \),[/tex] we take [tex]\( x_1 = 11 \) as \( q_1 \) and \( x_2 = 0 \) as \( q_2 \).[/tex]

Therefore,[tex]\( q_1 = 11 \, \text{C} \) and \( q_2 = 0 \, \text{C} \)[/tex]. But we need both charges to be positive, so this solution is not valid.

Let's check the other root [tex]\( x_2 = 0 \):[/tex]

If[tex]\( q_1 = 0 \), then \( q_2 = 11 \)[/tex] which is positive.

Thus, [tex]\( q_1 = 0 \, \text{C} \) and \( q_2 = 11 \, \text{C} \)[/tex] are the valid solutions.

Answer:

Light travels faster in a less dense medium

Explanation:

In a less dense medium, light will travel faster than it did in the more dense medium (an example of this transition can be light going from water to air).

This is because in a less dense medium there ares less particles, so light finds less resistence in its path.

The opposite happens in a more dense medium, there are more particles so light will travel slower because more particles will get in the way.

In summary, the more dense the medium, the more particles it has, and the slower that light will travel in it.

Answer:

Answer

Explanation:

If the satellite is in perfect orbit around the earth then gravitational pull of the earth on the satellite and moving forward satellite is in perfect balance i.e gravitational pull does not effect the orbit as the velocity of the satellite is good enough to counter the gravitational pull of the earth and inertia makes the satellite move forward with greater velocity, but when its velocity is suddenly reduced, the gravitational pull will counter on the orbiting satellite and it will crash onto the earth.

This is same as in well of death when motorist moves in circular motion inside the well with centripetal and centrifugal forces balanced, suddenly crashes on the ground when velocity of the motorist is not enough to balance forces.

the satellite's mass has no affect on its orbit around the earth rather the mass of the earth is only affecting factor .

Answer:8769 Kg/m^3

Explanation: First of all we have to calculate the volume of the cylinder so it is equal to:

Vcylinder= π*r^2*h where r and h are the radius and heigth, respectively.

Vcylinder= π*0.0275^2*0.048=1.14*10^-4 m^3

The density is defined by ρ=mass/volume

the we have

ρ=1Kg/1.14*10^-4 m^3=8769 Kg/m^3

Answer:

The current in the wire under the influence of the force is 216.033 A

Solution:

According to the question:

Length of the wire, l = 0.676 m

[tex]\theta = 57.7^{\circ}[/tex]

Magnetic field of the Earth, [tex]B_{E} = 5.29\times 10^{- 5} T[/tex]

Forces experienced by the wire, [tex]F_{m} = 6.53\times 10^{-3} N[/tex]

Also, we know that the force in a magnetic field is given by:

[tex]F_{m} = IB_{E}lsin\theta[/tex]

[tex]I = \frac{F_{m}}{B_{E}lsin\theta}[/tex]

[tex]I = \frac{6.53\times 10^{-3}}{5.29\times 10^{- 5}\times 0.676sin57.7^{\circ}[/tex]

I = 216.033 A

Answer with Explanation:

Newton's laws are applicable for inertial frames of reference which is a frame which is not accelerating when seen from the observer standing on earth.

For the person as he presses the brakes his frame is a decelerating frame of reference hence he cannot apply the newtons laws of motion as they are in their original form but if he analyses the motion he has to apply a correction known as  pseudo-force on the object he is analyzing. Pseudo Force has no basis in newton's laws but are a correction that needs to be applied if he wishes to analyse the motion from non inertial frame of reference

While as a person standing on earth outside the car since his frame is an inertial frame of reference he can apply newton's laws of motion without any correction.