A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 18.0 m/s, and the distance between them is 58.0 m . After t1 = 5.00 secs, the motorcycle starts to accelerate at a rate of 4.00 m/s^2 . The motorcycle catches up with the car at some time t2 . A) How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find t2-t1 Express the time numerically in seconds using three significant figures. B) How far does the motorcycle travel from the moment it starts to accelerate (at time t1 ) until it catches up with the car (at time t2)? Should you need to use an answer from a previous part, make sure you use the unrounded value. Answer numerically in meters using three significant figures.

Answer:

total number of electron in 1 litter is 3.34 × [tex]10^{26}[/tex] electron

Explanation:

given data

mass per mole = 18 g/mol

no of electron = 10

to find out

how many electron in 1 liter of water

solution

we know molecules per gram mole is 6.02 ×[tex]10^{23}[/tex] molecules

no of moles is 1

so

total number of electron in water is = no of electron ×molecules per gram mole × no of moles

total number of electron in water is = 10 × 6.02 ×[tex]10^{23}[/tex] × 1

total number of electron in water is = 6.02×[tex]10^{24}[/tex] electron

and

we know

mass = density × volume    ..........1

here we know density of water is 1000 kg/m

and volume = 1 litter = 1 × [tex]10^{-3}[/tex] m³

mass of 1 litter = 1000 × 1 × [tex]10^{-3}[/tex]

mass = 1000 g

so

total number of electron in 1 litter =  mass of 1 litter × [tex]\frac{molecules per gram mole}{mass per mole}[/tex]

total number of electron in 1 litter =  1000 × [tex]\frac{6.02*10{24}}{18}[/tex]

total number of electron in 1 litter is 3.34 × [tex]10^{26}[/tex] electron

Answer:

Refractive index of slab = 1.22

Explanation:

Using Snell's law as:

[tex]n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}[/tex]

Where,  

[tex]{\theta_i}[/tex]  is the angle of incidence  ( 27.0° )

[tex]{\theta_r}[/tex] is the angle of refraction  ( 21.8° )

[tex]{n_r}[/tex] is the refractive index of the refraction medium  (slab, n=?)

[tex]{n_i}[/tex] is the refractive index of the incidence medium (vacuum, n=1)

Hence,  

[tex]1\times {sin27.0^0}={n_r}\times{sin21.8^0}[/tex]

Refractive index of slab = 1.22

The index of refraction of the transparent material is found using Snell's law and is approximately 1.333.

To find the index of refraction of the transparent material, we can use Snell's law, which states the relationship between the angles of incidence and refraction at the boundary between two media with different refractive indices. Snell's law is given by the equation:

n1sin(θ1) = n2sin(θ2)

Where n1 is the refractive index of the first medium (vacuum in this case, which is 1), θ1 is the angle of incidence, n2 is the refractive index of the second medium (the transparent material), and θ2 is the angle of refraction.

Here, we have θ1 = 27.0° and θ2 = 21.8°. Plugging in the known values:

1 * sin(27.0°) = n2 * sin(21.8°)

n2 = sin(27.0°) / sin(21.8°)

Using a calculator, we find:

n2 ≈ 1.333

Thus, the index of refraction of the transparent material is approximately 1.333.

Answer:

Explanation:

Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula

[tex]\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]

where [tex]| \vec{A} |[/tex] its the magnitude and θ.

So, for our vectors, we will have:

[tex]\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )[/tex]

[tex]\vec{D}=  ( 2.121 m , -2.121 m )[/tex]

and

[tex]\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )[/tex]

[tex]\vec{E}= ( 2.71 m , 3.59 m )[/tex]

Now, we can take the sum of the vectors

[tex]\vec{R} = \vec{D} + \vec{E}[/tex]

[tex]\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )[/tex]

[tex]\vec{R} = ( 2.121 \ m  + 2.71 \ m , -2.121 \ m + 3.59 \ m ) [/tex]

[tex]\vec{R} = ( 4.831 \ m , 1.469 \ m ) [/tex]

This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem

[tex]|\vec{R}| = \sqrt{R_x^2 + R_y^2}[/tex]

[tex]|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}[/tex]

[tex]|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}[/tex]

[tex]|\vec{R}| = \sqrt{25.496 m^2}[/tex]

[tex]|\vec{R}| = 5.049 m[/tex]

To find the direction, we can use

[tex]\theta = arctan(\frac{R_y}{R_x})[/tex]

[tex]\theta = arctan(\frac{1.469 \ m}{4.831 \ m})[/tex]

[tex]\theta = arctan(0.304)[/tex]

[tex]\theta = 16\°54'33''[/tex]

As we are in the first quadrant, this is relative to the x axis.

Answer:

3469.788 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

First rocket

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=150\times t+\frac{1}{2}\times 5\times t^2\\\Rightarrow s=150t+2.5t^2\ m[/tex]

Second rocket

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=150\times t+\frac{1}{2}\times 15\times t^2\\\Rightarrow s=150t+7.5t^2\ m[/tex]

When this will collide the total distance they would have covered would be 6000 m.

[tex]6000=150t+2.5t^2+150t+7.5t^2\\\Rightarrow 6000=300t+10t^2\\\Rightarrow 10t^2+300t-6000=0[/tex]

[tex]t=5\left(\sqrt{33}-3\right),\:t=-5\left(3+\sqrt{33}\right)\\\Rightarrow t=13.72, -43.72[/tex]

Hence at 13.72 seconds they will collide assuming they are launched at the same time.

[tex]s=150t+7.5t^2\\\Rightarrow s=150\times 13.72+7.5\times 13.72^2\\\Rightarrow s=3469.788\ m[/tex]

The second rocket would have gone 3469.788 m when they collide

Final answer:

To find the distance covered by the second rocket at the point of collision, we analyze both rockets' motion based on their initial velocities and accelerations. By using the equations of motion, we can solve for the time of collision and calculate the distance each rocket has traveled, specifically looking for the second rocket's distance.

Explanation:

To find how far the second rocket has gone when they collide, we must analyze the motion of both rockets taking into account their initial velocities and accelerations. The first rocket has an initial velocity of 150 m/s and accelerates at 5 m/s2, while the second rocket starts with the same velocity but accelerates at 15 m/s2. Both rockets start 6000 meters apart. To find the point of collision, we use the equation of motion s = ut + (1/2)at2 for both, where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time until collision.

Let's denote the distance covered by the first rocket as s1 and the second as s2, with the total distance being s1 + s2 = 6000 m. By finding the time of collision and plugging it back into the equation of motion for the second rocket, we can find s2, the distance covered by the second rocket. The actual calculation involves setting up equations based on the motion of both rockets and solving for the time t, after which s2 can be calculated.

The spring constant of the spring is -1.376 N/m.

To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. In this case, the displacement of the spring is 1.2 cm, and the charges are 2.2 cm apart. The force exerted by the spring can be calculated using the equation F = kx, where F is the force, k is the spring constant, and x is the displacement.

Given that the displacement (x) is 1.2 cm and the force exerted is caused by the electric force between the charges, which is given by Coulomb's Law, we can write the equation:

F = kx = k(1.2 cm) = (k/100 cm) * 1.2 cm = k/83.33

Similarly, the electric force between the charges is given by Coulomb's Law: F = kq1q2/r^2, where k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between them. In this case, the charges are 2.2 cm apart and have magnitudes of 2.0 μC and -4.0 μC, respectively.

When the charges are 2.2 cm apart, we can calculate the electric force:

F = k(2.0 μC)(-4.0 μC)/(2.2 cm)^2 = (-8.0 kμC^2)/(4.84 cm^2) = (-8.0 k)/(4.84 cm^2) μC^2

Equating the force exerted by the spring to the electric force, we have:

k/83.33 = (-8.0 k)/(4.84 cm^2) μC^2

Solving for k:

k/83.33 = (-8.0 k)/(4.84 cm^2) μC^2

k * (4.84 cm^2)/(83.33) = -8.0 k * μC^2

(4.84 cm^2)/(83.33) = -8.0 μC^2

k = (-8.0 μC^2) * (83.33)/(4.84 cm^2) = -137.6 μC^2/cm^2 = -1.376 N/m

Therefore, the spring constant of the spring is -1.376 N/m.

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Answer:

Total distance covered during the journey is 235 m

Solution:

As per the question:

Initial velocity, v = 0 m/s

Acceleration, a = [tex]2 m/s^{2}[/tex]

Time, t = 5 s

Now,

For this, we use eqn 2 of motion:

[tex]d = vt + \farc{1}{2}at^{2}[/tex]

[tex]d = 0.t + \farc{1}{2}\times 2\times 5^{2} = 25 m[/tex]

The final speed of car after t = 5 s is given by:

v' = v + at

v' = 0 + 2(5) = 10 m/s

Now, the car travels at constant speed of 10 m/s for t' = 20 s with a = 0:

[tex]d' = vt + \farc{1}{2}at^{2}[/tex]

[tex]d' = 10\times 20 + \farc{1}{2}\times 0\times 20^{2} = 25 m[/tex]

d' = 200 m

Now, the car accelerates at a= - 5 [tex]m/s^{2}[/tex] until its final speed, v" = 0 m/s:

[tex]v"^{2} = v'^{2} + 2ad"[/tex]

[tex]0 = {10}^{2} + 2\times (- 5)d"[/tex]

[tex]100 = 10d"[/tex]

d" = 10 m

Total distance covered = d + d' + d"  = 25 + 200 + 10 = 235 m

Answer:

Part A: [tex]8.06\times10^{-3}\ m[/tex]

Part B: [tex]6.36\times 10^{-9}\ s[/tex]

Explanation:

Given:

Assumptions:

Part A:

Since the electron moves in the direction of the electric field, the electric force will act on it in the direction opposite to electric field. This electric force does work on it to make the electron come to rest.

Using the work-energy theorem, the work done by the electric field will be equal to the kinetic energy change of the electron.

[tex]\therefore -eEx = \dfrac{1}{2}m(v^2-v_o^2)\\\Rightarrow -eEx=-\dfrac{1}{2}mv_o^2\\\Rightarrow x=\dfrac{mv_o^2}{2eE}\\\Rightarrow x=\dfrac{9.1\times 10^{-31}\times (5.08\times 10^{6})^2}{2\times 1.6\times10^{-19}\times 9100}\\\Rightarrow x=8.06\times 10^{-3}\ m[/tex]

Hence, the electron comes to rest by travelling a distance of [tex]8.06\times 10^{-3}\ m[/tex].

Part B:

In this part, let us first find out the acceleration of the electron due to the electric force.

[tex]a = -\dfrac{eE}{m}\\\Rightarrow a= -\dfrac{1.6\times10^{-19}\times 9100}{9.1\times 10^{-31}}\\\Rightarrow a= -1.6\times 10^{15}\ m/s^2\\[/tex]

The electron moves with the above acceleration constantly as it moves in the uniform electric field.

Since the electron is supposed to move from a point and then again move back to the same point. This means the displacement of the electron is zero.

[tex]i.e.,\ s=0\\\Rightarrow v_ot+\dfrac{1}{2}at^2=0\\\Rightarrow (v_o+\dfrac{1}{2}at)t=0\\\Rightarrow \dfrac{(2v_o+at)}{2}t=0\\\Rightarrow t = 0\,\,\, or\,\,\, (2v_o+at)=0\\\Rightarrow t = 0\,\,\, or\,\,\, t=\dfrac{-2v_o}{a}\\[/tex]

Since the electron starts moving at t = 0 s.

[tex]\therefore t = \dfrac{-2v_o}{a}\\\Rightarrow t=\dfrac{-2\times 5.08\times 10^6}{-1.6\times 10^{15}}\\\Rightarrow t= 6.36\times 10^{-9}\ s[/tex]

Hence, the electron returns to the starting position after [tex]6.36\times 10^{-9}\ s[/tex].

Answer:

The electric field at a distance r = 0.02 m is 14062.5 N/C.

Solution:

Refer to fig 1.

As per the question:

Radius of sphere, R = 0.04 m

Charge, Q = [tex]5.0\times 10^{- 9} C[/tex]

Distance from the center at which electric field is to be calculated, r = 0.02 m

Now,

According to Gauss' law:

[tex]E.dx = \frac{Q_{enclosed}}{\epsilon_{o}}[/tex]

Now, the charge enclosed at a distance r is given by volume charge density:

[tex]\rho = \frac{Q_{enclosed}}{area}[/tex]

[tex]\rho = \frac{Q_{enclosed}}{\frac{4}{3}\pi R^{3}}[/tex]

Also, the charge enclosed Q' at a distance r is given by volume charge density:

[tex]\rho = \frac{Q'_{enclosed}}{\frac{4}{3}\pi r^{3}}[/tex]

Since, the sphere is no-conducting, Volume charge density will be constant:

Thus

[tex]\frac{Q_{enclosed}}{\frac{4}{3}\pi R^{3}} = \frac{Q'_{enclosed}}{\frac{4}{3}\pi r^{3}}[/tex]

Thus charge enclosed at r:

[tex]Q'_{enclosed} = \frac{Q_{enclosed}}{\frac{r^{3}}{R^{3}}[/tex]

Now, By using Gauss' Law, Electric field at r is given by:

[tex]4\pi r^{2}E = \frac{Q_{enclosed}r^{3}}{\epsilon_{o}R^{3}}[/tex]

Thus

[tex]E = \frac{Q_{enclosed}r}{4\pi\epsilon_{o}R^{3}}[/tex]

[tex]E = \frac{(9\times 10^{9})\times 5.0\times 10^{- 9}\times 0.02}{0.04^{3}}[/tex]

E = 14062.5 N/C

Answer:

[tex]t=6.19s[/tex]

Explanation:

With a initial velocity is zero, kinematics equation is:

[tex]y=1/2*g*t^2[/tex]

[tex]t=\sqrt{2y/g}=\sqrt{2*188/9.81}=6.19s[/tex]

Answer:

[tex]\frac{dv_{r}}{dt}=g-k_{r}v_{r}^2[/tex]

Explanation:

Second Newton's Law:

[tex]F=\frac{dp}{dt}=\frac{d(mv)}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt} \\[/tex]    (1)

m and v are the instantaneous mass and instantaneous velocity

The only force is the weight:

[tex]F=mg[/tex]          (2)

On the other hand we know:

[tex]\frac{dm}{dt}=k*m*v[/tex]        (3)

We replace (2) and (3) in (1), and we solve for dv/dt :

[tex]\frac{dv}{dt}=g-kv^2[/tex]

Answer:

The weight of Earth's atmosphere exert is [tex]516.6\times10^{17}\ N[/tex]

Explanation:

Given that,

Average pressure [tex]P=1.01\times10^{5}\ Pa[/tex]

Radius of earth [tex]R_{E}=6.38\times10^{6}\ m[/tex]

Pressure :

Pressure is equal to the force upon area.

We need to calculate the weight of earth's atmosphere

Using formula of pressure

[tex]P=\dfrac{F}{A}[/tex]  

[tex]F=PA[/tex]

[tex]F=P\times 4\pi\times R_{E}^2[/tex]

Where, P = pressure

A = area

Put the value into the formula

[tex]F=1.01\times10^{5}\times4\times\pi\times(6.38\times10^{6})^2[/tex]

[tex]F=516.6\times10^{17}\ N[/tex]

Hence, The weight of Earth's atmosphere exert is [tex]516.6\times10^{17}\ N[/tex]

Answer:

5.164 x 10^19 N

Explanation:

P = 1.01 x 10^5 Pa

R = 6.38 x 10^6 m

Area = 4 π R²

[tex]A= 4\times 3.14\times6.38\times10^{6}\times6.38\times10^{6}[/tex]

A = 5.112 x 10^14 m^2

Pressure is defined as the force exerted per unit area.

The formula for the pressure is

P = F / A

Where, F is the force, A be the area

here force is the weight of atmosphere.

F = P x A  

F = 1.01 x 10^5 x 5.112 x 10^14

F = 5.164 x 10^19 N

Answer:

potential energy of net is 8.58 kJ

Explanation:

given data

mass m = 70 kg

height h = 11 m

stop distance x = 1.50 m

to find out

potential energy of net

solution

we know here man eventually stop

so elastic potential energy of net = change in potential energy of man body

so equation will be

potential energy of net = m×g×h    ...................1

here m is mass and h is total height = ( h+ x)  

height = 11 + 1.5 = 12.5 m and g is 9.8

so put here value in equation 1

potential energy of net = m×g×h

potential energy of net = 70×9.8×12.5

potential energy of net = 8575 J

so potential energy of net is 8.58 kJ

Answer:

a) 200 m

b) 100 m/s

c) 709 m

d) -118.2 m/s

e) 26.24 s

Explanation:

The rocket flies upward with constant acceleretion.

The equation for position under constant acceleration is:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 = 0

V0 = 0

Y(t) = 1/2 * 25 * t^2

Y(t) = 12.5 * t^2

And speed under constant acceleration:

Vy(t) = Vy0 + a * t

Vy(t) = 25 * t

It burns for 4 s and runs out of fuel

Y(4) = 12.5 * 4^2 = 200 m

V(4) = 25 * 4 = 100 m/s

Form t = 4 the rocket will coast, it will be in free fall, affected only by gravity

It will be under constant acceleration. These new equations will have different starting constants.

Y(t) = Y4 + Vy4 * (t - 4) + 1/2 * g * (t - 4)^2

Vy(t) = Vy4 + g * (t - 4)

When it reaches its highest point it will have a speed of zero.

0 = Vy4 + g * (t - 4)

0 = 100 - 9.81 * (t - 4)

100 = 9.81 * (t - 4)

t - 4 = 100 / 9.81

t = 10.2 + 4 = 14.2 s

At that moment it will have a height of:

Y(14.2) = 200 + 100 * (14.2 - 4) - 1/2 * 9.81 * (14.2 - 4)^2 = 709 m

The rocket will fall and hit the ground:

Y(t) = 0 = 200 + 100 * (t - 4) - 1/2 * 9.81 * (t - 4)^2

0 = 200 + 100 * t - 400 - 4.9 * (t^2 - 8 * t +16)

0 = -4.9 * t^2 + 139.2 * t -278.4

Solving this equation electronically:

t = 26.24 s

At that time the speed will be:

Vy(t) = 100 - 9.81 * (26.24 - 4) = -118.2 m/s

Answer:

b) 2.2 kg

Explanation:

Net force acting on an object is the sum of the  two forces acting on the body.

The net force is calculated using the parallelogram law of vectors.

F =[tex]\sqrt{{A^{2}} + B^{2}+2 A B cos \theta}[/tex]

Here A = 20 N , B = 35 N and θ =80°

Net Force = F = 43.22 N

Acceleration = a = 20 m/s/s

Since F = ma, m = F/a = 43.22 / 20 = 2.161 kg = 2.2 kg

Answer:

The location of the center of gravity for the entire body, relative to the floor is 1.03 m

Explanation:

To find the center of gravity of a system of particles, we use that

[tex]R_{cog} = \frac{r_{1}*m_{1}+r_{2}*m_{2}+...+r_{n}*m_{n}}{M}[/tex]

where R is the vector center of gravity of the system, formed by n particles, and n masses.

In this case, for a person standing on the floor and being their body divided in three sectors, each one with a weight and an altitud in a specific point (center of gravity of the body sector), instead of mass we have every "particle" weight in Newtons (force instead of mass), being each "particle" in the formula, a sector of the body.

On the other hand, we use only magnitude for the calculation, because the gravity force is vertical to the floor, so instead of our vector formula, we use it in the vertical direction as a magnitude formula. Thus

[tex]Y_{cog} = \frac{y_{1}*W_{1}+y_{2}*W_{2}+y_{3}*W_{3}}{Wt}[/tex]

where Y is the center of gravity, y=1, 2, 3 is every "sector point" altitude from the floor, W=1, 2, 3 is every weight of a body "sector", and Wt is the sum of the three weights.

In this way we replace in our formula with the correspondent values

[tex]Y_{cog} = \frac{1.28m*438N+0.76m*144N+0.25m*87N}{669N}[/tex]

obtaining our result

[tex]Y_{cog}=1.03 m[/tex]

Answer:

2790 Pa

Explanation:

Given wavelength λ= 5μm

temperature T= 400 K

cross section of collision σ= 0.28 nm^2

molar mass = 39.9 g/mole

pressure = [tex]P= \frac{RT}{\sqrt{2}N_A\sigma\lambda }[/tex]

putting values we get

=[tex] \frac{8.314\times400}{\sqrt{2}\times6.022\times10^{23}\times0.28\times10^{-18}\times5\times10^{-6} }[/tex]

⇒P = 2790 J/m^3

the partial pressure are argon atoms expected= 2790 Pa

Answer:

0.06 Nm

Explanation:

mass of object, m = 3 kg

radius of gyration, k = 0.2 m

angular acceleration, α = 0.5 rad/s^2

Moment of inertia of the object

[tex]I = mK^{2}[/tex]

I = 3 x 0.2 x 0.2 = 0.12 kg m^2

The relaton between the torque and teh moment off inertia is

τ = I α

Wheree, τ is torque and α be the angular acceleration and I be the moemnt of inertia

τ = 0.12 x 0.5 = 0.06 Nm

Answer:

Power factor of the AC series circuit is [tex]cos\phi=0.5[/tex]

Explanation:

It is given that,

Impedance of the AC series circuit, Z = 60 ohms

Resistance of the AC series circuit, R = 30 ohms

We need to find the power factor of the circuit. It is given by :

[tex]cos\phi=\dfrac{R}{Z}[/tex]

[tex]cos\phi=\dfrac{30}{60}[/tex]

[tex]cos\phi=\dfrac{1}{2}[/tex]

[tex]cos\phi=0.5[/tex]

So, the power factor of the ac series circuit is [tex]cos\phi=0.5[/tex]. Hence, this is the required solution.

The power factor of an AC series circuit with an impedance of 60 Ohms and a resistance of 30 Ohms is 0.5, indicating that half of the power is effectively used.

The power factor in an AC circuit is a measure of how much of the power is being effectively used to do work. It's calculated as the ratio of the resistance R to the impedance Z. Given an AC series circuit with an impedance of 60 Ohms and a resistance of 30 Ohms, the power factor is the resistance divided by the impedance.

The calculation is straightforward: Power factor = R / Z = 30 Ohms / 60 Ohms = 0.5.

This means that the power factor for this particular AC circuit is 0.5, which indicates that only half of the total power is being used effectively while the other half is reactive power, which does not perform any work but is necessary for the functioning of the circuit's reactive components.

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Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive

Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:

[tex]F=\frac{k*q1*q2}{d^{2}}[/tex]

Replacing the dat we obtain F=82 nN.

The force is repulsive because the points charged have the same sign.

Answer:

1.86 75 X 10⁻⁵ C.

Explanation:

Capacitance of parallel plate capacitor is given by

C = K∈₀ A /d

C is capacitance , A is plate area , d is distance between plate and ∈₀ is permittivity of air which is 8.85 x 10⁻¹²  and K is dielectric constant

Using the data given in the question

C = [tex]\frac{6.7\times 8.85\times10^{-12}\times 1.4}{.04\times10^{-3}}[/tex]

C = .2075 X 10⁻⁵ F

Charge Q = CV

.2075 X 10⁻⁵ X 9 = 1.86 75 X 10⁻⁵ C.

The Earth moves 9.48 × 10⁸ kilometers in a year as it revolves around the Sun, calculated by multiplying the average speed of 30 km/s by the time of one year, 3.16 × 10⁷ seconds.

To calculate the distance that Earth moves in a year as it revolves around the Sun, we can use the formula for distance traveled, which is distance = speed × time. Given that Earth orbits around the Sun at an average speed of 30 kilometers per second (km/s) and the time it takes for one orbit is 3.16 × 107 seconds, the calculation would be:

30 km/s × 3.16 × 107 s = distance

Distance = 30 × 3.16 × 107 km

Distance = 94.8 × 107 km

Distance = 9.48 × 108 km

The Earth moves 9.48 × 108 kilometers in a year as it revolves around the Sun, using the correct number of significant figures.

Answer:

1.38 x 10^-18 J

Explanation:

q = - 1.6 x 10^-19 C

d = 5 x 10^-10 m

the potential energy of the system gives the value of work done

The formula for the potential energy is given by

[tex]U =\frac{Kq_{1}q_{2}}{d}[/tex]

So, the total potential energy of teh system is

[tex]U =\frac{Kq_{1}q_{2}}{d}+\frac{Kq_{2}q_{3}}{d}+\frac{Kq_{1}q_{3}}{d}[/tex]

As all the charges are same and the distance between the two charges is same so the total potential energy becomes

[tex]U =3\times \frac{Kq^{2}}{d}[/tex]

K = 9 x 10^9 Nm^2/C^2

By substituting the values

[tex]U =3\times \frac{9\times 10^{9}\times \ 1.6 \times 1.6 \times 10^{-38}}{5\times 10^{-10}}[/tex]

U = 1.38 x 10^-18 J

The work which must be done to bring three electrons from a great distance apart is 5.0×10⁻¹⁰m from one another is 1.38 x 10⁻¹⁸ J

q = -1.6 x 10⁻¹⁹C

d = 5 x 10⁻¹⁰m

Potential energy of the system =  Value of work done

Potential energy= U = Kq₁q₂/d

It is an equilateral triangle so the the charges and the distance between

the charges are the same.

Potential energy = 3 ₓ kq² / d

Substitute the values into the equation

3 × (9 × 10⁹ ˣ 1.6 ˣ 1.6 ˣ 10⁻³⁸) / (5ˣ 10⁻¹⁰)

U = 1.38 x 10⁻¹⁸ J

Read more about Work done here https://brainly.com/question/25573309

Answer:

[tex]10942249.24 m^{-1}[/tex]

Explanation:

Rydberg's formula is used to describe the wavelengths of the spectral lines of chemical elements similar to hydrogen, that is, with only one electron being affected by the effective nuclear charge. In this formula we can find the rydberg constant, knowing the wavelength emitted in the transcision between two energy states, we can have a value of the constant.

[tex]\frac{1}{\lambda}=Z^2R(\frac{1}{n^2_{1}}-\frac{1}{n^_{2}^2}})[/tex]

Where [tex]\lambda[/tex] it is the wavelength of the light emitted, R is the Rydberg constant, Z is the atomic number  of the element and [tex]n_{1} n_{2}[/tex] are the states where [tex]n_{1}<n_{2}[/tex].

In this case we have Z=1 for hydrogen, solving for R:

[tex]R=\frac{1}{\lambda}*(\frac{1}{n^2_{1}}-\frac{1}{n^_{2}^2}})^{-1}\\R=\frac{1}{658.9*10^{-9}m}*(\frac{1}{2^2}-\frac{1}{3^2}})^{-1}\\R=1.52*10^6m^{-1}*(\frac{36}{5})=1.09*10^7 m^{-1}=10942249.24m^{-1}[/tex]

This value is quite close to the theoretical value of the constant [tex]R=10967758.34 m^{-1}[/tex]

Answer:

Explained

Explanation:

This happens because the water cancels the focusing effect of our eye lens. A drop of water makes a lens because of the different refractive indices of water and air.

When we are underwater with naked eyes,  the front part of our eye lens does not have different refractive indices anymore… both the  inside lens and outside lens are essentially water ( lens inside our eyes are water only surrounded by a membrane)… hence, that part of the lens does not focus the light anymore. And thus, your eyes don’t work  well anymore. (It can only partly correct for this, by bending the lens more)

When we put swimming goggles on, we remove the water and again have air in front of our eye lenses, and they work normally now. Obviously, we have created an extra interface between water and air in  front of our swimming goggles. However, this surface(the goggle) is flat. And thus it does not function as a lens but only as a window.

Answer:

a) v1 = a1*t1 = 2162 m/s

b) v2 = v1 + a2*(t2-t1) = 2591 m/s

c) [tex]Dt = D1 + D2 = \frac{v1^{2}}{2*a1} + \frac{v2^{2}-v1^{2}}{2*a2}=51004.5m[/tex]

Explanation:

For any movement with constant acceleration:

Vf = vo + a*Δt.  Replacing the propper values, with get the answers for parts a) and b):

v1 = a1*t1 = 2162 m/s

v2 = v1 + a2*(t2-t1) = 2591 m/s

Using the formula for displacement we can calculate the total distance asked on part c):

[tex]V_{f}^{2}=V_{o}^2+2*a*D[/tex]  Solving for D and replacing the values for each part of the launch:

[tex]D=\frac{V_{f}^{2}-V_{o}^{2}}{2*a}[/tex]

D1 = 24863m

D2 = 26141.5m

Finally we add D1 + D2 for the total distance:

D = 51004.5m

Answer:

a) 5.65 s

b) 5cm/s

c) They will pass each other at both 1.1168 s and 5.84s

d)15.084cm and 38.7 cm

Explanation:

For part A, you need to keep in mind that acceleration is the rate of change of velocity per unit of time. For a constant acceleration, this can be told in this way:

[tex]a = \frac{v - v_o}{t}[/tex]

Reordering this equation, we can get v in terms of the initial velocity, the acceleration, and the time elapsed:

[tex]v = at + v_o[/tex]

Now, we can get the expressions for velocity of each toy car, and equalize them:

[tex]v_1 =a_1t + v_o_1\\v_2 =a_2t + v_o_2\\v_1 = v2\\a_1t +v_o_1 =a_2t + v_o_2\\(a_1 - a_2)t = v_o_2 - v_o_1\\t = \frac{v_o_2 - v_o_1}{a_1 - a_2} = \frac{5cm/s - (-3cm/s)}{2.3 cm/s^2 - 0 cm/s^2}= 3.47 s[/tex]

As toy car has no acceleration and, therefore, constant speed, both car will have the same speed when toy car 1 reaches this velocity = 5cm/s

c) The position of car 1, as it follows a constant acceleration motion, is given by this equation:

[tex]x_1 = \frac{1}{2}a_1t^2 + v_o_1t + x_o_1[/tex]

The position for car 2, as it has constant velocity, is given by this equation:

[tex]x_2 = v_2t + x_o_2[/tex]

We equalize both equation to find the time where the cars pass each other:

[tex]x_1 = x_2\\\frac{1}{2}a_1t^2 + v_o_1t + x_o_1 = v_2t+x_o_2\\\frac{1}{2} a_1t^2 + (v_o_1 - v_2)t + x_o_1 - x_o_2 = 0\\\frac{1}{2}2.3m/s^2t^2 +(-3cm/s-5cm/s)t+ 17cm - 9.5cm = 0\\1.15t^2 -8t + 7.5 = 0 | a = 1.15, b = -8, c = 7.5\\t = \frac{-b +-\sqrt{b^2 - 4ac}}{2a} = 5.84s | 1.1168 s[/tex]

The car will pass each other at both 1.1168s and 5.84s.

For the positions, we solve any of the position equation with the solutions:

[tex]x = v_2*t + x_o_2 = 5cm/s *5.84s + 9.5cm = 38.7 cm\\x = 5cm/s * 1.1168s + 9.5cm = 15.084 cm[/tex]

The two toy cars have equal speeds at approximately t = 3.48s, with both traveling at 5 cm/s. They pass each other at t ≈ 1.99s, with their location at approximately 15.4 cm.

Part A: Equal Speeds:

We know the first car has an initial velocity of -3 cm/s and an acceleration of 2.30 cm/s². Its velocity at any time t can be given by v1 = v0 + at, that is -3 + 2.30t.

The second car has a constant velocity of 5.0 cm/s.

For them to have equal speeds, v1 = v2,

so -3 + 2.30t = 5.

Solving for t gives t ≈ 3.48 s.

Part B: Speeds at Equal Times:

The common speed when both cars have equal speeds can be found by substituting the time back into either equation for velocity.

Doing so for the first car gives

-3 + 2.30(3.48) ≈ 5 cm/s.

Part C: Cars Passing Each Other:

The cars pass each other when their positions are equal. Using the equations for position x1 = x01 + v01t + (1/2)at² and x2 = x02 + v02t,

we get 17 + (-3)t + (1/2)(2.30)t² = 9.5 + (5)t.

Solving for t gives two possible times, but only the positive one is relevant, t ≈ 1.99 s.

Part D: Location at Passing Time:

Substituting the time when they pass each other back into the position equations, for the first car we have

17 + (-3)(1.99) + (1/2)(2.30)(1.99)² ≈ 15.4 cm.

The second car will be at the same position since they are passing each other.

Answer:

Option d

Explanation:

When we throw an object in the upward direction, we provide it with certain initial velocity due to which it covers a certain distance up to the maximum height.

While the object is moving in the upward direction, its velocity keeps on reducing due to the acceleration due to gravity which acts vertically downwards in the opposite direction thus reducing its velocity.

So, the maximum height attained by the object is the point where this upward velocity of the body becomes zero and after that the object starts to fall down.

Answer:

Part 1) Number of electrons in 1 liter of water equals[tex]N=3.346\times 10^{26}[/tex]

Part 2) Net charge of all the electrons equals [tex]Charge=53.61\times 10^{6}[/tex]

Explanation:

Since we know that the density of water is 1 kilogram per liter thus we infer that mass of 1 liter of water is 1 kilogram hence we need to find electron's in 1 kg of water.

Now since it is given that molar mass of water is 18.0 grams this means that 1 mole of water contains 18 grams of water.

Hence by ratio and proportion number of moles in 1 kg water equals

[tex]n=\frac{1000}{18}[/tex]

Now by definition of mole we know that 1 mole of any substance is Avagadro Number of particles.

Hence the no of molecules in 'n' moles of water equals

[tex]n'=\frac{1000}{18}\cdot N_a\\\\n'=\frac{1000}{18}\cdot 6.023\times 10^{23}\\[\tex][tex]\\n'=3.346\times 10^{25}[/tex]

Now since it is given that each molecule has 10 electron's thus the total number of electrons in n' molecules equals

[tex]N=10\times 3.346\times 10^{25}\\\\N=3.346\times 10^{26}[/tex]

Part 2)

We know that charge of 1 electron equals [tex]1.602\times 10^{-19}C[/tex] the the charge of electrons in 'N' quantity equals

[tex]Charge=1.6022\times 10^{-19}\times 3.346\times 10^{26}\\\\Charge=52.61\times 10^{6}Columbs[/tex]  

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

[tex]\frac{v_1}{v}[/tex] = e ( coefficient of restitution ) = [tex]\frac{1}{\sqrt{10} }[/tex]

and

[tex]\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }[/tex]

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

[tex]e = \sqrt{\frac{h_1}{6.1} }[/tex]

[tex]e = \sqrt{\frac{h_2}{h_1} }[/tex]

So on

[tex]e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }[/tex]

[tex](\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}[/tex]= .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

The ball can bounce approximately 8 times before it reaches the height of 2.38 meters after accounting for a 10% energy loss each collision.

This problem involves a ball undergoing inelastic collisions, losing 10% of its kinetic energy with each bounce, and determining how many times it can bounce to still reach a windowsill 2.38 m high.

First, let's calculate the initial potential energy (PEinitial) of the ball when it is dropped from a height (hinitial) of 6.10 m:

PEinitial = mghinitial

where, g = 9.8 m/s² (acceleration due to gravity)

As it falls, this entire potential energy converts into kinetic energy (KEinitial) at the ground:

KEinitial = PEinitial = mghinitial

Upon each bounce, the ball loses 10% of its kinetic energy. Therefore, it retains 90% of its kinetic energy:

KEnew = 0.9 × KEprevious

To find out how high it can bounce after each loss of energy, convert kinetic energy back into potential energy:

PE = KE = mgh

After each bounce, the height the ball can reach is calculated by applying the 90% retention factor:

hnew = 0.9 × hprevious

Starting with h0 = 6.10 m:

The number of bounces can be calculated using the formula: hn = 6.10 × (0.9)n, where n is the number of bounces. Set hn = 2.38 m and solve for n:

2.38 = 6.10 × (0.9)n

Divide both sides by 6.10:

0.39 ≈ (0.9)n

Taking the natural log (ln) of both sides:

ln(0.39) = n × ln(0.9)

Finally, solving for n:

n ≈ ln(0.39) / ln(0.9) ≈ 8

Therefore, the ball can bounce approximately 8 times and still reach the windowsill that is 2.38 m above the ground.

Answer:

Explanation:

Given

mass of each arrow=0.1 kg

velocity of arrow=30 m/s

One arrow is fired u=due to east and another towards south

Momentum of first arrow

[tex]P_1=0.1\left ( 30\hat{i}\right )=3\hat{i}[/tex]

[tex]P_2=0.1\left ( -30\hat{j}\right )=-3\hat{j}[/tex]

Total momentum P

[tex]P=P_1+P_2[/tex]

[tex]P=3\hat{i}-3\hat{j}[/tex]

magnitude [tex]|P|=\sqrt{3^2+3^2}=3\sqrt{2}[/tex]

direction

[tex]tan\theta =\frac{-3}{3}[/tex]

[tex]\theta =45^{\circ}[/tex] clockwise w.r.t to east

The magnitude of the total momentum of the two-arrow system is approximately 4.24 kg·m/s, and the direction is 45° south of due east.

The question involves calculating the total momentum of a system comprising two arrows fired in perpendicular directions. Momentum is a vector quantity, meaning it has both magnitude and direction. Since the two arrows have the same speed but are fired in perpendicular directions, their momenta are also perpendicular to each other. The momentum of each arrow can be calculated using the formula p = mv, where p is momentum, m is mass, and v is velocity.

The momentum of the first arrow (due east) is peast = m * v = 0.100kg * 30.0m/s = 3.0 kg·m/s east, and the momentum of the second arrow (due south) is psouth = m * v = 0.100kg * 30.0m/s = 3.0 kg·m/s south. To find the total momentum of the system, we need to find the resultant of these two momentum vectors.

The magnitude of the total momentum is found using the Pythagorean theorem: ptotal = √(peast² + psouth²) = √(3.0² + 3.0²) kg·m/s = √(9 + 9) kg·m/s = √18 kg·m/s ≈ 4.24 kg·m/s.

The direction of the total momentum with respect to due east can be found using trigonometry, specifically tan-1 (psouth/peast), which gives us 45° south of east, since the magnitudes are equal.