The total distance treaveled by a car moving in a straight line is as follows: After the first 7.0 minutes it has gone a total of 2.0 miles. After 14.0 minutes it has traveled a total of 4.5 miles. Finally at 21.0 minutes it has traveled a total of 6.0 miles. Find the average speed at: Time

Answer:

[tex]6.88\times 10^{-6}\ C/m^2[/tex]

Explanation:

Q = Total charge = [tex]6.3\times 10^{-8}\ C[/tex]

[tex]\rho[/tex] = Surface charge density

r = Radius = 2.7 cm

A = Surface area = [tex]4\pi r^2[/tex]

Charge is given by

[tex]Q=A\rho\\\Rightarrow Q=4\pi r^2\times \rho\\\Rightarrow \rho=\dfrac{Q}{4\pi r^2}\\\Rightarrow \rho=\dfrac{6.3\times 10^{-8}}{4\pi (2.7\times 10^{-2})^2}\\\Rightarrow \rho=0.00000687706544224\\\Rightarrow \rho=6.88\times 10^{-6}\ C/m^2[/tex]

The surface charge density is [tex]6.88\times 10^{-6}\ C/m^2[/tex]

Answer:

[tex]\theta=7^o[/tex]

Explanation:

Displacement

It is a vector that points to the final point where an object traveled from its starting point. If the object traveled to several points, then the individual displacements must be added as vectors.

The mail carrier leaves the post office and drives 2 km due north. The first displacement vector is

[tex]\vec r_1=<0,2>\ km[/tex]

Then the carrier drives 7 km in 60° south of east. The displacement has two components in the x and y axis given by

[tex]\vec r_2=<7cos60^o,-7sin60^o>\ km=<3.5\ ,-6.06>\ km[/tex]

Finally, he drives 9.5 km 35° north of east.

[tex]\vec r_3=<9.5cos35^o,9.5sin35^o>\ km=<7.78\ ,\ 5.45>\ km[/tex]

The total displacement is

[tex]\vec r_t=<0,2>\ km+<3.5\ ,-6.06>\ km+<7.78\ ,\ 5.45>\ km[/tex]

[tex]\vec r_t=<11.28,1.39>\ km[/tex]

The direction can be calculated with

[tex]\displaystyle tan\theta=\frac{1.39}{11.28}=0.1232[/tex]

[tex]\boxed{\theta=7^o}[/tex]

Answer:

a)  y = A x² , b)   A = - ½ g / v₀², c)    v₀ = 15.46 m / s

Explanation:

For this problem of two-dimensional kinematics, we will use that the time to reach the wall is the same

X axis

         x = v₀ₓ t

         t = x / v₀ₓ

Y Axis  

         y = [tex]v_{oy}[/tex] t - ½ g t²

As it shoots horizontally the vertical speed is zero

        y = - ½ g t²

We replace

         y = - ½ g (x / v₀ₓ)²

The initial speed is all horizontal

        v₀ₓ = v₀

         y = - ½ g / v₀²    x²

        y = A x²

b) the expression for the constant is

           A = - ½ g / v₀²

c) we look for the initial speed

           v₀² = - ½ g x² / y

   As the object falls below the exit point its height is negative

          v₀ = √ (- ½  9.8   3²/ (-0.21))

          v₀ = 15.46 m / s

Answer: v = 5.79 * 10^10m/s.

Explanation: By using the work-energy theorem, we know that the work done on the electron by the potential difference equals the kinetic energy of the electrons.

Mathematically, we have that

qV = 1/2mv²

q= magnitude of an electronic charge = 1.609*10^-16c

V= potential difference = 95v

m = mass of an electronic charge = 9.11* 10^-31kg.

v = velocity of electron.

Let us substitute the parameters, we have that

1.609*10^-16 * 95 = (9.11*10^-31 * v²) /2

1.609*10^-16 * 95 * 2 = 9.11*10^-31 * v²

305.71 * 10^-16 = 9.11 * 10^-31 * v²

v² = 305.71 * 10^-16/ 9.11 * 10^-31

v² = 3.355 * 10^21.

v = √3.355 * 10^21

v = 5.79 * 10^10 m/s

Final answer:

Using the principle of energy conservation and the equation KE = qV = ½ mv^2, the velocity of the electrons in a CRT with an accelerating potential of 95.0 V is calculated to be approximately 5.8 x 10^6 m/s.

Explanation:

To determine how fast electrons will be moving when they hit the screen in a Cathode-Ray Tube (CRT) with an accelerating potential of 95.0 V, one can use the principle of energy conservation. The amount of kinetic energy (KE) gained by an electron when it is accelerated through a potential difference (V) is equal to the change in electrical potential energy (PE), which is given by the equation KE = qV, where q is the charge of the electron (1.602 x 10-19 Coulombs) and V is the potential difference.

The formula for kinetic energy is KE = ½ mv^2, with 'm' representing the mass of the electron (9.109 x 10^-31 kg) and 'v' being the velocity of the electron. Setting these equations equal gives us qV = ½ mv^2. Solving for 'v' and plugging in the values for q, V, and m, we can find the velocity of the electrons when they hit the screen.

Calculation:

qV = ½ mv^2

2qV = mv^2

v = √(2qV/m)

v = √(2 * 1.602 x 10^-19 C * 95.0 V / 9.10^9 x 10^-31 kg)

v ≈ 5.8 x 10^6 m/s

Thus, the electrons would be traveling at approximately 5.8 x 10^6 m/s when they hit the screen.

Answer:

d = 3.53 m

Explanation:

The Coulomb Force is given as

[tex]\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r[/tex]

The x-component of the force is equal to

[tex]F_x = F\cos(\theta) = F\frac{x}{\sqrt{x^2 + y^2}} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{(5^2 + d^2)}\frac{d}{\sqrt{5^2 + d^2}} = \frac{1}{4\pi\epsilon_0}\frac{dq_1q_2}{(5^2 + d^2)^{3/2}}[/tex]

This is basically a function of (d). So, the maximum value of this function is the point where its derivative with respect to d is equal to zero.

[tex]\frac{dF_x}{dd} = \frac{kq_1q_2}{(d^2 + 5^2)^{3/2}} - \frac{3d^2kq_1q_2}{(d^2 + 5^2)^{5/2}} = 0\\3d^2 = d^2 + 5^2\\2d^2 = 25\\d = 3.53~m[/tex]

The value of d for which the x component of the force on the charge is the greatest, in accordance to Coulomb's Law, is when d equals 5 meters. At this distance, the x and y components of the force are equal, thus maximizing the x component.

The scenario you described involves the concept of Coulomb's Law which states the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Looking at your question, the value of d for which the x component of the force on 9×10−18 C is the greatest would be when d equals 5 meters. The charges would then form an equilateral triangle with the origin, meaning the x component and y component of the force would be equal, hence maximizing the x component.

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Answer:

Explanation:

Given

Work required [tex]W=1.35\times 10^{-20}\ J[/tex]

Work done to eject the sodium ion from interior of the cell is given by the product of charge and Potential difference between inner and outer surface of the cell.

[tex]W=q\times V[/tex]

Charge on sodium ion [tex]q=1.6\times 10^{19}\ C[/tex]

[tex]V=\frac{W}{q}[/tex]

[tex]V=\frac{1.35\times 10^{-20}}{1.6\times 10^{19}}[/tex]

[tex]V=0.0718\ V[/tex]            

The magnitude of the potential difference between the inner and outer surfaces of the cell is 0.0844 volt.

When potential difference is V and charge of ion is q. Then, work required to eject a postive charge from the interior of cell is computed as,

                   Work  = q * V

It is given that, work W = [tex]1.35*10^{-20}J[/tex]

Charge on sodium ion , [tex]q=1.6*10^{-19}C[/tex]

Substituting above values in above relation.

                   [tex]V=\frac{W}{q}\\ \\V=\frac{1.35*10^{-20} }{1.6*10^{-19} }\\ \\V=0.0844Volt[/tex]

Hence, the magnitude of the potential difference between the inner and outer surfaces of the cell is 0.0844 volt.

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Answer:

5.71428571422 m/s

Explanation:

u = Initial velocity = 20 m/s

v = Final velocity

s = Displacement

a = Acceleration

Time taken = 15-1 = 14 s

Distance traveled in 1 second = [tex]20\times 1=20\ m[/tex]

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 200-20=20\times 14+\frac{1}{2}\times a\times 14^2\\\Rightarrow a=\frac{2(180-20\times14)}{14^2}\\\Rightarrow a=-1.02040816327\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow v=20-1.02040816327\times 14\\\Rightarrow v=5.71428571422\ m/s[/tex]

The speed as she reaches the light at the instant it turns green is 5.71428571422 m/s

The final speed of the motorist as she reaches the light is 5.72 m/s

To calculate the speed of the motorist as she reaches the light, we need to first find the deceleration of the motorist.

Formula:

Where:

Given:

Therefore,

Then,

Where:

we use the formula below to calculate the deceleration of the motorist.

Where:

Given:

Substitute these values into equation 2

Solve for a

Finally, we use the formula below to get the speed of the motorist as she reaches the light.

Given:

Substitute these values into equation 3

Hence, The final speed of the motorist as she reaches the light is 5.72 m/s

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Answer:

a) [tex](x_2,y_2)=(-45.6,58.8)[/tex]

b) [tex]s=74.4097\ m[/tex]

Explanation:

Given:

a)

As we know that average velocity is total displacement per unit time.

Position of x-coordinate of rhino:

[tex]v_x=\frac{x_2}{t_2}[/tex]

[tex]-3.8=\frac{x_2}{12}[/tex]

[tex]x_2=-45.6\ m[/tex]

Position of y-coordinate of rhino:

[tex]v_y=\frac{y_2}{t_2}[/tex]

[tex]4.9=\frac{y_2}{12}[/tex]

[tex]y_2=58.8\ m[/tex]

b)

Now the distance from the origin:

[tex]s=\sqrt{x_2^2+y_2^2}[/tex]

[tex]s=\sqrt{(-45.6)^2+(58.8)^2}[/tex]

[tex]s=74.4097\ m[/tex]

The question concerns vector kinematics in physics. The rhinoceros's coordinates at 12 seconds are (-45.6,58.8), and its distance from the origin is approximately 74.2 meters.

The phenomenon described in your question essentially relates to Vector Kinematics. The rhinoceros is said to have a velocity in the x-direction of -3.8 m/s and in the y-direction of 4.9 m/s. These velocities are constant and last for 12 seconds.

To find the x- and y-coordinates, we can simply multiply the component velocities by time.

For the x-coordinate:
x = velocity_x * time = -3.8 m/s * 12 s = -45.6 m
And for the y-coordinate:
y = velocity_y * time = 4.9 m/s * 12 s = 58.8 m

So, for part (a) of your question, the rhino's coordinates at t2 = 12.0 s are (-45.6, 58.8).

Now, for part (b), to calculate the rhino's distance from the origin, we can use the Pythagorean theorem which is rooted in the geometrical interpretation of vectors. The distance from the origin (r) can be given by:
r = sqrt[x² + y²] = sqrt[(-45.6 m)² + (58.8 m)²] = approximately 74.2 m

Therefore, the rhino is approximately 74.2 m from the origin.

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Answer:

Average force =  2 kg m/s²

Explanation:

Given

mass = 5 kg

initial velocity = 0 m/s

final velocity = 4 m/s

time = 0.1 sec

Find

average force = ?

Formula

Average force = m (final velocity - initial velocity)/t₂- t₁

                        = (5kg)(0 m/s - 4 m/s)/0 - 0.1

                        = 5 kg (- 4 m/s)/(- 0.1 sec)

                         = 2 kg m/s²

Final answer:

The magnitude of the average force exerted on the ball by the wall is 400 N, computed using the impulse-momentum theorem and the ball's change in momentum during the collision.

Explanation:

To calculate the magnitude of the average force exerted on the ball, we need to first determine the change in momentum of the ball and then use the impulse-momentum theorem. The ball changes direction after hitting the wall, so the final velocity is -4 m/s (negative sign indicating the opposite direction) and the initial velocity is 4 m/s. The total change in velocity is thus -4 m/s - 4 m/s = -8 m/s. The change in momentum (impulse) is given by mass × change in velocity, which is 5 kg × -8 m/s = -40 kg·m/s. The impulse experienced by the ball equals the average force exerted on the ball multiplied by the time of contact, so:

Average Force = - Impulse / Time

Therefore, the magnitude of the average force is:

|Average Force| = 40 kg·m/s / 0.1 s = 400 N.

The negative sign indicates that the force is in the opposite direction of the initial motion, but since the question asks for the magnitude, we take the absolute value.

Answer:

Explanation: see attachment

Answer:

       [tex]\large\boxed{\large\boxed{\frac{a_1}{a_2}=1.2}}[/tex]

Explanation:

Newton's second law states the the net force exerted over a body, [tex]F[/tex], is proportional to the product of its mass, [tex]m[/tex] ,   and its acceleration,   [tex]a[/tex] :

     [tex]F=m\times a[/tex]

As a consequence of Newton's third law, action and reaction forces are equal in magnitude and opposite in direction. Hence, 20 N is the same magnitude of the forces over each sibling.

1. Sibling 1:

     [tex]20N=50kg\times a_1[/tex]

2. Sibling 2:

    [tex]20N=60kg\times a_2[/tex]

3. Ratio sibling 1: sibling 2:

                [tex]\frac{20N}{20N} =\frac{50kg\times a_1}{60kg\times a_2}[/tex]

                [tex]\frac{a_1}{a_2}=\frac{60}{50}=1.2[/tex]

Answer

given,

mass of the block, m = 0.5 Kg

displacement, x = 30 cm = 0.3 m

Spring constant, k = 2 N/m

a) Angular frequency

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

   [tex]\omega = \sqrt{\dfrac{2}{0.5}}[/tex]

  [tex]\omega = 2\ rad/s[/tex]

b) Period of oscillation

   [tex]T=\dfrac{2\pi}{\omega}[/tex]

   [tex]T=\dfrac{2\pi}{2}[/tex]

           T = 3.14 s

c) Position at  t = 2

 x = -A cos ω t

A = 0.3   ω = 2 rad/s

x = -0.3 cos (2 x 2)

x = 0.196 m

d) velocity at t= 2

 v = A ω sin ω t

 v = 0.3 x 2 x sin 4

 v = -0.454 m/s

e) acceleration at t= 2

   a = A ω² cos ω t

   a = 0.3 x 2² cos 4

   a = -0.784 m/s²

Answer:

934701926.438 Hz

Explanation:

Mass of molecule

[tex]m=3\times 16\times 1.67\times 10^{-27}\ kg[/tex]

Moment of inertia is given by

[tex]I=\dfrac{1}{12}ml^2\\\Rightarrow I=\dfrac{1}{12}\times 3\times 16\times 1.67\times 10^{-27}\times (2.5\times 10^{-10})^2\\\Rightarrow I=4.175\times 10^{-46}\ kgm^2[/tex]

E = Electric field = 8000 N/C

p = Dipole moment = [tex]1.8\times 10^{-30}\ Cm[/tex]

l = Length of rod = [tex]2.5\times 10^{-10}\ m[/tex]

Frequency of oscillations is given by

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{pE}{I}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{1.8\times 10^{-30}\times 8000}{4.175\times 10^{-46}}}\\\Rightarrow f=934701926.438\ Hz[/tex]

The frequency of oscillations is 934701926.438 Hz

Answer:

+5.4×10⁻⁷ C

Explanation:

Electric potential: This can be defined as the work done in bringing a unit charge from infinity to that point against the action of the field. The S.I unit of potential is volt (V)

The formula for potential is

V = kq/r............................ Equation 1

Where V = electric potential, k = proportionality constant, q = charge, r = distance.

making q the subject of the equation,

q = Vr/k............................ Equation 2

Given: V = 490 V, r = 10 m,

Constant: k = 9×10⁹ Nm²/C²

Substitute into equation 2

q = 490(10)/(9×10⁹)

q = 5.4×10⁻⁷ C

q = +5.4×10⁻⁷ C

Hence the charge is +5.4×10⁻⁷ C

Answer:

0.02 N.

Explanation:

Given:

m = 20 g

= 0.02 kg.

vi = 0.35 m/s

vo = 0.25 m/s

t = 0.1 s

Force, F = m * a

Where,

m = mass of the blood

a = acceleration.

Acceleration, a = (vi - vo)/t

Where,

vi = final velocity

vo = initial velocity

t = time.

a = (0.35 - 0.25)/0.1

= 0.1/0.1

= 1 m/s².

F = 0.02 * 1

= 0.02 N.

Answer:

[tex]\displaystyle V_t=36\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]

Explanation:

Electric Potential of Point Charges

The electric potential from a point charge Q at a distance r from the charge is

[tex]\displaystyle V=k\frac{Q}{r}[/tex]

Where k is the Coulomb's constant. The total electric potential for a system of point charges is equal to the scalar sum of their individual potentials. The potential is not a vector, so there is no direction or vectors to deal with.

We are required to compute the total electric potential in the center of the square. We need to know the distance from each corner to the center. The diagonal of the square is

[tex]d=\sqrt2 a[/tex]

where a is the length of the side.

The distance from any corner to the center is half that diagonal, thus

[tex]\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}[/tex]

The total potential in the center is  

[tex]V_t=V_1+V_2+V_3+V_4[/tex]

Please note all the potentials must be calculated including the sign of the charges. Since all the charges are equal to Q, and the distances are the same, the total potential is 4 times the individual potential of each charge.

[tex]V_t=4\times V[/tex]

[tex]\displaystyle V=9\times 10^9 \frac{Q}{\frac{a}{\sqrt{2}}}[/tex]

Operating

[tex]\displaystyle V=9\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]

Thus:

[tex]\displaystyle V_t=4\times 9\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]

[tex]\boxed{\displaystyle V_t=36\sqrt{2}\times 10^9 \frac{Q}{a}}[/tex]

Answer:

The charge on the droplet is [tex]3.106\times10^{-16}\ C[/tex].

Yes, quantization of charge is obeyed within experimental error.

Explanation:

Given that,

Radius = 1.6μm

Electric field = 46 N/C

Density of oil = 0.085 g/cm³

We need to calculate the charge on the droplet

Using formula of force

[tex]F= qE[/tex]

[tex]mg=qE[/tex]

[tex]V\times\rho\times g=qE[/tex]

[tex]q=\dfrac{V\times\rho}{E}[/tex]

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

Put the value into the formula

[tex]q=\dfrac{\dfrac{4}{3}\times\pi\times(1.6\times10^{-6})^3\times85\times9.8}{46}[/tex]

[tex]q=3.106\times10^{-16}\ C[/tex]

We need to calculate the quantization of charge

Using formula of quantization

[tex]n = \dfrac{q}{e}[/tex]

Put the value into the formula

[tex]n=\dfrac{3.106\times10^{-16}}{1.6\times10^{-19}}[/tex]

[tex]n=1941.25[/tex]

Yes, quantization of charge is obeyed within experimental error.

Hence, The charge on the droplet is [tex]3.106\times10^{-16}\ C[/tex].

Yes, quantization of charge is obeyed within experimental error.

Answer:

3.11 x 10^-16 C

Explanation:

radius of drop, r = 1.6 micro metre = 1.6 x 10^-6 m

Electric field, E = 46 N/C

density of oil, d = 0.085 g/cm³ = 85 kg/m³

Let the charge on the oil drop is q.

So, the weight of the drop is balanced by the electric force on the drop.

mass of drop x g = charge of the drop x electric field

m x g = q E

Volume x density x g = q E

[tex]\frac{4}{3} \pi\times r^{3}\times d\times g = q \times E[/tex]

[tex]\frac{4}{3} \pi\times 1.6^{3}\times 10^{-18}\times 85\times 9.8 = q \times 46[/tex]

q = 3.11 x 10^-16 C

Thus, the charge on the oil drop is 3.11 x 10^-16 C.  

Answer:

0.0167 m

Explanation:

f = Frequency = 18 GHz

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

When the speed of the wave is divided by the freqeuncy we get the wavelength

Wavelength is given by

[tex]\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{18\times 10^9}\\\Rightarrow \lambda=0.0167\ m[/tex]

The wavelength of the radio waves is 0.0167 m

Springfield has the smallest slope.

Answer: Option B.

Explanation:

Springfield is a city in the United States of America. This city is in the states of Massachusetts. It is besides the river Connecticut river which adds to the beauty of this state.

Springfield is the nick name of the city "The queen of the Ozarks". The other name of this city is "the cultural center of the Ozarks". It has a beautiful scenery and this scenic beauty adds to the tourist attraction to a lot of people all around the world.

A) The electric field inside the paint layer is zero

B) The electric field just outside the paint layer is [tex]3.2\cdot 10^7 N/C[/tex] (radially inward)

C) The electric field at 6.00 cm from the surface is [tex]1.2\cdot 10^7 N/C[/tex] (radially inward)

Explanation:

A)

We can solve the problem by applying Gauss Law, which states  that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:

[tex]\int EdS = \frac{q}{\epsilon_0}[/tex]

where

E is the magnitude of the electric field

dS is the element of the surface

q is the charge contained within the surface

[tex]\epsilon_0[/tex] is the vacuum permittivity

By taking a sphere centered in the origin,

[tex]\int E dS = E \cdot 4\pi r^2[/tex]

where [tex]4\pi r^2[/tex] is the surface of the Gaussian sphere of radius r.

In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than

[tex]R=9.0 cm = 0.09 m[/tex] (radius of the plastic sphere is half of the diameter)

Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:

[tex]q=0[/tex]

And therefore,

[tex]E4\pi r^2 = 0\\\rightarrow E = 0[/tex]

So, the electric field inside the plastic sphere is zero.

B)

Here we apply again Gauss Law:

[tex]E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]

In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so

[tex]r=R=0.18 m[/tex]

The charge contained within the Gaussian sphere is therefore

[tex]q=-29.0 \mu C = -29.0\cdot 10^{-6}C[/tex]

Therefore, the electric field is

[tex]E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.09)^2}=-3.2\cdot 10^7 N/C[/tex]

And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is

[tex]E=3.2\cdot 10^7 N/C[/tex]

C)

For this part again, we apply Gauss Law:

[tex]E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]

In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be

r = 9 cm + 6 cm = 15 cm = 0.15 m

While the charge contained within the sphere is again

[tex]q=-29.0 \mu C = -29.0\cdot 10^{-6}C[/tex]

Therefore, the electric field in this case is

[tex]E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.15)^2}=-1.2\cdot 10^7 N/C[/tex]

And again, this is radially inward, so according to the sign convention asked in the problem,

[tex]E=1.2\cdot 10^7 N/C[/tex]

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To find the electric field just inside the paint layer, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of the medium.

A) Inside the paint layer:

Since the paint is spread in a very thin uniform layer over the surface of the plastic sphere, we can consider a Gaussian surface just inside the paint layer, with a radius slightly smaller than the sphere's radius (let's call it r). The charge enclosed within this Gaussian surface is the charge on the sphere, which is -29.0 μC.

Gauss's law equation can be written as:

Φ = Q_enclosed / ε₀

The electric field just inside the paint layer (E₁) is given by:

E₁ * A = Q_enclosed / ε₀

E₁ * 4πr² = -29.0 μC / ε₀

E₁ = (-29.0 μC / ε₀) / (4πr²)

The direction of the electric field just inside the paint layer will be radially outward, opposite to the direction of the charge.

B) Just outside the paint layer:

The electric field just outside the paint layer (E₂) will be the same as the electric field inside the sphere when it was uncharged. This is because the charge on the paint layer is spread on the outer surface of the sphere.

So, E₂ = E₁ (since it's radially outward)

C) 6.00 cm outside the surface of the paint layer:

To find the electric field 6.00 cm outside the surface of the paint layer, we need to calculate the electric field due to the charge on the sphere.

E₃ = k * Q / r²

where k is the Coulomb's constant (k = 9.0 x 10^9 N m²/C²), Q is the charge on the sphere (-29.0 μC), and r is the distance from the center of the sphere to the point where we want to find the electric field (6.00 cm = 0.06 m + radius of the sphere).

Remember that the electric field due to a charged object decreases with the square of the distance.

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Answer:

The boiling point of Acetone is 329K (in 3 significant figures)

Explanation:

Boiling point of Acetone = 56°C = 56 + 273K = 329K (in 3 significant figures)

Answer: using the formula 0°C + 273.15 = 273.15K the boiling point in units of kelvin to significant figures is 329.15k.

Explanation: The boiling point of a substance ( acetone) is the temperature at which the vapour pressure of the liquid substance equals the pressure surrounding it. The boiling point of acetone serves as it's characteristic physical properties. This is measured in degree Celsius (°C ) which can be converted to units of Fahrenheit or kelvin. To convert degree Celsius to kelvin this formula is used: 0°C + 273.15 = 273.15K . Given that acetone has boiling point of 56°C,from the formula 0°C is substituted for 56°C. This gives us:

56°C + 273.15= 319.15k.

Also,measurements given in Kelvin will always be larger numbers than in Celsius and the Kelvin temperature scale does not use the degree (°) symbol because Kelvin is an absolute scale, based on absolute zero, while the zero on the Celsius scale is based on the properties of water. I hope this helps. Thanks

Answer:

W = 60.62 ft-lbs

Explanation:

given,

Horizontal force = 7 lb

distance of push, d = 10 ft

angle of ramp, θ = 30°

Work done on the box = ?

We know,

Work done is equal to force into displacement.

W = F.d cos θ

W = 7 x 10 x cos 30°

W = 70 x 0.8660

W = 60.62 ft-lbs

Hence, work done on the box is equal to W = 60.62 ft-lbs

Answer:

A day measured with respect to the sun differ from a day measured with respect to star by 4 minutes. This is because, it takes 4 minutes for the Earth to rotate the extra amount required for the Sun to return to the same place in the sky.

Explanation:

A day measured with respect to the sun differ from a day measured with respect to star by 4 minutes. This is because, it takes 4 minutes for the Earth to rotate the extra amount required for the Sun to return to the same place in the sky.

Each day that goes by, the Earth needs to turn a little further for the sun to return to the same place in the sky. And that extra time is about 4 minutes or 240 seconds.

Final answer:

A solar day is longer than a sidereal day by about 4 minutes, due to Earth's orbit around the Sun requiring additional rotation to align the Sun to the same position in the sky. The sidereal day, on the other hand, represents the true rotation period of Earth as it measures the time it takes for a distant star to reappear in the same position in the sky.

Explanation:

A day measured with respect to the Sun differs from a day measured with respect to the stars primarily in its length due to Earth's simultaneous rotation on its axis and revolution around the Sun. The solar day is the period during which Earth rotates on its axis so that the Sun appears at the same position in the sky on consecutive days. It is about 4 minutes longer than a sidereal day, which is the time it takes for a distant star to return to the same position in the sky.

This difference occurs because as Earth rotates, it also moves along its orbital path around the Sun, so it has to rotate a little more for the Sun to be in the same position overhead. The sidereal day is actually the true rotation period of Earth, being 235.9 seconds shorter than 24 hours, while the solar day includes an additional 4 minutes to account for Earth's movement in orbit.

Answer:

a)    vₓ = 6,457 m / s ,  v_{y}  = 0.518 m / s , b)    v = 6.478 m / s,      θ = 4.9°

Explanation:

a) This is a kinematic problem, let's use trigonometry to find the components of acceleration

           sin 31 = [tex]a_{y}[/tex] / a

           cos 31 = aₓ = a

           a_{y} = a sin31

           aₓ = a cos 31

Now let's use the kinematic equation for each axis

X axis

         vₓ = v₀ₓ + aₓ (t-t₀)

         vₓ = v₀ₓ + a cos 31 (t-t₀)

         vₓ = 2.6 + 0.45 cos 31  (20-10)  

         vₓ = 6,457 m / s

Y Axis

       v_{y} = v_{oy} + a_{y} t

       v_{y} = v_{oy} + a_{y}  sin31 (t-to)

       v_{y} = -1.8 + 0.45 sin31   (20-10)

       v_{y}  = 0.518 m / s

b) let's use Pythagoras' theorem to find the magnitude of velocity

       v = √ (vₓ² + v_{y}²)

       v = √ (6,457² + 0.518²)

       v = √ (41.96)

       v = 6.478 m / s

We use trigonometry for direction

       tan θ = v_{y} / vₓ

       θ = tan⁻¹  v_{y} / vₓ

       θ = tan⁻¹ 0.518 / 6.457

       θ = 4.9°

c) let's look for the vector at the initial time

        v₁ = √ (2.6² + 1.8²)

        v₁ = 3.16 m / s

        θ₁ = tan⁻¹ (-1.8 / 2.6)

        θ₁ = -34.7

We see that the two vectors differ in module and direction, and that the acceleration vector is responsible for this change.

          a = (v₂ -v₁) / (t₂-t₁)

Answer:

The two balls will hit the ground at the same time neglecting air resistance.

Explanation:

The vertical motion of the two ball are independent of each other. Also, the balls are falling from the same height in the same time. The ball projected horizontal neglecting air resistance is traveling with a constant velocity ( the same distance in the equal time) has only force of gravity acting on it. They will therefore hit the ground at the same time because they are acted upon by the same acceleration in the same time from the same height.

The vertical and horizontal motions of an object are independent. Hence, despite different trajectories, a ball dropped vertically and a ball launched horizontally from the same height will hit the ground simultaneously, as exemplified by Galileo's thought experiment.

In the realm of Physics, an interesting question has been asked: Which ball hits the ground first - a ball that is dropped vertically or a ball that is launched horizontally? According to the principle of independence of motion, the horizontal motion and the vertical motion of an object are independent of each other. It implies that the horizontal velocity of a body has no effect on its vertical velocity. Thus, the ball dropped directly and the ball launched horizontally would strike the ground at the same time. Both balls are subjected to the same gravitational acceleration (g), without any initial vertical velocity.

This concept is a perfect illustration of Galileo's thought experiment, whereby he proved that the time for the balls to hit the ground is determined by their vertical motion alone and that horizontal motion does not affect the descent time.

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Answer:

The speed of comet is 1442.77 m/s.

Explanation:

Given that,

Speed of comet's [tex]v_{c}= 5.3\times10^{4}\ m/s[/tex]

Distance from the sun [tex]r_{d}=9.8\times10^{10}\ m[/tex]

Distance [tex]r_{far}=3.6\times10^{12}\ m[/tex]

We need to calculate the speed of comet

Using conservation of angular momentum

[tex]L_{f}=L_{i}[/tex]

[tex]I\omega=I\omega[/tex]

Here. [tex]v = r\omega[/tex]

[tex]\omega=\dfrac{v}{r}[/tex]

[tex]mr_{far}^2\times\dfrac{v_{far}}{r_{far}}=mr_{d}^2\times\dfrac{v_{d}}{r_{d}}[/tex]

[tex]r_{far}\times v_{far}=r_{d}\times v_{d}[/tex]

Put the value into the formula

[tex]3.6\times10^{12}\times v_{far}=9.8\times10^{10}\times5.3\times10^{4}[/tex]

[tex]v_{far}=\dfrac{9.8\times10^{10}\times5.3\times10^{4}}{3.6\times10^{12}}[/tex]

[tex]v_{far}=1442.77\ m/s[/tex]

Hence, The speed of comet is 1442.77 m/s.

Final answer:

The speed of Halley's comet when it is 3.6 x 10^12 m away from the sun can be found using the conservation of angular momentum, resulting in a speed of 1.44 x 10^3 m/s.

Explanation:

To find the speed of Halley's comet at a distance of 3.6 × 1012 m from the sun, we will use the conservation of angular momentum. The orbital momentum of an object around the sun is given by L = mrv, where m is the mass of the comet, r is the radius of the orbit (distance from the sun), and v is the velocity of the comet along that radius.

We are given the comet's speed at a distance of 9.8 × 1010 m is 5.3 × 104 m/s. When the comet is at a distance of 3.6 × 1012 m from the sun, we want to find its new speed, v'. Because angular momentum is conserved, the initial and final angular momentum should be equal:

L initial = L final
m r v = m r' v'

Since the mass of the comet, m, cancels out, we are left with:

r v = r' v'

Plugging in the known values, we get:

9.8 × 1010 m × 5.3 × 104 m/s = 3.6 × 1012 m × v'

Therefore:

v' = (9.8 × 1010 × 5.3 × 104) / (3.6 × 1012)

v' = 1.44 × 103 m/s

The speed of Halley's comet at a distance of 3.6 × 1012 m from the sun is thus 1.44 × 103 m/s.

Answer:

Explanation:

Given

Astronaut in space gives an equal push to baseball and bowling ball.

Since the mass of the bowling ball is more than the baseball so inertia associated with a bowling ball is more as compared to baseball

Force applied on baseball and bowling ball is the same so the acceleration of baseball will be more because the mass of baseball is less.

[tex]Force=mass\times acceleration[/tex]

If both are traveling with the same speed then momentum associated with them is given by-product of mass and velocity

Since the mass of the bowling ball is more, therefore, the momentum of bowling ball is more as compared to baseball

Answer:

Explanation:

I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air. Before , the ball gets separated from my hand , both me the ball will be moving with velocity of 50 m /s in horizontal direction .

As soon as ball is separated from the hand , it acquires addition velocity in upward direction and acceleration in downward direction . This will give relative velocity to the ball with respect to me . So I will see the ball going in upward direction under  gravitational acceleration . It appears as if I am sitting at rest and ball is going in upward direction under deceleration . My motion at 50 m/s will have no effect on the motion of ball in upward direction , according to first law of Newton . It is so because ball too will be moving in forward direction with the same speed which will not be visible to me because I too am moving with the same speed.

If I  am  sitting at rest at home and I threw a ball straight up into the air , I will have the same experience of seeing ball going in similar way as described above.

Answer:

a)-296 m/s

b)-176 m

c) 198 m

Explanation:

given data:

[tex]v_{oy}\\[/tex]=-20 m/s

[tex]v_{x}[/tex]= 15 m/s

t = 6 s

To find

a)[tex]H_{1}[/tex]

b)[tex]H_{2}[/tex]

c) d

d) find the velocity of observer at rest in basket and rest on ground

a) according to kinematic equation the displacement is given by:

[tex]H_{1}[/tex]=[tex]v_{oy}t\\[/tex]-[tex]\frac{1}{2} gt^{2}[/tex]..............(1)

putting values in 1

[tex]H_{1}[/tex]=-296 m/s

negative sign is due to direction and also we choose to take off point to be the origin

b) while the stone was travelling [tex]H_{1}[/tex] for 6 s so the ballon was also travelling displacement  [tex]H_{b}[/tex] with [tex]v_{oy}\\[/tex]

[tex]H_{b}[/tex] =[tex]v_{oy}t\\[/tex]

     =-6*20

     =-120 m

the height above the earth is given by

[tex]H_{2}[/tex]=[tex]H_{1}[/tex]-[tex]H_{2}[/tex]= -176 m

c) the horizontal displacement is given by

x=[tex]v_{x}t\\[/tex]=15*6=90 m

so the distance between balloon and stone is

d=[tex]\sqrt{H_{1}^{2}+x^{2} }[/tex]=198 m

d) the velocity component of the balloon :

1) relative to person rest in the balloon are given by

The vertical component:

        [tex]v_{yr} =v_{by} +v_{py}[/tex]=[tex](-gt)-0=9.8*6=-58.8 m/s[/tex]

The horizontal component:

         [tex]v_{xr} =v_{bx} +v_{px}= 15-0=15m/s[/tex]

2) relative to person rest in the earth are given by

The vertical component:

        [tex]v_{yr} =v_{by} -v_{py}=(-gt)-20=9.8*6-20=-78.8 m/s[/tex]

The horizontal component:

        [tex]v_{xr} =v_{bx} -v_{px}= 15-0=15m/s[/tex]

Final answer:

To calculate the speed of a muon before it decays, the distance the muon travels and the time it takes are used in the formula L = v x t. For a distance of 9.5 cm and a time of 2.20 microseconds, the speed is found to be approximately 43,182 meters per second.

Explanation:

The question is seeking to calculate the speed of a muon before it decays. Assuming the mean lifetime of a muon is 2.20 microseconds (us) and the muon makes a track 9.5 cm long in this time, we can calculate the muon's speed. Let's consider the formula:

L = v x t

where L is the distance traveled, v is the velocity, and t is the time. Given L = 9.5 cm (or 0.095 m) and t = 2.20 x 10-6 s, we plug in the values:

v = L/t

v = 0.095 m / 2.20 x 10-6 s

v = 4.31818 x 104 m/s

The muon's speed was approximately 43,182 m/s.