Is it possible for an object to be (a) slowing down while its acceleration is increasing in magnitude; (b) speeding up while its acceleration is decreasing? In both cases, explain your reasoning.

Answer:

The voltage across the capacitor decreases by a factor of 2

Explanation:

When we put a dielectric material on a capacitor, we generate a polarization on it that changes the capacitance of the capacitor. The dielectric constant is the ratio between the initial capacitance (Co) without the dielectric and the new capacitance (C) with the dielectric:

[tex]k=\frac{C}{C_0} [/tex]

So, if k is equal to 2:

[tex] 2= \frac{C}{C_0}[/tex]

[tex]C=2*C_0 [/tex]

So, the capacitance is doubled with the dielectric

Because the battery is disconnected, the charge on the plates of capacitor must be constant because conservation of charge.

The effective electric field of a capacitor is the electric filed when we put a dielectric on it, and it is:

[tex]E_{eff}=\frac{E}{k}=\frac{E}{2} [/tex]

With E the initial electric field, so the electric field is halved

Finally, because we know electric field and voltage (V) on a parallel capacitor with distance between plates (d) are related by:

[tex] E=\frac{V}{d}[/tex]

Because electric field and voltage are directly proportional and d remains constant if Electric field is halved, then voltage is halved too. The voltage across the capacitor decreases by a factor of 2.

Answer:

(a) [tex]t=\frac{H}{v_0}[/tex]

(b) [tex]H=\frac{v_0^2}{g}[/tex]

Explanation:

Let the two balls collide at a height x from the ground. Therefore, ball 2 travels a distance of (H-x) before colliding with ball 1.

Using the following Newton's law of motion,

[tex]S=ut+\frac{1}{2}at^2[/tex]

where,

[tex]S[/tex] = displacement

[tex]u[/tex] = initial velocity

[tex]a[/tex] = acceleration

[tex]t[/tex] = time

we can write the equations of motion of the two balls(ball 1 and ball 2 respectively):

[tex]x=v_0t-\frac{1}{2}gt^2[/tex]     ......(1)   ([tex]a=-g[/tex], ball is moving against gravity)

[tex]H-x=\frac{1}{2} gt^2[/tex]      .......(2)    (initial velocity is zero; [tex]a=+g[/tex])

Substituting [tex]x[/tex] from equation (1) in (2),

[tex]H-v_0t+\frac{1}{2}gt^2=\frac{1}{2}gt^2[/tex]

or, [tex]t=\frac{H}{v_0}[/tex]      ......(a)

(b) Now, it is said that the collision will occur when ball 1 is at it's highest point. That is, it's final velocity must be zero.

This time we shall have to use another equation of motion given by,

[tex]v^2=u^2+2aS[/tex]

where, [tex]v[/tex] = final velocity

therefore, we get for ball 1,

[tex]0=v_0^2-2gx[/tex]       ([tex]u=v_0,v=0,a=-g[/tex])

or, [tex]x=\frac{v_0^2}{2g}[/tex]

Putting the value of [tex]x[/tex] in equation (2) and rearranging, we get,

[tex]\frac{g}{2v_0^2}H^2-H+\frac{v_0^2}{2g}=0[/tex]

which is a quadratic equation, whose solution is given by,

[tex]H=\frac{+1\pm\sqrt{(-1)^2-(4\times\frac{g}{2v_0^2} \times\frac{v_0^2}{2g}) } }{2\times\frac{g}{2v_0^2} }[/tex]

[tex]=\frac{v_0^2}{g}[/tex]

(a) The time at which the balls collide is H/[tex]v_{0}[/tex]

(b) The height H is equal to [tex]\frac{v_{0} ^{2} }{g}[/tex]

Let the balls collide at a height x above the ground.

Then the distance traveled by the ball thrown above is x.

And the distance traveled by the ball dropped from height H is (H-x).

(i) Both the balls will take the same time to travel respective distances in order to collide.

[tex]H-x=\frac{1}{2}gt^{2}[/tex]

[tex]x = v_{0}t - \frac{1}{2}gt^{2}[/tex]

We get:

[tex]x=v_{0}t-(H-x)[/tex]

[tex]t=\frac{H}{v_{0}}[/tex] , is the time after which the balls collide.

(ii) Let the ball thrown up attains its maximum height x at the time of thecollision

[tex]v^{2} = u^{2}-2gx[/tex] here v is the final velocity which is 0 when the ball attains maximum height

[tex]0=v_{0} ^{2}-2gx[/tex]

[tex]x=\frac{v_{0} ^{2} }{2g}[/tex] is the maximum height attained.

Now, the ball thrown downward travels distance (H-x) just before collision:

[tex]H-x=\frac{1}{2}gt^{2}[/tex]

[tex]H-\frac{v_{0} ^{2} }{2g}=\frac{1}{2}g\frac{H^{2} }{v_{0} ^{2} }[/tex]

Solving the quadratic equation we get:

[tex]H=\frac{v_{0} ^{2} }{g}[/tex]

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Answer:

Explanation:

Given

mass of object [tex]m=5\ kg[/tex]

Object is attached to the ceiling of an elevator by a rope

Suppose T is the Tension in the rope so

acceleration of the elevator [tex]a=4\ m/s^2[/tex]  (upward)

From Free Body diagram we can write as

[tex]T-mg=ma[/tex]

[tex]T=m(g+a)[/tex]

[tex]T=5(9.8+4)[/tex]

[tex]T=69\ N[/tex]

Answer:

0.906

Explanation:

Let g = 9.81 m/s2. We can calculate the rate of change in potential energy when m = 201kg of water is falling down a distance of h = 131m per second

[tex]\dot{E_p} = \dot{m}gh = 201*9.81*131 = 258307 J/s (W) = 258.307 kW[/tex]

So the efficiency of the water turbine is the ratio of output power over input power:

[tex]\frac{234}{258.307} = 0.906[/tex]

This question involves the concepts of efficiency and potential energy.

The efficiency of the turbine is "90.6%".

The efficiency of the turbine can be given by the following formula:

[tex]Efficiency = \frac{Turbine\ shaft\ power}{Potential\ Energy\ power\ of\ Water}\\\\[/tex]

where,

Turbine shaft power = 234 KW

Potential energy power of water = mgh/t = (201 kg/s)(9.81 m/s²)(131 m)

Potential energy power of water = 258307.11 W = 258.31 KW

Therefore,

[tex]Efficiency=\frac{234\ KW}{258.31\ KW}[/tex]

Efficiency = 0.906 = 90.6%

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Answer:

0.43 s

Explanation:

We have the following parameters:

Initial velocity, u = 7.4 m/s

Acceleration of gravity, g = 9.8 [tex]m/s^2[/tex]

Distance, s = 43 in + 10 ft = 1.092 m + 3.048 m = 4.14 m

Time, t = ?

Using the equation of motion [tex]s=ut +\frac{1}{2}gt^2[/tex], we have

[tex]4.14 = 7.4t + 0.5\times9.8t^2[/tex]

[tex]4.9t^2 + 7.4t - 4.14 =0[/tex]

Using the quadratic formula [tex]\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex] where a = 4.9, b = 7.4 and c = - 4.14, and solving for the positive value of t only, we have

[tex]t = 0.43[/tex] s

To solve this problem we will apply the concepts of Boiling Point elevation and Freezing Point Depression. The mathematical expression that allows us to find the temperature range in which these phenomena occur is given by

Boiling Point Elevation

[tex]\Delta T_b = K_b m[/tex]

Here,

[tex]K_b[/tex] = Constant ( Different for each solvent)

m = Molality

Freezing Point Depression

[tex]\Delta T_f = K_f m[/tex]

Here,

[tex]K_f =[/tex] Constant ( Different for each solvent)

m = Molality

According to the statement we have that

[tex]\Delta T_f = T_0 -T_f = 0-(-2.05)[/tex]

[tex]\Delta T_f = 2.05\°C[/tex]

From the two previous relation we can find the ratio between them, therefore

[tex]\frac{\Delta T_b}{\Delta T_f} = \frac{K_b}{K_f}[/tex]

[tex]\frac{\Delta T_b}{\Delta T_f} = \frac{0.512}{1.86}[/tex]

[tex]\frac{\Delta T_b}{\Delta T_f} = 0.275[/tex]

We already know the change in the freezing point, then

[tex]\Delta T_b = 0.275 (\Delta T_f)[/tex]

[tex]\Delta T_b = 0.275 (2.05)[/tex]

[tex]\Delta T_b = 0.5643\°C[/tex]

The temperature difference in the boiling point is 100°C (Aqueous solution), therefore

[tex]T_b -100 = 0.5643[/tex]

[tex]T_b = 100.56\°C[/tex]

Therefore the boiling point of an aqueous solution is [tex]100.56\°C[/tex]

Final answer:

The boiling point of the aqueous solution is 99.44 °C.

Explanation:

The boiling point of an aqueous solution can be determined using the formula:

Tbp = Tbpsolvent + (Kbp * m)

where:

Tbp is the boiling point of the solution

Tbpsolvent is the boiling point of the pure solvent

Kbp is the ebullioscopic constant for the solvent

m is the molality of the solution

In this case, we are given that the freezing point of the solution is -2.05 degrees C. We can use the formula for freezing point depression to find the molality:

AT = Kf * m

Rearranging the formula, we can solve for m:

m = AT / Kf

Substituting the given values:

m = (-2.05 °C) / (1.86 °C kg.mol-¹)

Rounding to 2 decimal places:

m = -1.10 mol/kg

Now, we can use the formula for boiling point elevation to find the boiling point:

Tbp = Tbpsolvent + (Kbp * m)

Substituting the given values:

Tbp = 100.00 °C + (0.512 °C/m * -1.10 mol/kg)

Calculating:

Tbp = 100.00 °C - 0.56 °C = 99.44 °C

Therefore, the boiling point of the aqueous solution is 99.44 °C.

Answer:

A) r•s = -21.99 units

B) rXs = 22.77 units (In the direction of k)

Completed question

Two vectors, r and s, lie in the xy plane. Their magnitudes are 4.25 and 7.45 units, respectively, and their directions are 312° and 86.0°, respectively, as measured counterclockwise from the positive x axis. What are the values of the following products?

A) r•s =

B) (r)X(s)=

Explanation:

A) dot product of r and s can be expressed mathematically as;

r•s = |r||s|cos(A) .......1

Where A is the angle between the two vectors.

|r| and |s| are the magnitude of vector r and s.

r•s = 4.25×7.45 ×cos(86-312) = -21.99 units

B) cross product of r and s can be expressed mathematically as;

rXs = |r||s|sin(A) .......2

Where A is the smallest angle between the two vectors

|r| and |s| are the magnitude of vector r and s.

A = 312-86 = 226

A = 360-226 = 134 smallest

rXs = 4.25 × 7.45 × sin134.

rXs = 22.77 units

In the direction of k

(a) -22.00 units

(b) 22.77 units

Given vectors are r and s

Where;

r = |r| = 4.25   and ∠r = 312°    measured anticlockwise

s = |s| = 7.45   and ∠s = 86°    measured anticlockwise

First, let's calculate the angle between vectors r and s by representing them in the figure below;

                                y |        /s

                                   |      /

                                   |    /

                                   |  /  

                                   |/ )86°                                x

                                   |\) 48°

                                   |  \

                                   |    \

                                   |      \

                                   |        \ r

To get the acute angle between r and the +x axis, subtract the reflex angle of r (312°) from 360° as follows;

360 - 312 = 48°

As shown in the diagram, the angle between vectors r and s is 48° + 86° = 134°

Now,

(a) The (r)(s) represents the dot or scalar product of the two vectors and it is given as;

(r) (s) = r x s cos θ            ---------------------------(i)

Where;

r = magnitude of vector r = 4.25

s = magnitude of vector s = 7.45

θ is the angle between the two vectors r and s = 134°

Substitute these values into equation (i) as follows;

(r) (s) = 4.25 x 7.45 cos 134°

(r) (s) = 4.25 x 7.45 x -0.6947

(r) (s) = 31.66 x -0.6947

(r) (s) = -22.00 units

(b) The (r) X (s) represents the vector product of the two vectors and it is given as;

(r) (s) = r x s sin θ            ---------------------------(ii)

Where;

r = magnitude of vector r = 4.25

s = magnitude of vector s = 7.45

θ is the angle between the two vectors r and s = 134°

Substitute these values into equation (ii) as follows;

(r) (s) = 4.25 x 7.45 sin 134°

(r) (s) = 4.25 x 7.45 x 0.7193

(r) (s) = 31.66 x 0.7193

(r) (s) = 22.77 units

Answer:

Visible Light and Radio waves

Explanation:

The earth's atmosphere is transparent to a few windows in the electromagnetic spectrum. it is completely transparent to allow observation from the ground in visible light rang 380 to 740 nano meters. Also in the range of radio wave as communication are done from space to ground in the form of radio waves.

it is Partially transparent to Microwave and infrared range.

The atmosphere is transparent to certain regions of the electromagnetic spectrum, such as visible light, radio waves, and parts of the infrared and microwave spectra.

The Earth's atmosphere is transparent to certain regions of the electromagnetic spectrum, allowing observations to be made from the ground. These regions include the visible light spectrum, the radio waves spectrum, and parts of the infrared and microwave spectra.

In these regions, electromagnetic waves can pass through the Earth's atmosphere relatively easily, allowing ground-based observations to be conducted.

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Answer:

a = (m₂-m₁) / (m₂ + m₁ + ½ m)

a = (m₂-m₁) / (m₂ + m₁ + 1)

Explanation:

An Atwood machine consists of two masses of different m1 and m2 value that pass through a pulley, in this case with mass. Let's use Newton's second law for this problem.

Assume that m₂> m₁, so the direction of descent of m₂ is positive, this implies that the direction of ascent of m₁ is positive

Equation of the side of m₁

              T₁ - W₁ = m₁ a

Equation of the side of m₂

              W₂ - T₁ = m₂ a

Mass pulley equation m; by convention the counterclockwise rotation is positive

              τ = I α

              T₁ R - T₂ R = I (-α)

The moment of inertia of a disk is

              I = ½ m R²

Angular and linear acceleration are related

             a = α R

             α = a / R

The rotation is clockwise, so it is negative

We replace

             (T₁ –T₂) R = ½ m R² (-a / R)

             T₁ -T₂ = - ½ m a

Let's write our three equations together

              T₁ - m₁ g = m₁ a

              m₂ g - T₂ = m₂ a

              T₁ –T₂ = -½ m a

Let's multiply the last equation by (-1) and add

               m₂ g - m₁ g = m₂ a + m₁ a + ½ m a

                a = (m₂-m₁) / (m₂ + m₁ + ½ m)

calculate

               a = (m₂ - m₁)/ (m₁ +m₂ + 1)

Based on the calculations, the acceleration of this system is equal to 0.58 [tex]m/s^2[/tex].

Given the following data:

First of all, we would determine the moment of inertia of this disk by using this formula:

[tex]I=\frac{1}{2} mr^2\\\\I=\frac{1}{2} \times 2 \times (0.248)^2\\\\I=0.0615\;Kgm^2[/tex]

Next, we would use a free body diagram to determine the tensional forces acting on the disk by applying Newton's Second Law of Motion as follows:

For the first force:

[tex]F_1g-F_{T1}=m_1a\\\\m_1g-F_{T1}=m_1a\\\\1.61(9.8)-F_{T1}=1.61a\\\\15.8-F_{T1}=1.61a\\\\F_{T1}=15.8-1.61a[/tex]

For the second force:

[tex]F_{T2}-F_2g=m_2a\\\\F_{T2}-m_2g=m_2a\\\\F_{T2}-1.38(9.8)=1.38a\\\\F_{T2}-13.5=1.38a\\\\F_{T2}=13.5+1.38a[/tex]

For the torque, we have:

[tex]\sum T =I\alpha \\\\F_{T1}r-F_{T2}r=I\alpha\\\\(F_{T1}-F_{T2})r=I\alpha\\\\(15.8-1.61a-[13.5+1.38a])0.248=0.0615 \times \frac{a}{0.248} \\\\(15.8-1.61a-13.5-1.38a)0.248=0.0615 \times \frac{a}{0.248} \\\\(2.3-2.99a)0.248=0.0615 \times \frac{a}{0.248}\\\\0.5704-0.7415a=\frac{0.0615a}{0.248}\\\\0.1415-0.1839a=0.0615a\\\\0.1839a+0.0615a=0.1415\\\\0.2454a=0.1415\\\\a=\frac{0.1415}{0.2454}[/tex]

Acceleration, a = 0.58 [tex]m/s^2[/tex]

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Answer:

0.212V

Explanation:

The induced emf in circular loop is [tex]\epsilon=-d\frac{\phi_B}{dt}=-NA\frac{dB}{dt}[/tex]

N = Number of loops = 15

A = Cross sectional Area = [tex]\pi[/tex][tex]0.03^{2}[/tex] = 0.0009[tex]\pi[/tex]

dB = change in magnetic Field = 0-0.5 = -0.5

dt = time taken = 0.1sec

[tex]\epsilon=-15*0.0009\pi *\frac{-0.5}{0.1}[/tex] = 0.212V

Explanation:

A.

Displacement,S; Θ = S/r

= 180/360 * 2π * 3500

= 10995.6 m

Θ = 10995.6/3500

= 3.142

B.

Angular speed,w = Θ/t

= 3.142/1.503102

= 2.09 rad/s

Velocity = w * r

= 2.09 × 3500

= 7315 m/s.

C.

The same as B.

The airplane's Motion in a Circle displacement is 7.0km, its average velocity is 0.0467 km/s, and its average speed is 0.0737 km/s.

The airplane is flying half a circle, which is essentially semi-circular motion. Under this condition, we can calculate the values asked in the question as follows:

(a) Displacement: Displacement is a vector quantity and represents the shortest distance between the initial and the final points of an object's path. But since the airplane is rounding half the circle, the displacement is essentially the diameter of the circular path. The diameter will be two times the radius, therefore 2*3.50km = 7.00 km.

(b) Average velocity: Average velocity is the total displacement divided by the total time. So, the average velocity would be 7.00km / (1.50*10^2) s = 0.0467 km/s.

(c) Average speed: Average speed is defined as the total distance traveled by the object divided by the total time taken. Here, the airplane travels half the circumference of the circle in the given time. The formula for the circumference of a circle is 2*pi*r, so half the circle's circumference will be pi*3.50 km. Then, Average speed = (pi*3.50 km) / (1.50*10^2) s = 0.0737 km/s.

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Answer:

(a). differential relation becomes dρ/ρ = -τρV2 dV/V

(b). fraction change in density; dρ/ρ = -8.7 ˣ 10⁻⁶

(c).  dρ/ρ = -8.7 ˣ 10⁻²

Explanation:

Let us begin,

(a).  given from the question we have that dp = -ρVdV

where dρ = ρ τ dp, i.e.

dp = dρ/ρτ ...............(1)

replacing value of dp we have,

-ρVdV = dρ/ρτ

so that dρ = -τp2 VdV

finally, dρ/ρ = -τp V2 dV/V

(b). from the question here, we were given Velocity to be = 10 m/s

density (ρ) =  1.23 kg/m3

pressure (p) =  1.01 x 10⁵ N/m2

from formula,

dρ/ρ = τs ρ V2 dV/V .............(2)

but τs = 1/γp = 1/(1.4× 1.01×10⁵) = 7.07 ˣ 10⁻⁶ m²/N

substituting value of τs  into equation (2) we have

dρ/ρ = τs ρ V2 dV/V =  (7.07 ˣ 10⁻⁶) ˣ (1.23) ˣ (10ˣ2) (0.01) = -8.7 ˣ 10⁻⁶

dρ/ρ = -8.7 ˣ 10⁻⁶

(c). from we question we know that dρ/ρ has a large ratio of (1000/10)²

so dρ/ρ = -8.7 ˣ 10⁻⁶ × (1000/10)² = -8.7 ˣ 10⁻²

dρ/ρ = -8.7 ˣ 10⁻².

comparing this result with part (b). we can see that when we increase the velocity of a factor 100, there is an increased factorial change in the density by a factor 104.

Answer:

The mass of the asteroids is 0.000334896182184 times the mass of the Earth.

39929.4542466 m

Explanation:

Total mass of the asteroids

[tex]m_a20000\times 10^{17}=2\times 10^{21}\ kg[/tex]

[tex]m_e[/tex] = Mass of Earth = [tex]5.972\times 10^{24}\ kg[/tex]

The ratio is

[tex]\dfrac{m_a}{m_e}=\dfrac{2\times 10^{21}}{5.972\times 10^{24}}\\\Rightarrow \dfrac{m_a}{m_e}=0.000334896182184[/tex]

The mass of the asteroids is 0.000334896182184 times the mass of the Earth.

Volume is given by

[tex]V=\dfrac{m}{\rho}\\\Rightarrow \dfrac{4\pi}{3\times 8} d^3=\dfrac{m}{\rho}\\\Rightarrow d^3=\dfrac{3\times 8}{4\pi}\dfrac{m}{\rho}\\\Rightarrow d=(\dfrac{3\times 8}{4\pi}\dfrac{m}{\rho})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{3\times 8}{4\pi}\dfrac{10^{17}}{3000})^{\dfrac{1}{3}}\\\Rightarrow d=39929.4542466\ m[/tex]

The diameter is 39929.4542466 m

Answer:

Explanation:

Given

Cannon is fired with a velocity of [tex]u=72.50\ m/s[/tex]

Using Equation of motion

[tex]y=ut+\frac{1}{2}at^2[/tex]

where

[tex]y=displacement[/tex]

[tex]u=initial\ velocity[/tex]

[tex]a=acceleration[/tex]

[tex]t=time[/tex]

after time [tex]t=3.3 s[/tex]

[tex]y=72.50\times 3.3-\frac{1}{2}\times 9.8\times (3.3)^2[/tex]

[tex]y=239.25-53.36[/tex]

[tex]y=185.89\ m[/tex]

So after 3.3 s cannon ball is at a height of 185.89 m

Answer:

Volume of 22 drop will be 0.68 ml

Explanation:

We have given volume of one drop = 0.031 ml

We know that 1 liter = 1000 ml

So [tex]1ml=10^{-3}L[/tex]

So 0.031 ml will be equal to [tex]0.031\times 10^{-3}L[/tex]

We have to find the volume of 22 drop

For finding volume of 22 drop we have to multiply volume of one drop by 22

So volume of 22 drop will be [tex]=22\times 0.031\times 10^{-3}=0.682\times 10^{-3}L=0.68mL[/tex]

So volume of 22 drop will be 0.68 ml

Answer:

0 MN/C, 2.697 MV

4.1953 MN/C, 1.2586 MV

19.2642857143 MN/C, 2.697 MV

Explanation:

k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]

Electric field at r = 8 cm

E = 0 (inside)

Electric potential is given by

[tex]V=\dfrac{kq}{R}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV[/tex]

Electric potential is 2.697 MV

r = 30 cm

[tex]E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3^2}\\\Rightarrow E=4195333.33\ MN/C=4.1953\ MN/C[/tex]

Electric field is 4.1953 MN/C

[tex]V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3}\\\Rightarrow V=1258600\ V=1.2586\ MV[/tex]

Electric potential is 1.2586 MV

r = R = 14 cm

[tex]E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14^2}\\\Rightarrow E=19264285.7143\ N/C=19.2642857143\ MN/C[/tex]

The electric field is 19.2642857143 MN/C

[tex]V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV[/tex]

The potential is 2.697 MV

To calculate the electric field and electric potential at different distances from the center of a spherical conductor, use the equations for electric field and electric potential due to a point charge. At a distance of 8.0 cm, the electric field is 7.87 * 10^3 N/C and the electric potential is 5.25 kV. At a distance of 30.0 cm, the electric field is 1.26 * 10^3 N/C, but the electric potential cannot be calculated without a reference point. At 14.0 cm, the electric field and electric potential are zero.

To calculate the electric field and electric potential at different distances from the center of a spherical conductor with a radius of 14.0 cm and a charge of 42.0 µC, we can use the equations for electric field and electric potential due to a point charge.

(a) At a distance of 8.0 cm from the center, the magnitude of the electric field can be calculated using the equation:

E = k * (Q/r²)

Substituting the given values, we get:

E = (9 * 10^9 Nm²/C²) * (42 * 10^-6 C) / (0.08 m)² = 7.87 * 10^3 N/C

To calculate the electric potential at this distance, we can use the equation:

V = k * (Q/r)

Substituting the given values, we get:

V = (9 * 10^9 Nm²/C²) * (42 * 10^-6 C) / (0.08 m) = 5.25 kV

(b) At a distance of 30.0 cm from the center, the magnitude of the electric field can be calculated in the same way:

E = (9 * 10^9 Nm²/C²) * (42 * 10^-6 C) / (0.3 m)² = 1.26 * 10^3 N/C

But the electric potential at this distance cannot be calculated using the given information, as we need the reference point for potential measurement.

(c) At a distance of 14.0 cm (the radius of the conductor) from the center, the electric field and electric potential are zero, as the conductor is in electrostatic equilibrium.

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Answer:

Explanation:

Given

Dimension of Plastic sheet is [tex]2\times 4\ m^2[/tex]

Charge on sheet [tex]Q=-10\ \mu C[/tex]

Charge density [tex]\sigma =\frac{q}{A}[/tex]

[tex]\sigma =\frac{-10\times 10^{-6}}{8}=1.25\times 10^{-6}\ C/m^2[/tex]

Sphere has mass of  [tex]m=4\ \mug[/tex]

If sphere is suspended motionless then its weight is balanced by repulsion force

Repulsive force [tex]F_r=qE[/tex]

where E=Electric field due to sheet

[tex]E=\frac{q}{2\epsilon _0}[/tex]

[tex]E=\frac{-10\times 10^{-6}}{2\times 8.85\times 10^{-12}}[/tex]

[tex]E=5.647\times 10^{5}\ N/C[/tex]

[tex]F_r=q\times 5.647\times 10^{5}[/tex]

[tex]F_r=mg[/tex]

[tex]q=\frac{mg}{E}[/tex]

[tex]q=\frac{4\times 10^{-6}\times 9.8}{5.647\times 10^{5}}[/tex]

[tex]q=6.9417\times 10^{-11}\ C[/tex]

To solve this problem we will make a diagram in the Cartesian plane that will allow us to find and understand more accurately the displacement and the angle of rotation.

According to Pythagoras, the distance traveled would be equivalent to

[tex]d = \sqrt{(2.7)^2+(3.5)^2}[/tex]

[tex]d = 4.4 miles[/tex]

The individual had a displacement of 4.4 thousand from the starting point.

Now the angle [tex]\theta[/tex] plus the previously given angle will allow us to find the direction of travel.

[tex]tan\theta = \frac{\text{Opposite side}}{\text{Adjacent side}}[/tex]

[tex]tan\theta = \frac{3.5}{2.7}[/tex]

[tex]\theta = tan^{-1} (\frac{3.5}{2.7})[/tex]

[tex]\theta = 52.35\°[/tex]

[tex]\angle =[/tex] [tex]\theta + 35 = 52.35+35 = 87.35\°[/tex]

Therefore the net direction of the man is S 87.35° W

To solve this problem we will apply the concept related to the Poisson ratio for which the longitudinal strains are related, versus the transversal strains.  First we need to calculate the longitudinal strain as follows

[tex]\epsilon_x = \frac{l_f-l_i}{l_i}[/tex]

[tex]\epsilon_x = \frac{(9.80554)-(9.8)}{9.8}[/tex]

[tex]\epsilon_x = 0.0005653[/tex]

Second we will calculate the lateral strain as follows

[tex]\epsilon_y = \frac{a_f-a_i}{a_i}[/tex]

[tex]\epsilon_y = \frac{2.59952-2.6}{2.6}[/tex]

[tex]\epsilon_y = -0.0001846153[/tex]

The Poisson's ratio is the relation between the two previous strain, then,

[tex]\upsilon = -\frac{\epsilon_y}{\epsilon_x}[/tex]

[tex]\upsilon = -\frac{(-0.0001846153)}{0.0005653}[/tex]

[tex]\upsilon = 0.3265[/tex]

Therefore the Poisson's ratio for the material is 0.3265

To find the temperature at 2000 feet above sea level, we need to use the average lapse rate of 3.5°F per 1000 feet. By calculating the temperature decrease with increasing altitude, we can determine that the temperature at 2000 feet would be 93°F.

To find the temperature at 2000 feet, we need to use the average lapse rate. The average lapse rate tells us how much the temperature decreases with increasing altitude. In this case, the average lapse rate is 3.5°F per 1000 feet. So for every 1000 feet increase in altitude, the temperature decreases by 3.5°F.

Since we want to find the temperature at 2000 feet, which is 2000 feet above sea level, we need to divide 2000 by 1000 to find how many intervals of 1000 feet we have. There are 2 intervals. So, we multiply the lapse rate of 3.5°F by 2 to get a temperature decrease of 7°F.

Therefore, the temperature at 2000 feet would be 100°F - 7°F = 93°F.

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If the temperature at the surface of Earth (at sea level) is 100°F, at 2000 feet if the average lapse rate is 3.5°F/1000 the temperature is 93°F.

If the temperature at the surface of Earth at sea level is 100°F, we can calculate the temperature at 2000 feet using the average lapse rate, which is given as 3.5°F per 1000 feet.

The temperature change is 3.5°F/1000 feet × 2000 feet = 7°F.

Therefore, the temperature at 2000 feet above sea level would be the surface temperature minus the temperature change due to the lapse rate:

100°F - 7°F

= 93°F.

Complete question:

A parallel-plate capacitor has plates with an area of 405 cm² and an air-filled gap between the plates that is 2.25 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery.   (a) How much energy is stored in the capacitor? (b) The separation between the plates is now increased to 4.50 mm.   How much energy is stored in the capacitor now?

Answer:

The energy stored in the capacitor when the plates is increased to 4.50 mm is 1.32 X 10⁻⁵ J

Explanation:

Given:

Area of the plates = 405 cm² = 405 X (10⁻²)² m²= 405 X 10⁻⁴m² = 0.0405m²

Energy stored in a capacitor = CV²/2

Where;

V is the voltage across the plates = 575 V

C is the capacitor =?

C = Kε(A/d)

K is constant = 1.0

ε is permittivity of free space = 8.885 X 10⁻¹²

d is the diameter of the two plates = 2.25 mm = 0.00225m

C = 1.0 x 8.885 X 10⁻¹² x (0.0405/0.00225)

C = 1.5993 X 10⁻¹⁰ F

(a) Energy stored in a capacitor = 0.5 X 1.5993 X 10⁻⁹ X 575²

= 2.64 X 10⁻⁵ J

(b) The separation between the plates is now increased to 4.50 mm.

C = Kε(A/d)

New diameter, d = 4.5 mm = 0.0045 m

C = 1.0 x 8.885 X 10⁻¹² x (0.0405/0.0045)

C = 7.9965 X 10 ⁻¹¹ F

Energy stored in a capacitor = 0.5 X 7.9965 X 10 ⁻¹¹ X  575²

= 1.32 X 10⁻⁵ J

Therefore, the energy stored in the capacitor when the plates is increased to 4.50 mm is 1.32 X 10⁻⁵ J

Answer:

Mass of the silver will be equal to 46.70 gram

Explanation:

We have given heat required to raise the temperature of silver by 24°C is 269 J , so [tex]\Delta T=24^{\circ}C[/tex]

Specific heat of silver = 0.240 J/gram°C

We have to find the mass of silver

We know that heat required is given by

[tex]Q=mc\Delta T[/tex], here m is mass, c is specific heat of silver and [tex]\Delta T[/tex] is rise in temperature

So [tex]269=m\times 0.240\times 24[/tex]

m = 46.70 gram

So mass of the silver will be equal to 46.70 gram

Explanation:

It is known that inside a sphere with uniform volume charge density the field will be radial and has a magnitude E that can be expressed as follows.

           E = [tex]\frac{q}{4 \p \epsilon_{o}R^{3}}r[/tex]

    V = [tex]-\int_{0}^{r}E. dr[/tex]

        = [tex]-\int_{0}^{r}\frac{q}{4 \p \epsilon_{o}R^{3}}r dr[/tex]

        = [tex](\frac{q}{4 \p \epsilon_{o}R^{3}})(\frac{r^{2}}{2})[/tex]

        = [tex]\frac{-qr^{2}}{8 \pi \epsilon_{o}R^{3}}[/tex]

At r = 1.45 cm = [tex]1.45 \times 10^{-2}[/tex]  (as 1 m = 100 cm)

     V = [tex]\frac{2.50 \times 10^{-15} \times 1.45 \times 10^{-2}}{8 \times 3.1416 \times 8.85 \times 10^{-12} \times 2.36 \times 10^{-2}}[/tex]

         = [tex]6.905 \times 10^{-6}[/tex] mV

Thus, we can conclude that value of V at radial distance r = 1.45 cm is  [tex]6.905 \times 10^{-6}[/tex] mV.

Answer:

a. 1.174 rad[/tex] or 67.3 degree

b. t = 49.28 s

Explanation:

Let [tex]v_v[/tex] be the vertical component of the boat velocity with respect the the river, pointing North. Let [tex]v_h[/tex] be the horizontal component of the boat velocity with respect to the river, pointing West, aka upstream. Since the total velocity of the boat is 4.4m/s

[tex]v_v^2 + v_h^2 = 4.4^2 = 19.36[/tex]

The time it takes for the boat to cross 200m-wide river at [tex]v_v[/tex] rate is

[tex]t = 200 / v_v[/tex] or [tex]v_v = 200 / t[/tex]

This is also the time it takes for the boat to travel 35m upstream, horizontally, at the rate of [tex] v_h - 0.99[/tex] m/s

[tex]t = \frac{35}{v_h - 0.99}[/tex]

[tex]v_h - 0.99 = 35/t[/tex]

[tex]v_h = 35/t + 0.99[/tex]

We can substitute [tex]v_v,v_h[/tex] into the total velocity equation to solve for t

[tex]\frac{200^2}{t^2} + (\frac{35}{t} + 0.99)^2 = 19.36[/tex]

[tex]\frac{40000}{t^2} + \frac{35^2}{t^2} + 2*0.99*\frac{35}{t} + 0.99^2 = 19.36[/tex]

From here we can multiply both sides by [tex]t^2[/tex]

[tex]40000 + 1225 + 69.3t + 0.9801t^2 = 19.36t^2[/tex]

[tex]18.38 t^2 - 69.3t - 41225 = 0[/tex]

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

[tex]t= \frac{69.3\pm \sqrt{(-69.3)^2 - 4*(18.3799)*(-41225)}}{2*(18.38)}[/tex]

[tex]t= \frac{69.3\pm1742.31}{36.7598}[/tex]

t = 49.28 or t = -45.51

Since t can only be positive we will pick t = 49.28

[tex]v_h = 35 / t + 0.99 = 35 / 49.38 + 0.99 = 1.7 m/s[/tex]

The angle, relative to the flow of river direction is

[tex]cos(\alpha) = \frac{v_h}{v} = \frac{1.7}{4.4} = 0.3864[/tex]

[tex]\alpha = cos^{-1}(0.3864) = 1.174 rad[/tex] or 67.3 degree

Answer:

  P₂= 141.15 kPa  

Explanation:

Given that

Volume ,V= 3 L

V= 0.003 m³

Initial temperature ,T₁ = 273 K

Initial pressure ,P₁ = 105 kPa

Final temperature ,T₂ = 367 K

Given that volume of the cylinder is constant .

Lets take final pressure = P₂

We know that for constant volume process

[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}\\P_2=P_1\times \dfrac{T_2}{T_1}\\P_2=105\times \dfrac{367}{273}\ kPa\\\\P_2=141.15\ kPa[/tex]

Therefore the final pressure = 141.15 kPa

P₂= 141.15 kPa

Answer:

130.165636364°C

Explanation:

P = Pressure

V = Volume

n = Number of moles

R = Gas constant = 0.082 L atm/mol K

From ideal gas law we have

[tex]PV=nRT\\\Rightarrow n=\dfrac{PV}{RT}\\\Rightarrow n=\dfrac{22\times 25}{0.082\times (25+273.15)}\\\Rightarrow n=22.496451696\ moles[/tex]

[tex]PV=nRT\\\Rightarrow T=\dfrac{PV}{nR}\\\Rightarrow T=\dfrac{48\times 15.5}{22.496451696\times 0.082}\\\Rightarrow T=403.315636364\ K[/tex]

The initial temperature is [tex]403.315636364-273.15=130.165636364\ ^{\circ}C[/tex]

Answer: 130 degrees Celsius

Explanation:

P1V1 / T1 = P2V2 / T2

Let the subscript 1 represent the initial 15.5L of gas and the subscript 2 represent the gas at the final volume of 25L. Rewrite the temperature in Kelvin by adding 273.

P1=48atm, V1=15.5L, P2=22atm, V2=25L, T2=298K, and T1 is unknown.

Substitute the known values into the combined gas law equation.

P1V1 / T1=P2V2 / T2 → (48atm)(15.5L) / T1 = (22atm)(25L) / 298K

Solve the equation for T1, simplify, and round to the nearest degree.

T1 = P1V1T2 / P2V2

T1 = (48atm)(15.5L)(298K)(22atm)(25L)

T1 = 403K

To get the temperature in Celsius, subtract 273.

T1 = 130∘C

Answer

given,

length of tube = 120 cm

a) Open at both ends.

distance between two successive nodes or anti nodes = λ/2

for first possibility =  λ/2 = 120

                          λ  =2 x 120 = 240 cm

for second Possibility, distance between two node or antinode acting symmetrically

                       λ = 120 cm

for three nodes in the tube, distance between them is equal to

[tex]\dfrac{3}{2}\lambda = 120[/tex]

[tex]\lambda = 80 cm[/tex]

b) Open at one end

 First possibility, The distance between one node or anti node

  [tex]\dfrac{\lambda}{4} = 120[/tex]

             λ = 480 cm

 Second Possibility, distance between two node or anti node is equal to

[tex]\dfrac{3}{4}\lambda = 120[/tex]

    λ = 160 cm

Third possibility, distance between three nodes and three anti nodes is equal to

[tex]\dfrac{5}{4}\lambda = 120[/tex]

   λ = 100 cm

From the work theorem, this is defined as the amount of force applied on an object displaced on a longitudinal unit. Mathematically this is

[tex]W = \vec{F} \times \vec{d}[/tex]

Here,

F = Force vector

d = Displacement vector

Of all the options presented, only in the last one there is no change in distance, so the work done there is zero.

The correct option is 5.

Work in Physics is when a force causes displacement. Of the choices, the example of a runner stretching by pushing against a wall does not involve work, as there is no displacement.

In the context of Physics, 'work' is defined as a force causing displacement on an object. Essentially, work is done when a force acts upon an object to cause or prevent motion. Among the options provided, 5. A runner stretches by pushing against a wall does not involve 'work'. This is because, despite the runner exerting a force against the wall, there is no displacement of the wall in response to this force.

Work (W) is calculated by multiplying the force (F) that is applied to an object and the distance (d) that the object is moved, i.e., W = F * d. In this case, since the displacement (d) is zero, the work done is also zero.

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Answer: Height of building = 17.69m, velocity of brick = 18.6m/s

Explanation: From the question, the body has a zero initial speed, thus initial velocity (u) all through the motion is zero.

By ignoring air resistance makes it a free fall motion thus making it to accelerate constantly with a value of [tex]a = 9.8m/s^{2}[/tex].

Time taken to fall = 1.90s

a)

thus the height of the building is calculated using the formulae below

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

but u = 0 , hence

[tex]H = \frac{1}{2} gt^{2}[/tex]

[tex]H = \frac{1}{2} *9.8* 1.9^{2} \\\\H = 17.69m[/tex]

b)

to get the value of velocity (v) as the brick hits the ground, we use the formulae below

[tex]v^{2} = u^{2} + 2aH[/tex]

but u= 0, hence

[tex]v^{2} = 2gH\\[/tex]

[tex]v^{2} = 2 * 9.8 * 17.69\\v = \sqrt{2 *9.8* 17.69} \\v = 18.62m/s[/tex]

find the attachment in this answer for the accleration- time graph, velocity- time graph and distance time graph

a. The height of building, in meters, is equal to 17.69 meters.

b. The magnitude of the brick’s velocity just before it reaches the ground is equal to 18.72 m/s.

Given the following data:

We know that acceleration due to gravity (a) for an object in free fall is equal to 9.8 meter per seconds square.

a. To determine the height of building, in meters, we would use the second equation of motion:

Mathematically, the second equation of motion is given by the formula;

[tex]S = ut + \frac{1}{2} at^2[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]S = 0(1.90) + \frac{1}{2} \times 9.8 \times 1.90^2\\\\S = 0 + 4.9 \times 3.61[/tex]

Distance, S = 17.69 meters.

b. To determine the magnitude of the brick’s velocity just before it reaches the ground, we would use the third equation of motion;

[tex]V^2 = U^2 + 2aS[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]V^2 = 0^2 + 2 \times 9.8 \times 17.69\\\\V^2 = 350.26\\\\V = \sqrt{350.26}[/tex]

V = 18.72 m/s

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Answer:

A) The car should be traveling at 31.9 m/s.

B) The speed of the car just before it lands on the other side is 37.0 m/s.

Explanation:

Hi there!

A) Please see the attached figure for a better description of the problem. When the car reaches the other side of the river, its position vector will be r1 in the figure. The components of this vector are r1x and r1y.

If we place the origin of the frame of reference at the edge of the cliff, the components of the vector r1 will be:

r1x = 61.0 m

r1y = -20.0 m + 2.1 m = -17.9 m

The equations for the x and y-components of the position vector of the car are the following:

x = x0 + v0 · t

y = y0 + 1/2 · g · t²  

Where:

x = horizontal position at a time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

y = vertical position at a time t.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Using the equation of the y-component of r1, we can find the time it takes the car to reach the other side of the river. We have to find the time at which the vector r1y is -17.9 m:

y = y0 + 1/2 · g · t²   (y0 = 0 because the origin of the frame of reference is located at the edge of the cliff).

y = 1/2 · g · t²

-17.9 m = -1/2 · 9.8 m/s² · t²

-17.9 m / -4.9 m/s² = t²

t = 1.91 s

Now, using the equation of the x-component, we can find the initial velocity. We know that at t = 1.91 s, the horizontal component of the vector r1 is 61.0 m:

x = x0 + v0 · t (x0 = 0 because the origin of the frame of reference is located at the edge of the cliff).

x = v0 · t

61.0 m = v0 · 1.91 s

v0 = 61.0 m / 1.91 s = 31.9 m/s

The car should be traveling at 31.9 m/s.

B) The equation of the velocity vector of the car is the following:

v = (v0, g · t)

The horizontal component of the velocity vector is v0, 31.9 m/s.

Let's calculate the value of the vertical component:

vy = g · t

vy = -9.8 m/s² · 1.91 s

vy = -18.7 m/s

Then, the velocity vector of the car just before it lands on the other side is the following:

v = (31.9, -18.7) m/s

The magnitude of this vector is calculated as follows:

|v| = √[(31.9 m/s)² + (-18.7 m/s)²]  

|v| = 37.0 m/s

The speed of the car just before it lands on the other side is 37.0 m/s.

Final answer:

To safely clear the river, the car must travel at 23.4 m/s at the cliff's edge, and it will land with a speed of 22.6 m/s on the other side.

Explanation:

A) To clear the river and land safely on the other side, the car should be traveling at a speed of 23.4 m/s as it leaves the cliff. This is calculated using the principles of projectile motion and conservation of energy.

B) The speed of the car just before it lands safely on the opposite side would be 22.6 m/s. This speed is also determined by energy considerations, such as the conversion of potential energy to kinetic energy.