A uniformly charged thin rod of length L and positive charge Q lies along the x-axis with its left end at the origin as shown in Figure 1.

a. Set up a correct integral expression for the potential at point A,which lies a distance H above the right end of the rod. Point A has coordinates (L, H). You need to give appropriate limits of integration and expressions for r and dq

b. Set up a correct integral expression for the potential at point B on the x-axis, a distance D from the left end of the rod with the appropriate limits of integration.You need to give appropriate limits of integration and expressions for r and dq.

Answer:

6.2 × 10^-11 m

Explanation:

1 eV = 1.602 × 10-19 joule

2.00 × 104 ev. = 2.00 × 10^4 eV × 1.602 × 10^-19 joule/1eV

= 3.2 × 10^-15 J

From;

E= hc/λ

λ = hc/E

λ = 6.6 × 10^-34 × 3 × 10^8/3.2 × 10^-15

λ = 6.2 × 10^-11 m

Answer: The focal length of the lens is 2.60 cm

Explanation:

The equation for lens formula follows:

[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]

where,

f = focal length = ? cm

v = image distance = 2.9 cm

u = Object distance = -25 cm

Putting values in above equation, we get:

[tex]\frac{1}{f}=\frac{1}{2.9}-\frac{1}{(-25)}\\\\\frac{1}{f}=\frac{1}{2.9}+\frac{1}{(25)}\\\\\frac{1}{f}=\frac{25+2.9}{2.9\times 25}\\\\f=\frac{72.5}{27.9}=2.60cm[/tex]

Hence, the focal length of the lens is 2.60 cm

Answer:

1140 J, 6840 J, 10260 J

Explanation:

Lo x 2 Lo x 3 Lo, Lo = 0.2 m,  K = 150 J/(s · m · C˚) , T = 35 ˚C, T' = 16 ˚C,

time, t = 6 s

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 1140 J[/tex]

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 6840 J[/tex]

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{2\times 0.2}\\\\H = 10260 J[/tex]

Answer:

The electron’s wavefunction has at least one node (i.e., at least one place in space where it goes to zero).

Explanation:

We know that the p-orbitals have nodes. A node is a region where the probability of finding an electron goes down to zero.

P orbitals are oriented along the x,y,z Cartesian axes and are known to have angular nodes along the axes.

Hence, if an electron in a hydrogen atom is in a p state, the electron’s wavefunction has at least one node

Answer:

A) v_{f1} = -3.2 m / s,  B) LEFT , C) v_{f2} = -0.12 m / s,  D) LEFT

Explanation:

This is a collision exercise that can be solved using momentum conservation, for this we define a system formed by gliders, so that the forces during the collision are internal and the moment is conserved.

Let's use the subscript 1 for the lightest glider m1 = 0.160 kg and vo1 = 0.820 m / s

subscript 2 for the heaviest glider me² = 0.820 kg and vo2 = -2.27 m / s

Initial instant. Before the crash

          p₀ = m₁ v₀₁ + m₂ v₀₂

Final moment. After the crash

          p_f = m₁ v_{f1} + m₂ v_{f2}

          p₀ = p_f

          m₁ v₀₁ + m₂ v₀₂ = m₁ v_{f1} + m₂ v_{f2}

 as the shock is elastic, energy is conserved

         K₀ = K_f

         ½ m₁ v₀₁² + ½ m₂ v₀₂² = ½ m₁ [tex]v_{f1}^2[/tex] + ½ m₂ [tex]v_{f2}^2[/tex]

         m₁ (v₀₁² - v_{f1}²) = m₂ (v_{f2}² -v₀₂²)

let's make the relationship

         (a + b) (a-b) = a² -b²

         m₁ (v₀₁ + v_{f1}) (v₀₁-v+{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)

let's write our two equations

         m₁ (v₀₁ -v_{f1}) = m₂ (v_(f2) - v₀₂)                                  (1)

         m₁ (v₀₁ + v_{f1}) (v₀₁-v_{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)

we solve

         v₀₁ + v_{f2} = v_{f2} + v₀₂

we substitute in equation 1 and obtain

         M = m₁ + m₂

         [tex]v_{f1} = \frac{m_1-m_2}{M} v_o_1 + 2 \frac{m_2}{M} v_f_2[/tex]

         [tex]v_f_2 = \frac{2m_1}{M} v_o_1 + \frac{m_2-m_1}{M} v_o_2[/tex]vf2 = 2m1 / mm vo1 + m2-m1 / mm vo2

we calculate the values

         m₁ + m₂ = 0.160 +0.3000 = 0.46 kg

         v_{f1} = [tex]\frac{ 0.160 -0.300} {0.460} \ 0.820 + \frac{2 \ 0300}{0.460} \ (-2.27)[/tex]

         v_{f1} = -0,250 - 2,961

          v_{f1} = - 3,211 m / s

         v_{f2} = [tex]\frac{2 \ 0.160}{0.460} \ 0.820 + \frac{0.300 - 0.160}{0.460 } \ (-2.27)[/tex]

         v_{f2} = 0.570 - 0.6909

         v_{f2} = -0.12 m / s

now we can answer the different questions

A) v_{f1} = -3.2 m / s

B) the negative sign indicates that it moves to the left

C) v_{f2} = -0.12 m / s

D) the negative sign indicates that it moves to the LEFT

Answer:

A gold medal has the (minimum) dimensions of:

diameter = 60mm

thickness = 3mm

So we will work with those dimensions.

The medal is then a cyinder of diameter

D = 60mm = 6cm

and height:

H = 3mm = 0.3cm

Remember that the volume of a cylinder is:

V = pi*(D/2)^2*H

where pi = 3.14

Then the volume of a medal is:

V = 3.14*(6cm/3)^2*0.3cm = 3.768 cm^3

The density of the gold in g/cm^3 is:

d = 19.3 g/cm^3

And remember that:

density = mass/volume

So, if the volume is 3.768 cm^3

Then the mass will be:

mass = density*volume =  19.3 g/cm^3*3.768 cm^3 = 72.7 g

So, a single gold medal would weight 72.7 grams

And each gram of gold costs $40

Then the total cost of the gold medal would be:

value = $40*72.7 = $2,908

Now, if yo think that in the Olympics there are 35 sports (a lot with a large number of players) and near 50 disciplines, they need a lot of gold medals.

And each gold medal costs $2,908

So the total cost (only for the gold medals, ignoring the others) would be to high.

This is why the gold medals are made mostly of silver.

Answer:

[tex]\mathbf{F_1=4.41*10^4\ N}[/tex]

[tex]\mathbf{F_2 = 1.176*10^5 \ N}[/tex]

Explanation:

The missing image of the figure slide is attached in below.

However, from the model, it is obvious that it is in equilibrium.

As a result, the relation of the force and the torque is said to be zero.

i.e.

[tex]\sum F = 0[/tex] and [tex]\sum \tau = 0[/tex]

From the image, expressing the forces through the y-axis, we have:

[tex]F_1+F_2 = W_B + W_P \\ \\ \implies 9.8(1500+15000) \\ \\ \implies \mathtt{1.617\times 10^5 \ N}[/tex]

Also, let the force [tex]F_1[/tex] be the pivot and computing the torque to determine [tex]F_2[/tex]:

Then:

[tex]F_1(0)+F_2(20.0) = 10.0W_B + 15.0W_P[/tex]

[tex]F_2 = \dfrac{((10*1500)+(15*15000))*9.8}{20.0}[/tex]

[tex]F_2 = 117600 \ N[/tex]

[tex]\mathbf{F_2 = 1.176*10^5 \ N}[/tex]

For the force equation:

[tex]F_1+F_2=1.617*10^5 \ N;[/tex]

where:

[tex]F_2 = 1.176*10^5 \ N[/tex]

Then:

[tex]F_1+1.176*10^5 \ N=1.617*10^5 \ N[/tex]

[tex]F_1=1.617*10^5 \ N-1.176*10^5 \ N[/tex]

[tex]F_1=44100\ N[/tex]

[tex]\mathbf{F_1=4.41*10^4\ N}[/tex]

Answer: [tex]29.85\ W/m^2[/tex]

Explanation:

Given

Power [tex]P=60\ W[/tex]

Distance from the light source [tex]r=0.4\ m[/tex]

Intensity is given by

[tex]I=\dfrac{P}{4\pi r^2}[/tex]

Inserting values

[tex]\Rightarrow I=\dfrac{60}{4\pi (0.4)^2}\\\\\Rightarrow I=\dfrac{60}{2.010}\\\\\Rightarrow I=29.85\ W/m^2[/tex]

Answer:

29.85 W/ m^2

Explanation:

Answer:

p = 1.2 kg-m/s

Explanation:

The question is, "An object of mass 0.5kg is moving along a rectilinear path with constant acceleration of 0.3m / s2. If it started from rest and the magnitude of its momentum in kg * m / s after 8s is".

Mass of the object, m = 0.5 kg

Acceleration of the object, a = 0.3 m/s²

We need to find the momentum after 8 seconds.

We know that,

[tex]p=F\times t[/tex]

i.e.

p = mat

So,

[tex]p=0.5\times 0.3\times 8\\\\p=1.2\ kg-m/s[/tex]

So, the momentum of the object is 1.2 kg-m/s.

Full Question:

What is the temperature of a system in thermal equilibrium with another system made up of water and steam at one atmosphere of pressure?

A) 0°F

B) 273 K

C) 0 K

D) 100°C

E) 273°C

Answer:

The correction Option is D) 100°C

Explanation:

The temperature above is referred to as the critical point.

it is the highest temperature and pressure at which water (which has three phases - liquid, solid, and gas) can exist in vapor/liquid equilibrium. If the temperature goes higher than 100 degrees celsius, it cannot remain is liquid form regardless of what the pressure is at that point.

There is also a condition under which water can exist in its three forms: that is  

- Ice (solid)

- Liquid (fluid)

- Gas (vapor)

That state is called triple point. The conditions necessary for that to occur are:

Cheers

Cheers

Answer:

-45,597.07

Explanation:

if not in scientific calculator and yung answer nung sa scientific sa comment na lang dinadownload ko ka eh

Answer:

P = 6.63 kW

Explanation:

Given that,

Current, I = 260 A

Voltage of the battery, V = 25.5 V

We need to find the power supplied to the starter motor. We know that,

P = VI

Put all the values,

P = 25.5 × 260

P = 6630 W

or

P = 6.63 kW

So, the power supplied to the motor is 6.63 kW.

Answer:

The power is 6.63 kW.

Explanation:

Current, I = 260  A

Voltage, V = 25.5 V

Power of an electrical appliance is given by

P = V I

P = 25.5 x 260

P = 6630 W

1 kW = 1000 W

So, the power is

P = 6.63 kW

Answer:

[tex]\sigma=0.014\ C/m^2[/tex]

Explanation:

Given that,

The radius of sphere, r = 5 cm = 0.05 m

Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C

We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

[tex]\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2[/tex]

So, the surface charge density on the sphere is [tex]0.014\ C/m^2[/tex].

Answer:

 T = 5.45 10⁻¹⁰   [tex]\sqrt{(R_e + h)^3}[/tex]

Explanation:

Let's use Newton's second law

          F = ma

force is the universal force of attraction and acceleration is centripetal

          G m M / r² = m v² / r

          G M / r = v²

as the orbit is circular, the speed of the satellite is constant, so we can use the kinematic relations of uniform motion

          v = d / T

the length of a circle is

          d = 2π r

we substitute

        G M / r = 4π² r² / T²

        T² = [tex]\frac{4\pi ^2 }{GM} \ r^3[/tex]

the distance r is measured from the center of the Earth (Re), therefore

        r = Re + h

where h is the height from the planet's surface

let's calculate

         T² = [tex]\frac{4\pi ^2}{ 6.67 \ 10^{-11} \ 1.991 \ 10^{30}}[/tex]   (Re + h) ³

         T = [tex]\sqrt{29.72779 \ 10^{-20}} \ \sqrt[2]{R_e+h)^3}[/tex]

         T = 5.45 10⁻¹⁰   [tex]\sqrt{(R_e + h)^3}[/tex]

Answer:

2.70

Explanation:

pH = -log[H+]

pH = -log[2.0x10^-3]

pH = 2.70

Answer:

XL sleep usual Addison officer at home and ear is not a short time to be be free and ear is a short time to make a short time

Explanation:

so that I can take the class on Monday and ear is not a short time to be be free and ear is not a short time to be be free and ear is not a short time to be be free and ear is not a short time to time for a day or night and ear buds is Anshu and duster and duster fgor a day or night is Anshu and duster for a day or not a week of computer science from your computer and I am in the same as I am a short of ti and you can be the first time I will be be

Explanation:

The initial kinetic energy [tex]KE_0[/tex] is

[tex]KE_0 = \frac{1}{2}m_0v_0^2[/tex]

When its mass and velocity are doubled, its new kinetic energy KE is

[tex]KE = \frac{1}{2}(2m_0)(2v_0)^2 = \frac{1}{2}(2m_0)(4v_0^2)[/tex]

[tex]\:\:\:\:\:\:\:= 8 \left(\frac{1}{2}m_0v_0^2 \right)= 8KE_0[/tex]

Therefore the kinetic energy will increase by a factor of 8.

Answer:

v₂ = 53.23 m/s

Explanation:

Given that,

The mass of a golf club, m₁ = 158 g = 0.158 kg

The initial speed of a golf club, u₁  =  48.2 m/s

The mass of a golf ball, m₂ = 46 g = 0.046 kg

It was at rest, u₂ = 0

Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s

We use the conservation of energy to find the speed of the golf ball just after impact as follows :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s[/tex]

So, the speed of the golf ball just after the impact is equal to 53.23 m/s.

Answer:

f = 7.06 N

Explanation:

The maximum frictional force on the knee joint of the person can be given by the following formula:

[tex]f = \mu R = \mu W \\[/tex]

where,

f = maximum frictional force = ?

μ = static friction coefficient = 0.016

W = Weight load on knee = mg

m = mass supported by knee = 45 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]f = \mu mg\\f = (0.016)(45\ kg)(9.81\ m/s^2)\\[/tex]

f = 7.06 N

Answer:

Objects that are said to be undergoing free fall, are not encountering a significant force of air resistance; they are falling under the sole influence of gravity.

Explanation:

Under such conditions, all objects will fall with the same rate of acceleration, regardless of their mass.

Answer:

1) attached below

2) assumption that the earth is spherical

Explanation:

1) Four illustrations of a globe

attached below

2) Reason for distortions at areas that do not fall on lines of tangency or secancy

The reason for distortion on areas outside the lines of tangency or secancy is because of the assumption that the earth is spherical which is not true hence map projections on the areas that fall on the lines of tangency do not experience distortion and are true

Answer:

wareffctgggyyggghhhh

Answer:

towards the center

Explanation:

that is the solution above

Answer:

1 eV = 1.60 * 10^-19 J      work done in accelerating electron thru 1 V

KE (total energy) = 1350 ^ 1 eV     (note proton goes from +  to -)

KE = 1.60 * 10^-19 * 1350 = 2.16 * 10^-16 Joules

1/2 m v^2 = KE = 2.16 * 10^-16 J

v^2 = 4.32 * 10E-16 / 1.67 * 10-27 = 2.59 * 10^11

v = 5.09 * 10^5 m/s

The proton's kinetic energy, in joules is  2.16 *[tex]10^{-16}[/tex] J. The proton's velocity is 5.09 * [tex]10^{5}[/tex]m/s.

When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.

Uniform motion an object is said to have uniform motion when object cover equal distance in equal interval of time within exact fixed direction. For a body in uniform motion, the magnitude of its velocity remains constant over time.

1 eV = 1.60 * [tex]10^{-19} J[/tex]     work done in accelerating electron throw 1 V

K.E (total energy) = 1350 ^ 1 eV     (note proton goes from +  to -)

K.E = 1.60 * [tex]10^{-19}[/tex]J * 1350 = 2.16 * [tex]10^{-16}[/tex] Joules

1/2 m v² = KE = 2.16 *[tex]10^{-16}[/tex] J

Velocity of proton is,

v² = 4.32 * 10[tex]e^{-16}[/tex] / 1.67 * [tex]10{-27}[/tex] = 2.59 * [tex]10^{11}[/tex]

v = 5.09 * [tex]10^{5}[/tex]m/s

The proton's kinetic energy, in joules is  2.16 *[tex]10^{-16}[/tex] J. The proton's velocity is 5.09 * [tex]10^{5}[/tex]m/s.

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Answer:

D) the two spheres remain of equal size.

Explanation:

Answer:

The number of turns of wire needed is 573.8 turns

Explanation:

Given;

maximum emf of the generator, = 190 V

angular speed of the generator, ω = 3800 rev/min =

area of the coil, A = 0.016 m²

magnetic field, B = 0.052 T

The number of turns of the generator is calculated as;

emf = NABω

where;

N is the number of turns

[tex]\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s[/tex]

[tex]N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns[/tex]

Therefore, the number of turns of wire needed is 573.8 turns

Answer:

C.  Rotational motion

Explanation:

The kinematics of rotational motion describes the relationships between the angle of rotation, angular velocity, angular acceleration, and time. It only describes motion—it does not include any forces or masses that may affect rotation (these are part of dynamics). Recall the kinematics equation for linear motion: v = v+at (constant a).

Rotational motion is mutually exclusive with each of the others. Hence, option (C) is correct.

"The motion of an object around a circular route, in a fixed orbit, is referred to as rotational motion."

Rotational motion dynamics are identical to linear or translational dynamics in every way. The motion equations for linear motion share many similarities with the equations for the mechanics of rotating objects. Rotational motion only takes stiff bodies into account. A massed object that maintains a rigid shape is referred to as a rigid body.

Curvilinear motion is the movement of an object along a curved route. Example: A stone hurled at an angle into the air.

The motion of a moving particle that follows a predetermined or known curve is referred to as curvilinear motion. Two coordinate systems—one for planar motion and the other for cylindrical motion—are used to examine this type of motion.

Learn more about motion here:

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Answer:

100j

Explanation:

Answer:

i think the answer is 0.001m³