The image of your face will be located 66.5 cm behind the front **surface** of the Christmas tree ball.

Surface is the outer layer of a **physical object**, such as a planet or a star. It is also the top layer of an object, such as the ground or a body of water. Surface can refer to the interface between two objects, such as the atmosphere and the ocean, or the interface between two materials, such as the air and the ground. The term can also refer to the top layer of a material, such as the skin of a human or the bark of a tree. In some cases, surface can refer to the way a material or object looks or feels, such as the smoothness of a rock or the texture of a fabric.

This is because of the principles of **spherical mirrors**; the image is located twice the distance behind the mirror as the object is in front of it. Therefore, the distance between your face and the image of your face is 58.0 cm (29.0 cm + 66.5 cm). The answer is expressed to two significant figures, and the appropriate units are included.

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Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.

The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b).

1. By what angle has planet Y rotated through during this time?

The **angle **of the **planet **is mathematically given as

**dY= 704 degrees**

With **Kepler's third rule**, which states that a planet's orbit squared is a function of cubed radius, we can prove that this is the case.

Generally, the equation for **the period **is mathematically given as

(periodX / periodY)^2 = (radius X / radius Y)^3

Therefore

(pX / pY)^2 = 4^3

(pX / pY)^2 = 64

\sqrt{(pX / pY )^2}= \sqrt{64}

(pX / pY=8

In conclusion, Because it takes 8 times longer to **complete **one orbit on planet X, planet Y travels 8 times farther than planet X does in the same time **period**...

planet Y travels ;

dY=8 * 88.0

**dY= 704 degrees**

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Thanks to the direction finding feature in gmaps application, which most of us use, we can find our way. Here, the maps application offers us alternative routes. Among these suggestions, I want to choose the path that will have the least fuel and do this based on calculations. For example, one of the two directions may be short, but if that short route is also uphill, it will not be an economical route. In my opinion the most important factor is elevation. If we take elevation into account other factors such as friction, where assuming the same asphalt type is often used in the same area for friction, I think the correct result will be achieved. In your opinion, what are the input data required to find the least energy path, what assumptions can be made and what are the necessary formulations and calculations?

In **my opinion**, I think that the **input data **that are required to find the **least energy path **are:

This refers to the **use of a diagram **to represent the **features of a place **that shows its **physical landforms **to help in navigation.

Hence, we can see that when using maps like gmaps, it is important to consider both **elevation and distance **to be able to find the path that uses the **least energy.**

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ATTA-KAY PHYSICS 121 3. A ball A is left to roll down an inclined plane of inclination 30º. Just at the moment a second ball B is pushed up the plane with a velocity of 45ms¹. The balls met at a point where the velocity of B is 1.8 times the velocity of A. Calculate (a) the velocities of A and B when they meet. (b) when the two balls meet. (c) where the two ball meet. (g = 10 ms²)

vr>vs because the rolling ball acquires **rotational **as well as** translational kinetic energy.**

The accelerating force acting on the ball as it goes along a smooth plane is mgsin. Its acceleration is therefore equal to gsin. The mgsin acts down the plane as the ball travels down the rough inclined plane, but friction develops that acts up the plane.

Since both balls' potential energy is lost at the same rate, their KEs are actually equal at the base of the planes. However, a ball sliding down a smooth plane has only** translational kinetic energy,** but a ball rolling down a rough plane contains both** translational and rotational kinetic energy** at the bottom of the plane. As a result, the ball's translational KE will be lower than its** translational Kinetic energy.**

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A closed curve encircles several conductors. The line integral around this curve is (image attached below)

a) What is the net current in the conductors?

b) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral?

The net current in the **conductors **and the value of the line **integral**

This is further explained below.

What is the net current in the conductors?Generally,

To put it another way, the total **current **In flowing across a surface S (contained by C) is **proportional **to the line integral of the magnetic B-field (in tesla, T).

[tex]\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \iint_S \mathbf{J} \cdot \mathrm{d}\mathbf{S} = \mu_0I_\mathrm{enc}[/tex]

[tex]I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A[/tex]

B)

In conclusion, It is **possible **for the line integral to go around the loop in either direction (clockwise or counterclockwise), the vector area dS to point in either of the two normal **directions** and Ienc, which is the net current **passing **through the surface S, to be positive in either direction—but both directions can be chosen as positive in this example. The right-hand rule solves these ambiguities.

The resultant remains the same **at 3.2 *10^4 Tm**

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The mass of Saturn is 5.68 x 1026 kg, and use an orbital radius of 3.00 x 105 km. (G = 6.67 × 10-11 N ∙ m2/kg2). Find the orbital speed of an ice cube in the rings of Saturn.

The **orbital speed **of an **ice cube **in the rings of Saturn is determined as **355,366.5 m/s.**

The** orbital speed** of an astronomical body or object is the speed at which it orbits around the center of mass of the most massive body.

The **orbital speed** of ice cube in the rings of Saturn is calculated as follows;

v = √GM/r

where;

G is universal gravitation constantM is mass of Saturnr is the distance of the ice cubev = √(6.67 x 10⁻¹¹ x 5.68 x 10²⁶)/(3 x 10⁵)

v = 355,366.5 m/s

Thus, the **orbital speed **of an **ice cube **in the rings of Saturn is determined as **355,366.5 m/s.**

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Three ropes A, B and C are tied together in one single knot K.

If the tension in rope A is 65.3 N, then what is the tension in rope B?

The **tension **in the** rope B** is determined as **10.9 N.**

tanθ = (6 - 4)/(5 - 0)

tan θ = (2)/(5)

tan θ = 0.4

θ = arc tan(0.4) = 21.8 ⁰

Angle between B and Cθ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰

Apply **cosine rule** to determine the tension in rope B;

A² = B² + C² - 2BC(cos A)

B = C

A² = B² + B² - (2B²)(cos A)

A² = 2B² - 2B²(cos 43.6)

A² = 0.55B²

B² = A²/0.55

B² = 65.3/0.55

B² = 118.73

B = √(118.73)

B = 10.9 N

Thus, the **tension **in the **rope B** is determined as **10.9 N**.

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Hello!

This is an example of a force summation in the vertical direction.

We have the tension of rope A upward (+), and the equal vertical components of the tensions of rope B and C downward (-).

These forces sum to zero, since the knot is stationary.

[tex]\Sigma F = T_A - T_{By} - T_{Cy} \\\\0 = T_A - T_{By} - T_{Cy}[/tex]

Ropes 'B' and 'C' form equivalent angles from the vertical. (If you were to draw a line from rope A down). We can use right-triangle trig to determine the angle:

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

The ropes are 5 m long and 2 m tall, which are the opposite and adjacent sides respectively:

[tex]tan^{-1}(\frac{5}{2}) = 68.2^o[/tex]

The vertical components are the adjacent sides from this angle, so, we would use cosine.

[tex]0 = T_A - T_Bcos\theta - T_Ccos\theta[/tex]

Rope 'B' and 'C' have the same tensions since they form the same angle with the vertical and are the same length, so we can call them 'T'.

[tex]0 = T_A - 2Tcos\theta[/tex]

Solving for 'T':

[tex]2Tcos\theta = T_A \\\\T = \frac{T_A}{2cos\theta}\\\\T = \frac{65.3}{2cos(68.2)} = \boxed{87.92 N}[/tex]

A hypothetical planet has a mass 2.81 times that of Earth, but the same radius.

What is g near its surface?

The **acceleration due to gravity **near the surface of the planet is **27.38 m/s².**

g = GM/R²

where;

G is universal gravitation constantM is mass of the planetR is radius of the planetg is acceleration due to gravity = ?g = (6.626 x 10⁻¹¹ x 2.81 x 5.97 x 10²⁴) / (6371 x 10³)²

g = 27.38 m/s²

Thus, the **acceleration due to gravity **near the surface of the planet is **27.38 m/s².**

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A 2.30 mH toroidal solenoid has an average radius of 6.20 cm and a cross-sectional area of 2.80 cm2.

a) How many coils does it have? In calculating the flux, assume that B is uniform across a cross section, neglect the variation of B with distance from the toroidal axis.

b) At what rate must the current through it change so that a potential difference of 2.60 V is developed across its ends?

(a) The** number of turns** of the coil is determined as **1,596 turns**.

(b) The rate of change of **current **is determined as **1,130.43 A/s.**

L = N²μA/l

where;

L is inductance N is number of turnsA is areal is average length = 2πrN²μA = LI

N² = LI/μA

N² = (2.3 x 10⁻³ x 2π x 0.062)/(4π x 10⁻⁷ x 2.8 x 10⁻⁴)

N² = 2,546,428.6

N = √2,546,428.6

N ≈ 1,596 turns

Rate of current changeL = (emf)/I

I = (emf)/L

I = (2.6)/(2.3 x 10⁻³)

I = 1,130.43 A/s

Thus, the** number of turns** of the coil is determined as **1,596 turns**.

The rate of change of **current **is determined as **1,130.43 A/s.**

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What is the minimum work needed to push a 950- kg car 700 m up along a 8.5 ∘ incline? Ignore friction.

Express your answer with the appropriate units.

The **minimum work **needed to pus the cart up the inclined plane is **960000 J.**

The **work done** on a inclined plane is given below as:

Distance = 700 m

The force on an inclined plane, F = mgsinθ

where;

m is mass in kg

g = 9.81 m/s²

θ = 8.5°

Work done = 950 * 9.81 * sin 8.5 * 700

Work done = 960000 J

Therefore, **minimum work **needed is** 960000 J.**

In conclusion, the **work done** is a product of force and distance.

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a 5.5kg bowling ball has a weight on earth closest to what in N

The **weight **of the **body **is obtained as **53.9 N.**

The term **weight **refers to the product of the **mass **and the **acceleration **due to gravity.

Now we have the mass of the body as **5.5kg **and the acceleration due to gravity as** 9.8 m/s^2.**

It the follows that the weight is;

W = mg = 5.5kg * 9.8 m/s^2 = **53.9 N**

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Calculate the de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s (about 60 mi/h)

The **de Broglie wavelength** of a 0.56 kg ball moving with a constant velocity of 26 m/s is **4.55×10⁻³⁵ m**.

The **wavelength** that is incorporated with the **moving object** and it has the relation with the **momentum** of that object and mass of that object. It is **inversely proportional** to the momentum of that moving object.

λ=h/p

Where, λ is the de **Broglie wavelength**, h is the **Plank constant**, p is the **momentum** of the moving object.

Whereas, p=mv, m is the **mass** of the object and v is the **velocity** of the moving object.

Therefore, λ=h/(mv)

λ=(6.63×10⁻³⁴)/(0.56×26)

λ=4.55×10⁻³⁵ m.

The **de Broglie wavelength** associated with the object weight 0.56 kg moving with the **velocity** of 26 m/s is **λ=4.55×10⁻³⁵ m**.

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A person standing at the edge of a cliff throws one ball straight up and another ball straight down, each at the same initial speed. Neglecting air resistance, which ball hits the ground below the cliff with the greater speed?

Since both balls have the same conditions at the cliffs edge (in downward motion), they will have the same speed just before they reach the ground. Ball B will reach the ground quicker since Ball A had to travel up and then back down again to reach the edge of the cliff.

Define the term work and state its unit. An ant is dragging a house-fly and the elephant is pushing a big tree which is not moving. Who is doing work, the ant or the elephant? Justify your answer. 922.5 205

Ant is performing a work

what is work?

Work is the force applied on an individual with respect to displacement.

Work = Force × displacement

Unit is Nm

Elephant is pushing bt there is no displacement occurred so the work of elephant is zero.

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The **ant** works, but the **elephant** does not.

Work done = Force × Displacement.

If there are **ants** and houseflies,

**Ants** drag the house bug, so they use specific force to move the house bug from one point to another, so we can say they work.

In the case of the **elephant **and the tree,

When the** elephant** pushes the tree (applying a force), the tree does not move, i.e., there is no displacement, so there is work.

Work done = Force × Displacement

= Force × 0

= 0

Therefore,

The **ant** works, but the **elephant** does not.

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A 1300 kg steel beam is supported by two ropes. (Figure

1)

What is the tension in rope 1?

What is the tension in rope 2?

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is[tex]R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0[/tex]

where [tex]R_1,R_2[/tex] are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is[tex]R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0[/tex]

where [tex]m=1300\,\rm kg[/tex] and [tex]g=9.8\frac{\rm m}{\mathrm s^2}[/tex].

Eliminating [tex]R_2[/tex], we have

[tex]\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)[/tex]

[tex]R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2[/tex]

[tex]R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2[/tex]

[tex]-R_1 \sin(50^\circ) = -\dfrac{mg}2[/tex]

[tex]R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}[/tex]

Solve for [tex]R_2[/tex].

[tex]\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0[/tex]

[tex]\dfrac{R_2}2 = -mg\cot(110^\circ)[/tex]

[tex]R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}[/tex]

A canon ball is shot out of a cannon at an angle of 45 degrees. What is the initial velocity of the cannon ball if its initial horizontal velocity is 8 m/s?

**Answer:**

11.31 [m/s].

**Explanation:**

1. the required velocity can be calculated according to

[tex]V=\frac{V_{horizontal}}{sin45};[/tex]

2. according to the formula above:

V=8*1.41≈11.3137085 [m/s].

In the given figure, weight of stone inside water

is 9N and water displaced by stone is 2N then,

i)What is the actual weight of stone?

ii) Which principle is the

experiment based on?

The actual **weight** of the stone is **11 N**. It is based on the **Archimedes** principles.

Actual weight= weight inside water+ weight of water displaced

= 9N + 2N = **11N**

Thus, we can conclude that the **actual weight** of the object is **11N**.

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By how many newtons does the weight of a 100-kg person decrease when he goes from sea level to mountain top at an altitude of 5000 m? The mean radius of the earth is 6.38 × 106 m.

The weight of a 100-kg person **decreases by 4 N** when he goes from sea level to mountain top at an altitude of 5000 m.

The** weight **of a person is determined by the mass of the body and the acceleration due to gravity.

The **acceleration due to gravity, g **is dependent on the mass of the earth, M the radius of the earth and the gravitational force constant , G.

Mathematically, the acceleration due to gravity at the mountain top is determined using the formula:

g = GM/r²where:

G = 6.67 × 10⁻¹¹ Nm²/kg²

M = 5.9736 x 10²⁴ kg

r = 638000 + 5000 = 6385000

g = (6.67 × 10⁻¹¹ * 5.9736 x 10²⁴ )(6385000)²

g = 9.77 m/s²

His weight at the mountain top will be:

weight = 100 * 9.77

weight = 977 N

Weight at sea level = 100 * 9.81 = 981 N

Decrease in weight = 981 - 977

Decrease in **weight** = 4 N

In conclusion, the **weight** of the man varies according to his distance from the earth.

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A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall.

Calculate the tension in the cable.

The** tension, T **in the cable is equal to** 323.5 N.**

**Tension** is force exerted by a cable or string on another object usually a weight suspended from the cable or string

The** tension** in the cable is found this:

Angle of the boom with horizontal, θ = tan⁻¹(5/10) = 26.56°

The angle of cable with horizontal, B = tan⁻¹(4/10) = 21.80

Taking moments about the pivot:

175.5 * cos 26.56 + 94.7 * cos 26.56 * 0.5 = T (sin(26.56 + 21.80) * 1

T = 241.68/0.747

**T = 323.5 N**

In conclusion, the **tension** in the cable is determined by taking moments about the pivot.

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Light of intensity I0 is polarized vertically and is incident on an analyzer rotated at an angle from the vertical. Find the angle if the transmitted light has intensity

I = (0.750)I0, I = (0.500)I0, I = (0.250)I0, and I = 0.

(Enter your answers in degrees.)

a.** θ = 41. 4°**

b.** θ = 60°**

c. **θ = 75. 5°**

d. **θ = 90°**

From the given information, we would be using the **Malus' law**

It is given as;

I = I0 cos²θ

Where I0 is the intensity of the polarized light after passing through P

a. To find the angle, compare with the given equation

I = (0.750)I0

I = I0 cos θ

then

cos θ = 0. 750

θ = [tex]cos^-^1(0. 750)[/tex]

θ = 41. 4°

b. I = (0.500)I0

cos θ = 0. 500

θ = [tex]cos^-^1(0. 500)[/tex]

θ = 60°

c. I = (0.250)I0

cos θ = 0. 250

θ = [tex]cos^-^1 (0. 250)[/tex]

θ = 75. 5°

d. I = 0

cos θ = 0

θ = [tex]cos^-^1 (0)[/tex]

θ = 90°

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How big is this restoring force compared with the tensile force stretching the spring?

A. Bigger

B. Not enough info

C. Smaller

D. Same size

The** restoring force** on the spring is found to have exactly the **same **magnitude as the** stretching force. Option D**

The restoring force is the force that seeks to restore the spring to its equilibrium position. It has the same magnitude as the stretching force but acts in opposite direction.

Thus, the** restoring force** on the spring is found to have exactly the **same **magnitude as the** stretching force.**

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An object with a density of 941.0 kg/m3 and a mass of 1039.0 kg is thrown into the ocean. Find the volume that sticks out of the water. (use ρseawater = 1024 kg/m3)

The **volume** that **sticks **out of the** water** is **83 m³**.

To find the answer, we need to know about the **archimedes principle**.

= 1024 - 941 = **83 m³**

Thus, we can conclude that the **volume** that sticks out of the** water** is **83 m³**.

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Haley is trying to pull an object upward. The below forces are acting on the object.

Fp= 5500 N

Fg= 6000 N

Which represents the net force?

The **net force** is represented by **↓ 500N.**

The** net force **is the force that has the same effect in **magnitude **and **direction **as two or more forces acting together.

Now we have the **forces**;

Thus we can obtain the **net force **as;

5500 N - 6000 N

= **- 500 N**

Therefore the **net force** is represented by **↓ 500N.**

Missing parts:

Haley is trying to pull an object upward. The below forces are acting on the object.

Fp = 5500N

Fg = 6000N

Which represents the net force?

← 500N

→ 500N

↑ 500N

↓ 500N

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Using a schematic diagram, explain the steps of the laser technique.

The **steps **of the use of the **laser technique **is explained below:

This refers to the **type of surgery **that makes use of special **light beams **in order to cut open the **human **body in a surgical procedure.

Hence, we can see that the **laser technique **is considered safer than conventional surgical methods.

**Laser techniques** include:

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the temperature at which the velocity of sound in air is twice its velocity at 15°C

With the use of below** formula, **at **879 °C**, velocity will be double the velocity at 15 °C.

The velocity of **sound waves in air** is proportional to the square root of **Thermodynamic temperature**. That is, V = K[tex]\sqrt{T}[/tex]

Given that the temperature at which the velocity of sound in air is twice its velocity at 15°C, Let us make use of the **formula**;

(v2/v1) = √(T2 / T1)

Where

v2 = final velocityv1 = initial velocityT2 = final absolute temperatureT1 = initial temperature.Recall that absolute temperature = °C + 273.

If v2 = 2 × v1 and temperature in degree Celsius = 15°C, then,

Temperature in Kelvin K = 15 + 273 = 288

Substitute all the **parameters** into the formula

(2 × v1)/v1 = √(T2/288)

2 = √ (T2 /288)

**Square both sides**

4 = (T2/288)

T2 = 4 × 288

T2 = 1152K

Temperature in degrees Celsius = 1152 - 273 = 879 °C.

Therefore, at 879 °C, velocity will be double the velocity at 15 °C.

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Discuss the aspects of either the Gemini Program or the Soyuz Program.

**Answer:**

The Gemini program

**Explanation:**

The Gemini Program was the second human spaceflight program hosted by Nasa in the year 1961. Taking place between mission Mercury and Apollo, the Gemini spacecraft carried two people to space and marked the foundation to the upcoming Apollo mission to Moon. It was a series of missions into the outer orbital which took place between 1965 and 1966. Prior to the Gemini missions, NASA had little to no information about space and space traveling. It was crucial for them to get acquainted with life outside before establishing successful Moon landings. And the series of Gemini missions helped them do just that.

How much work must be done to stop a 975- kg car traveling at 105 km/h ?

Express your answer to two significant figures and include the appropriate units.

The **amount** of **work done** to stop a 975- kg car traveling at 105 km/h is **414,808.34J**.

The **amount** of **work done** by a moving **object** can be calculated using the following formula:

W (**Kinetic energy**) = ½ mv²

Where;

m = massv = velocityAccording to this question, a car of **975 kg** is traveling at **105 km/h**. This **speed** in m/s is **29.17m/s**.

K.E = ½ × 975 × 29.17²

K.E = **414,808.34J**

Therefore, the **amount** of **work done** to stop a 975- kg car traveling at 105 km/h is **414,808.34J**.

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Three equal positive charges 'q' are at the corners of an equilateral triangle of side 'a'.

a. Assuming that the three charges together create an electric field, find the location of a point other than the obvious one where the electric field is zero.

b. What is the magnitude and direction of the electric field at the top corner due to the two charges at the base?

(a) The location of a point where the **electric field** is **zero **is at the **center **of the triangle which is equal to** ¹/₆√3a**.

(b) The magnitude and direction of the **electric field **at the** top corner** due to the two charges at the base is **1.732 kq/a².**

The **electric field** is **zero **at the center of the equilateral triangle whose magnitude is equal to √3a/6.

E = E₁ + E₂

where;

E₁ is electric field at the left base cornerE₂ is electric field at the right base cornerE = kq/a²[(cos 60i + sin 60j) + (-cos 60i + sin 60j)]

E = kq/a²[2(sin 60j)] = 1.732 kq/a²

Thus, the location of a point where the **electric field** is **zero **is at the **center **of the triangle which is equal to** ¹/₆√3a**.

The magnitude and direction of the **electric field **at the** top corner** due to the two charges at the base is **1.732 kq/a².**

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A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale.

Evaluate the magnitude of the force on the left hand pole.

The **Force** on the left hand pole, **F' **= 0.167N

**Force** is an agent which produces a change in the motion or state of an object.

**Force **is a vector quantity.

The general **force **is calculated as follows:

F = mg/sinθ

m = 17.1 g = 0.0171 kg

g = 9.81 m/s²

θ = 45°

F = 0.0171 * 9.81/sin45

F = 0.237 N

**Force** on the left hand pole, F' = Fcosθ

F' = 0.237 * cos 45

**F' **= 0.167N

In conclusion, the **force** on the left hand pole is the horizontal component of force.

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Earth travels around the Sun each year in sn elliptical path, as opposed to a perfect curcle. This means that the speed if earth and its dustance from the Sun change over rhe course if a year. What does this sayabout the magnitude of the centripetal acceleration if earth over the course of a year

A change in the **linear speed** of the Earth around the sun will cause a change in the magnitude of the **centripetal acceleration.**

**Centripetal acceleration **is** **the acceleration of a body moving a circular path.

The relationship between **centripetal acceleration** and **speed**;

a = v²/r

where;

v is linear speeda is centripetal accelerationr is radius of the pathSince the **centripetal acceleration** is directly proportional to square of **linear speed**, a change in the **linear speed** of the Earth around the sun will cause a change in the magnitude of the **centripetal acceleration.**

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What is the x-component of a vector with a magnitude of 115 km at an angle of 22°?

The x-component of a **vector** are < 106.6, 43.07 >

Depending on the **angle** we are provided, the x-component of a **vector **can either be cos or sin. Cos always corresponds to the right triangle's side that contacts the specified angle.

If a **vector** v with magnitude ||v|| makes an angle θ with the positive **x-axis **then,

v = ||v|| cos θi + ||v|| sin θj

= < ||v|| cos θ , ||v|| sin θ >

Magnitude p = 115 km

Angle = 22°

p = ||p|| < cos θ, sin θ >

p = 115 < cos 22°, sin 22° >

p = 115 < 0.927, 0.3746 >

p = < 106.6, 43.07 >

Therefore, the x-component of a** vector** are < 106.6, 43.07 >

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