You look at yourself in a shiny 8.8-cm-diameter Christmas tree ball.
If your face is 29.0 cm away from the ball's front surface, where is your image? Follow the sign conventions. Express your answer to two significant figures and include the appropriate units.

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

g = GM/R²

where;

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) 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;

E = 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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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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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;

v = √(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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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

θ = 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]

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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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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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;

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

This is further explained below.

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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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].

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,

[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;

[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) 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;

N²μA = 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

L = (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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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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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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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.

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:

where:

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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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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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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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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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

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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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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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 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;

Since 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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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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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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