The magnitude of charge A is half the magnitude of charge B. However, the electrostatic force experienced by charge A and by charge B is the same. It can be concluded that the electric field strength of charge A is ________ the electric field strength of charge B. one-fourth one-half two times four times

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 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 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 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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Answer:

Higher levels of organization are built from lower levels.

Molecules combine to form cells.

Cells combine to form tissues.

Tissues combine to form organs.

Organs combine to form organ systems,

and organ systems combine to form organisms.

Explanation:

Hope it helps.

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 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 tension in the cable is 380.55 N and the vertical component of the force exerted by the hi.nge on the beam is 11.45 N.

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (40 x 9.8)/(2 x sin31)

T = 380.55 N

T + F = W

F = W - T

F = (9.8 x 40) - 380.55

F = 11.45 N

Thus, the tension in the cable is 380.55 N and the vertical component of the force exerted by the hi.nge on the beam is 11.45 N.

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(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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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 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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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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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 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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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]

C. The center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.

The center of mass of an object is the unique point where the weighted relative position of the distributed mass sums to zero.

Center of gravity is the point from which the weight of a body or system may be considered to act.

Thus, the center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.

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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 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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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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Both forming images when placed in their respective places best describes  Becquerel’s experiments.

This was conducted by  Henri Becquerel in which he sought to know how uranium salts are affected by light.

He discovered that the salts emits a penetrating radiation and formed an image in the presence of light but didn't form any in darkness.

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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 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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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]

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].

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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Answer:

Explanation:

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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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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