true or false please help me now.
Calibration graphs can be used to determine unknown concentrations in electrochemical ​

(a) The magnetic field the line produced at ground level is 3.75 x 10⁻⁶ T.

(b) The lines magnetic field as a percent of the earth's magnetic field is 7.5%.

(c) No, since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.

B = (μ x I) / (2πr)

B = (4π x 10⁻⁷ x  150) / (2π x 8)

B = 3.75 x 10⁻⁶ T

x = 3.75 x 10⁻⁶ /0.5G

x = (3.75 x 10⁻⁶ ) / (0.5 x 10⁻⁴)

x = 0.075

x = 0.075 x 100% = 7.5%

Thus, we can conclude that, since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.

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Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

[tex]\Sigma \tau = 0[/tex]

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
[tex]\tau = r \times F[/tex]

Doing the summation using their respective lever arms:

[tex]0 = L Tsin\theta - dF_g[/tex]

[tex]dF_g = LTsin\theta[/tex]

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

[tex]tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o[/tex]

Now, let's solve for 'T'.

[tex]T = \frac{dMg}{Lsin\theta}[/tex]

Plugging in the values:
[tex]T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}[/tex]

The solution for the acceleration of gravity is given as

This is further explained below.

Generally,

Mass of earth [tex]$M=5.97 \times 10^{24} \mathrm{~kg}$[/tex]

Radius of earth [tex]$R=6371 \mathrm{~km}$[/tex]

Gravitational const. [tex]$G=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$[/tex]

height [tex]$h_{1}=7900 \mathrm{~m}=7.9 \mathrm{~km}$$$[/tex]

[tex]R+h_{1}=6371+7.9 &\\\\R+h_{1}=6378.9 \mathrm{~km} \\\\&R+h_{1}=6378.9 \times 10^{3} \mathrm{~m}\end{aligned}[/tex]

In conclusion, acceleration due to gravity at this point will be

[tex]g_{1}=\frac{G M}{\left(\bar{R}+\overline{h_{1}}\right)^{2}}$\\\\$g_{1}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(6378.9 \times 10^{3}\right)^{2}}$\\\\$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$[/tex]

for [tex]$h_{2}=7900 \mathrm{~km}$[/tex]

[tex]R+h_{2}=6371+7900\\\\R+h_{2}=14271 \mathrm{~km}\\\\$g_{2}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(14271 \times 10^{3}\right)^{2}}$\\\\$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$[/tex]

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The initial speed of the rock as it left the astronaut's hand is 168 m/s.

The speed of the satellite is 60.2 m/s.

g = GM/R²

where;

g = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(6.32 x 10⁴)²

g = 0.132 m/s²

v² = u² - 2gh

0 = u² - 2gh

u² = 2gh

u = √2gh

u = √(2 x 9.8 x 1440)

u = 168 m/s

v = √GM/r

v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(1.45 x 10⁵)]

v = 60.2  m/s

Thus, the initial speed of the rock as it left the astronaut's hand is 168 m/s.

The speed of the satellite is 60.2 m/s.

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The speed of a satellite in a circular orbit around the Earth is 4,188 m/s.

The speed of the satellite is calculated as follows;

v = √GM/r

where;

v = √[(6.67 x 10⁻¹¹ x 5.98 x 10²⁴) / (3.57 x 6.37 x 10⁶)]

v = 4,188 m/s

Thus, the speed of a satellite in a circular orbit around the Earth is 4,188 m/s.

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Answer:

d

Explanation:

Gravitational force is a type of force that pulls objects to the Earth's surface.

The acceleration due to gravity is 3 m/s^2 and the work done is -4.3 J.

Now we must use the formula;

h = ut + 1/2gt^2

Since it was dropped from a height u = 0 m/s

h = height

u = initial velocity

g = acceleration due to gravity

t = time

h = 1/2gt^2

g = 2h/t^2

g = 2 * 2.85 /(1.38)^2

g = 5.7/1.9

g = 3 m/s^2

The work that must be done is against gravity hence;

W = -(mgh)

W = - (0.5 kg *  3 m/s^2 * 2.85 m)

W = - 4.3 J

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The resultant force is 8N  

Given that mass is 2kg , v= 40m/s, u =20m/s and we need to calculate resultant force
F=ma

m is given
so for a
v-u/t=a { first equation of motion }

40-20/4= 4
so a=4

F = ma =2*4 = 8N
The difference between the forces that are acting on an object as part of a system is known as the resultant force.
v = u + at is the first equation of motion. Here, v denotes the end speed, u the starting speed, an acceleration, and t the passage of time. The first equation of motion is provided by the velocity-time relation, which may be used to calculate acceleration.

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Force necessary to support the object on piston 2 is 24× 10⁴ N.

To find the answer, we need to know about the force and pressure on piston 1 and piston 2.

= 991 Kg × 9.8 = 9414.5N

= 9414.5N × π(9.46² cm² /4)

= 9414.5N × π(9.46²× 10^(-4)m²/4)

= 66.2 N/m²

= 66.2/ π(1.87² cm² /4)

= 66.2/ π(1.87²×10^(-4)m² /4)

= 24× 10⁴ N

Thus, we can conclude that force necessary to support the object on piston 2 is 24× 10⁴ N.

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The speed of the space craft relative to the earth is given as: 0.024c. This is solved using the the equation for time dilation.

Time dilation is the "slowing down" of a clock as determined by an observer in relative motion with regard to that clock under the theory of special relativity.

The formula is given as :

Δt = [Δr]/ √ 1 - (v²/c²)

Thus,

v = c√1 - (Δr/Δt)²

= c √(1 - (3600/3601)²

v = 0.024c

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3.00 m is the magnitude of the resultant vector

137.1° is the directional angle of the resultant vector.

1) To find the magnitude of the resultant

Discover each vector's individual components first. 

We must consider the signs of the components because the angles in the figure are measured in various ways. 

In this case, both the y component of vector C and the x component of vector A are negative. 

The vectors' components are as follows, using a little trigonometry:  

Magnitude of A = 5 m

[tex]A_{x}[/tex] = - (5.00m) cos20° = -5 ×0.408 = -4.698 m

[tex]A_{y}[/tex] = + (5.00m)sin20° = +5 × 0.342 = +1.710 m

Magnitude of B = 5m

[tex]B_{x}[/tex] = +(5.00m)cos60° = 5 × 0.5 = +2.5m

[tex]B_{y}[/tex] = +(5.00m)sin60° = 5 × √3/2 = +4.33 m

Cx = 0

Cy = -4.00m

The sum of all three vectors, which we refer to as R, produces components.

Rₓ = Aₓ + Bₓ + Cₓ

   = -4.698 + 2.5 + 0

   = -2.198 m

[tex]R_{y}[/tex] = [tex]A_{y} + B_{y} + C_{y}[/tex]

    = +1.71 +4.33 - 4.00

    = 2.040 m

R = [tex]\sqrt{R_{x^{2} }+R_{y^{2} }[/tex]

  = [tex]\sqrt{(-2.198)^{2 } + (2.040)^{2} }[/tex]

  = 3.00 m

2) To find the directional angle of resultant

tan θ = 2.040/-2.198 = -0.928

θ = -42.9°

Such a vector would be in the so-called "fourth quadrant," as it is well known. However, we discovered that R has a negative x component and a positive y component, indicating that such a vector must reside in the "second quadrant.

The calculator accidentally returned an angle that is 180 degrees off, thus we must add 180 degrees to the naïve angle in order to get the right angle.

Therefore, R's actual direction is determined by-

θ = -42.9° + 180° = 137.1°

Hence, the magnitude of resultant vector is 3.00 m and directional angle is 137.1°

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Three moons can fit inside the volume of the sun.

The moon is a non luminous body found in the space. It could cause a solar eclipse when it comes between the sun and the earth.

Since the Earth’s diameter is about 8,000 miles and the Moon’s diameter is about 2,000 miles, to obtain the number of moons that could fit inside the sun we have;

8,000 miles/ 2,000 miles = 3

Hence, three moons can fit inside the volume of the sun.

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The altitude or height of the pole vaulter as she crosses the bar is 4.04 m.

The height of the pole vaulter is determined from the change in kinetic energy which is equal to the potential energy at that height.

h = (v - u)²/2g

h = (10 - 1.1)²/2 * 9.8

h = 4.04 m.

In conclusion, the height is determined from the potential energy at that height.

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Solutions for the three problems are is mathematically given as

(a) First cosmic speed (arbitral velocily)

[tex]v=\sqrt{\frac{G M}{r}}\\v=\sqrt{\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24}}{5870 \times 10^{3}}}[/tex]

v=7338.9349

(b) Second cosmic speed (escape velo.)

$$

\begin{gathered}

[tex]V=\sqrt{\frac{2 G M}{r}}\\\\V=\sqrt{2} \sqrt{\frac{G M}{r}}\\\\=V\sqrt{2} \times 7338.9349 \\[/tex]

[tex]V=14677.86986 \mathrm{~m} / \mathrm{s}[/tex]

(c) In conclusion, in a circular orbit, the gravitational force is gets balanced by centripetal force

[tex]&m_{q \omega \omega^{2}}=\frac{G M M}{r^{2}} \\&r^{3}=\frac{G M}{\omega^{2}}=\frac{G M}{4 \pi^{2}} T^{2} \\&r^{3}=\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24} \times(16.6 \times 3600)^{7}}{4 \pi^{2}} \\[/tex]

[tex]&r=30581248.06 \times 10^{4} \mathrm{~m} / \mathrm{s}[/tex]

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The image is present at 20cm from the crown glass spherical surface.

To find the answer, we need to know about the lens formula.

=> (1/V)+5=10

=> 1/V= 5

=> V=0.2m = 20cm

Thus, we can conclude that the image is present at 20cm.

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Sound will travel fastest in air at 15°C.

The speed of sound in air, given in the range of 100 degrees Celsius and 0 degree Celsius include;

Air at 100°C 387 m/s

Air at 0°C 331 m/s

From the date above, the speed of sound in air increases with increases in temperature. Thus, Sound will travel fastest in air at 15°C.

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The tangential speed of the wheel is determined as 4.786 m/s.

The tangential speed of the wheel is calculated as follows;

v = ωr

where;

v = (2.17 x 2π rad)/s x 0.351 m

v = 4.786 m/s

Thus, the tangential speed of the wheel is determined as 4.786 m/s.

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Answer:

The type of lens involved is a convex lens and the object is positioned at the center of the curvature of the convex lens.

According to the question, a real, inverted, and same-sized image of the object is formed.

A concave lens is a diverging lens and always forms virtual images. But convex lenses are converging lenses and are the only type of lens that produce real and inverted images of the corresponding objects.

When an object is placed at the center of the curvature of a convex lens, its corresponding image is formed on the opposite side of the convex lens. The image formed is real and inverted.

The distance of the image from the lens is equal to the distance of the object from the lens.

For a convex lens, the distance of the center of curvature from the lens is double the focal length of the lens.

That's why the convex lens and center of curvature are the correct answer to this question.

Explanation:

The lowest carrier frequency for which each component of the modulated wave S(t) is uniquely determined by m(t) is

"-w,w -1kHz,1kHz" are the frequency components of the detector output in this case.

This is further explained below.

Generally, the equation for the modulated wave containing frequency components  is  mathematically given as

[tex]Fc,Fc+w,Fc-W = > 1.5k,2.5k,0.5kHz[/tex]

[tex]-Fc,-Fc+W,-Fc-wW = > -1.5k,-0.5k,-2.5kHz[/tex]

"-w,w -1kHz,1kHz" are the frequency components of the detector output in this case.

b)

In conclusion,, then the modulated wave contains frequency components as

[tex]Fc,Fc+W,Fc-W = > 0.75k,1.75k,-0.25kHz[/tex]

[tex]-Fc,-Fc+w,-Fc-W = > -0.75k,0.25k,-1.75kHz[/tex]

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5.77 ×[tex]10^1^4[/tex] Hz is the green photon's frequency .

The distance between similar points (adjacent crests) in adjacent cycles of a waveform signal that is propagated in space is known as the wavelength. A wave's wavelength is often measured in meters (m), centimeters (cm), or millimeters (mm) (mm). The relationship between frequency and wavelength is inverse.

Wavelength of green light = 520 nm

f = c / λ

where, f = Frequency

            c = Speed of light = 3 × [tex]10^8[/tex] m/s

            λ = Wavelength of light

∴ f = c / λ

  f = [tex]\frac{3*10^8}{520 * 10^-^9}[/tex]

    = 5.77 ×[tex]10^1^4[/tex] Hz

Therefore,  5.77 ×[tex]10^1^4[/tex] Hz is the green photon's frequency .

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Acceleration is the rate of change of velocity. Usually, acceleration means the speed is transforming, but not always. When an object moves in a circular path at a steady speed, it is still accelerating, because the focus of its velocity is changing.

a = 4,552 m / s²,   b)  a = 2,588 m / s²

Newton's second law is

F = ma

a = F / m

in this case the force remains constant

indicate us

* for a mass m₁

a₁ = F/m₁

a₁ = 12, m/ s²

* for a mass m₂

a₂= 3.3 m / s²

acceleration  m = m₂-m₁

substitute

[tex]$\begin{aligned}&a=\frac{F}{m_{2}-m_{1}} \\&1 / a=\frac{m_{2}}{F}-\frac{m_{1}}{F}\end{aligned}$[/tex]

let's calculate

1/a=1/3.3 - 1/12  = 0.21969

 a = 4,552 m / s²

a= 4,552m/s²

Speed ​​estimates the rate of movement of an object, that is, the distance traveled per unit of time. Acceleration calculates the rate of change of velocity, that is, the change in velocity between two different moments

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The change in momentum in time interval, given the data will be F × Δt

Momentum is defined as the product of mass and velocity. It is expressed as

Momentum = mass × velocity

This is defined as the change in momentum of an object.

Impulse = change in momentum

But

Impulse = force × time

Therefore

Force × time = change in momentum

Force × time = change in momentum

F × Δt = change in momentum

Change in momentum = F × Δt

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323.5 N is the tension in the cable.

Given

Mass of crate(M) = 175.5 kg

Mass of boom(m) = 94.7 kg

The tension, T in the cable can be calculated by taking moments of force about the central point of marked X.

The Angle of the boom with the horizontal can be calculated by

tanθ = 5/10

θ = tan⁻¹(5/10) = 26.56°

Angle of the boom with horizontal is 26.56°

The angle of cable with horizontal can be calculated by

tan B = 4/10

B = tan⁻¹(4/10) = 21.80°

Angle of cable with horizontal is 21.80°

Taking moments of force about the point X

(Mcosθ + mcosθ) 0.5 = T(sin(θ +B)1

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

By calculating, we get

Tension(T) = 241.68/0.747

Tension(T) = 323.5 N

Hence, 323.5 N is the tension in the cable.

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The velocity with which the particle Q is fired is 15m/s upwards.

The vector quantity velocity (v), represented by the equation v = s/t, quantifies displacement (or change in position, s) over the change in time (t). Speed (or rate, r) is a scalar number, denoted by the equation r = d/t, that quantifies the distance traveled (d) over the change in time (t).

Assume the upward direction is positive and the downward direction is negative.

Velocity P, = 40m/s

Distance traveled by P,

Using the first equation of motion for particle P,

v = u + at

⇒ 0 = 40 + (-10)t

⇒ t = 4s

This is the time it takes for P to rise

Now, the maximum height(s) reached by the particle P is,

Using the second equation of motion,

s = ut + 1/2at²

⇒ s = 40×4 + 1/2 × (-10) × 4²

⇒ s = 80m

a) A particle P falls when Q is shot after 4 seconds from the initial time:

V² = U² + 2aH₁

⇒ H₁ = 15² - 0/ 2(-10)

⇒ H₁ = 11.25m

Particles P and Q meet at a distance from the ground (H₂).

Height, H₂ = s - H

⇒ H₂ = 80 - 11.25

         = 68.75m

Particles P and Q meet at a distance of 68.75m from the ground.

v = u + at₁

⇒ 15 = 0 + (-10) t₁

⇒ t₁ = 1.5s

It is equal to P's fall time and Q's rise time.

b) For particle Q

H₂ = u₂t₁ + 1/2at₁

⇒ 11.25 = u₂ × 1.5 + 1/2 (-10) × 1.5

⇒ u₂ = 15 m/s

Therefore,

The velocity with which the particle Q is fired is 15m/s upwards.

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(a) The electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

(b) The total flux is 1.24 Vm.

(c) The total amount of charge that is inside the box is  1.1 x 10⁻¹¹ C.

The area of the left face is calculated as follows;

A1 = 0.03 m x 0.02 m = 0.0006 m²

Ф1 = EA1

Ф1 = (400 V/m)( 0.0006 m²) = 0.24 Vm

Let the dimension of the right face = 5 cm by 2 cm

A2 = 0.05 m x 0.02 m = 0.001 m²

Ф2 = EA2

Ф2 = (1000 V/m)( 0.001 m²) = 1 Vm

Ф = Ф1 + Ф2

Ф = 0.24 Vm + 1 Vm = 1.24 Vm

Ф = Q/ε

Q = εФ

Q = (8.85 x 10⁻¹²)(1.24)

Q = 1.1 x 10⁻¹¹ C

Thus, the electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

The total flux is 1.24 Vm and the total amount of charge that is inside the box is  1.1 x 10⁻¹¹ C.

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The magnitude of the force on the left-hand pole is mathematically given as

f'=0.167N

Q=45 degrees

Generally, the equation for Force is  mathematically given as

F=mg/sinФ

Therefore

F(17.1*10^{-3})*9.8/sin 45

F=0.237N

Considering horizontal axis or plane

f'-fcos=0

Therefore

f'=0.237*cos45

f'=0.167N

In conclusion, the slope

tanФ=30/30

tanФ=1

Ф=45

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Answer: The speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth is 4188.11 m/s.

Explanation: To find the answer, we need to know about the equation of motion of a satellite around earth.

                 [tex]F_g=\frac{GMm}{r^2}[/tex]

                 [tex]F_c=\frac{mv^2}{r} \\[/tex]

                    [tex]\frac{GMm}{r^2} =\frac{mv^2}{r} \\thus,\\v=\sqrt{\frac{GM}{r} }[/tex]

                  [tex]r=3.57 R_E=3.57*6.37*10^3=22.74*10^3 km\\M=5.98*10^24kg\\G=6.67*10^{-11}Nm^2/kg^2[/tex]

                  [tex]v=\sqrt{\frac{6.67*10^{-11}*5.98*10^{24}}{22.74*10^6m} } =4188.11 m/s[/tex]

Thus, we can conclude that the speed of satellite will be 4188.11 m/s.

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In a circular orbit with a radius that is 3.57 times the mean radius of the Earth, a satellite moves at a speed of 132.43km/s.

In order to get the solution, we must understand the satellite's planetary motion equation.

                          [tex]F_g=\frac{GMm}{r^2}[/tex]

                         [tex]F_c=\frac{MV^2}{r}[/tex]

                          [tex]\frac{GMm}{r^2}=\frac{mV^2}{r}\\V=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10{-11}*5.98*10^{24}}{22.74*10^3} } \\V=132.43*10^3m/s[/tex]

As a result, we may estimate that the satellite will move at a speed of 132.43 km/s.

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Answer:

It will still hover until the spaceship "hits" or exerts a force on it.

Explanation:

Remember, if there is no net force, there is no acceleration or movement.

In this case, our ball is hovering in the spaceship, and in space, we can assume there is no [tex]F_g[/tex], and we can assume there is no [tex]F_N[/tex], nor no forces acting against it.

So, the ball would not move.

However, once the spaceship starts accelerating, the ball would still hover until the spaceship exerts a force on it.

This is because of the same thing as explained above, no forces acting on it, therefore, no acceleration.

Think about it this way.

Imagine you jumped up, then someone threw a ball at you. Now let's imagine you can't move until you hit the floor, meaning that in an ideal situation only  [tex]F_g[/tex] is acting on you. Now again, let's imagine time slows really down for you, but not the ball. Before the ball comes and hits you, you are "hovering" like a ball. But after the ball hits you, you move a little because the ball exerted a force on you.

If you did not understand what I meant above, just forget about it, and think about the fact that if there is a Net force (all the force values added up), then there is acceleration and movement.

A grandfather paradox is a situation where individual travels to the past and then introduces a change which affects or contradicts the present.

A paradox is a situation or statement which involves two contradictions.

A grandfather paradox is a situation which is defined by the ability of an individual to travel to a time in the past usually before the birth of their grandfather and then introduces a change which affects or contradicts the present. For example, killing the grandfather to prevent their birth.

In conclusion, a grandfather paradox is is an event which contradicts the present as a result of a change done to the past.

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(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.

(b) The net work done on the crate while it is on the rough surface is 23.7 J.

(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.

F(net) = F - μFf

F(net) = 280 - 0.351(92 x 9.8)

F(net) = -36.46 N

W = F(net) x L

W = -36.46 x 0.65

W = - 23.7 J

a = F(net)/m

a = -36.46/92

a = - 0.396 m/s²

v² = u² + 2as

v² = 0.845² + 2(-0.396)(0.65)

v² = 0.199

v = √0.199

v = 0.45 m/s

Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.

The net work done on the crate while it is on the rough surface is 23.7 J.

The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.

Learn more about work done here: https://brainly.com/question/8119756

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