Select the choice from the list below that most accurately completes the statements below. The first phrase will go in the first blank of the statement and the second phrase will go in the second blank.

A 1: greater than 2: because of Newton's Second Law
B 1: less than 2: because of Newton's Third Law
C 1: equal to 2: because of Newton's Third Law.
D 1: equal to 2: because of Newton's Second Law
E 1: less than 2: because of Newton's Second Law
F 1: greater than 2: because of Newton's Third Law.

You push a box horizontally along level ground but friction is causing the box to slow down. The magnitude of the friction force from the ground on the box is _________the magnitude of your force. You push a box horizontally along level ground at constant velocity. The magnitude of the friction force from the ground on the box is____________the magnitude of your force. You are standing on a hill where the ground is sloped. The magnitude of the normal force from the ground on you is___________the magnitude of the normal force from you on the ground. You are standing on a hill where the ground is sloped. The magnitude of the normal force from the ground on you is _____________the magnitude of your weight. You are facing your (very large and very strong) friend, Commander Worf. Worf gets very angry and pushes you to the ground. Worf remains stationary (a = 0). The magnitude of the force from Worf on you is __________the magnitude of the force from you on Worf. You are standing at a place where the ground is level. The magnitude of your weight is__________the magnitude of the normal force from the ground on you.

Answer:

Same

Explanation:

Mass is the quantity of matter in a certain object.

WHEREVER you take a 2kg object, the mass will remain 2kg. All that changes is the Weight ..Weight the force which the centre of a Planet uses to pull everything towards itself.

On earth, it is 9.81 whereas on the Moon it is 1.6

Here, we are required to determine the mass of a 2kg object on the Earth and on the moon.

The mass of an object is simply defined as the quantity of matter present in the object and this quantity remains constant.

The mass of an object is simply defined as the quantity of matter present in the object and this quantity remains constant.In essence, if the mass of an object is 2kg, this mass remains constant on planet earth and the moon.

PS: The weight of an object which is a measure of the acceleration due to gravity which differs with location and position is the constant which may be different about the object on the earth and the moon respectively.

The acceleration due to gravity on earth and the moon is 9.8m/s² and 1.6m/s² respectively.

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Answer:

a)     Eₓ = 0  ,   b)    c)    Ex_total = 3.96 10⁻⁶ N

Explanation:

For this problem we will look for the expression for the electric field of a ring of charge, at a point on its central axis.

From the symmetry of the ring, the resulting field is in the direction of the axis, we suppose that it corresponds to the x and y axis and the perpendicular direction but since the two sides of the annulus are offset.

       dE = k dq / r²

The component of this field in the direction to x is

       d Eₓ = dE cos θ

the distance of ring to the point

        r² = x² + a²

where a is the radius of each ring

the cosine is the adjacent leg between the hypotenuse

       Cos θ = x /r

we substitute

   d Ex = k / (x² + a²)  dq      x /√ (x² + a²)

we integrate

      Ex = ∫  k / (x² + a²)^{3/2}  x dq

      Ex = k x/ (x² + a²)^{1.5}   Q

the total field is the sum of the fields created by each ring, since they are on the same line (x-axis), you can perform an algebraic sum

     Eₓ_total = Eₓ₁ + Eₓ₂

a) at the origin x = 0

           in field is zero, since the numerator e makes zero

              Eₓ = 0

b) for point x = 40.0cm = 0.400 m

we substitute the values ​​in our equation

       Ex_total = k x [Q (x² + 0.1²)^{1.5} + 2Q / (x² + 0.2²2) ^3/2]

        Ex_total = k Q x [1 / (x² + 0.01)^1.5 + 2 / (x² + 0.04)^1.52]

let's calculate

       Ex_total = 9 10⁹ 30 10⁻⁹ 0.40 [1 / (0.4 2 + 0.01) 3/2 + 2 / (0.4 2 + 0.04) 3/2] 10⁻⁹

       Ex_total = 108 [14.2668 + 22.36] 10⁻⁹

       Ex_total = 3.96 10⁻⁶ N

Answer:

C requires the most work and A requires the least work

Answer:

Which scenario requires the most work?

✔ C

Which scenario requires the least work?

✔ A

Answer:

Each kilogram of the cannonball has a total energy of 52396.454 joules.

Explanation:

From Principle of Energy Conservation we understand that energy cannot be destroyed nor created, but transformed. In this case non-conservative forces can be neglected, so that total energy of the cannonball ([tex]E[/tex]) is the sum of gravitational potential ([tex]U_{g}[/tex]) and translational kinetic energies ([tex]K[/tex]), all measured in joules. That is:

[tex]E = U_{g}+K[/tex] (Eq. 1)

By applying definitions of gravitational potential and translational kinetic energies, we proceed to expand the expression:

[tex]E = m\cdot g \cdot y + \frac{1}{2} \cdot m \cdot v^{2}[/tex] (Eq. 2)

Where:

[tex]m[/tex] - Mass of the cannonball, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]y[/tex] - Height of the cannonball above ground level, measured in meters.

[tex]v[/tex] - Speed of the cannonball, measured in meters per second.

As we do not know the mass of the cannonball, we must calculated the unit total energy ([tex]e[/tex]), measured in joules per kilogram, whose formula is found by dividing (Eq. 1) by the mass of the cannonball. Then:

[tex]e = g\cdot y + \frac{1}{2}\cdot v^{2}[/tex]

If we know that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y =122\,m[/tex] and [tex]v = 320\,\frac{m}{s}[/tex], the unit total energy of the cannonball is:

[tex]e = \left(9.807\,\frac{m}{s^{2}}\right)\cdot (122\,m)+\frac{1}{2}\cdot \left(320\,\frac{m}{s} \right)^{2}[/tex]

[tex]e = 52396.454\,\frac{J}{kg}[/tex]

Each kilogram of the cannonball has a total energy of 52396.454 joules.

Answer:

1233 N

Explanation:

force=mass×accerelation

The initial angular acceleration is 5.78 rad/s^2.

Since

A gymnast on the uneven parallel bars is at rest, tipped at a 45∘ angle from the vertical. The distance from her hands to her feet is 1.8 m.

So,

[tex]Torque \tau = r \times F\\\\ \tau = l\alpha\\\\so, r \times F = l\alpha\\\\mg\ cos\ 45 \times r = (1/3)mR^2 \times \alpha\\\\9.8\times cos\ 45 \times 0.9 = (1/3)\times (1.8)^2 \times \alpha\\\\\alpha = 5.78\ rad/s^2[/tex]

hence, The initial angular acceleration is 5.78 rad/s^2.

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Answer:

13.72 m/s

Explanation:

Here, we are to determine the final velocity of the coconut. Applying first equation of free fall;

v = u + gt

where: v is the final velocity of the object, u is the initial velocity, g is the force of gravity and t is the time.

Since the coconut falls in the direction of gravity, then g = 9.8 m/[tex]s^{2}[/tex].

t = 1.4 s, u = 0 m/s

So that;

v = 0 + 9.8(1.4)

  = 13.72

The coconut hits Chief Boolie's toe with a velocity of 13.72 m/s

Answer:

A) 4.035 × 10^(9) J

B) 9.29 × 10^(9) J

Explanation:

We are given;

Capacitance of the original capacitor; C = 1.33 F

Potential difference given to the original capacitor; V = 77.9 kV = 77.9 × 10³ V

A) The formula for Potential energy (U) for the original capacitor is given as:

U = ½CV²

Plugging in the relevant values, we have;

U = ½ × 1.33 × (77.9 × 10³)²

U = 4.035 × 10^(9) J

B) We are told that the capacitor with dielectric constant of 427, was replaced with one possessing a dielectric constant of 983.

Thus;

U = ½ × 1.33 × (983/427) × (77.9 × 10³)²

U = 9.29 × 10^(9) J

Answer:

0.65 m

Explanation:

See attachment for calculation

Answer:

Explanation:

Key point: the vertical motion is free fall, the horizontal motion remains at uniform speed of 100 m/s (because there is no force in the horizontal direction if we neglect air resistance.)

The time of free fall can be calculated first:

h = 1/2 g t^2 => t = sqrt(2 h /g) = sqrt(2 * 300 m / 9.81 s) = 7.82 seconds.

In that time, the horizontal displacement will be

s = v_horizontal * t = 100 m/s * 7.82 s = 782 m.

So the box will land 782 m further than the point it was dropped from (and 300 m lower of course)

Explanation:

The density of a substance can be found by using the formula

[tex]density = \frac{mass}{volume} \\ [/tex]

From the question

mass = 249 g

volume = final volume of water - initial volume of water

volume = 19 - 15 = 4 mL

We have

[tex]density = \frac{249}{4} \\ [/tex]

We have the final answer as

Hope this helps you

I don't see any units of measurement or speed I don't see anything

Answer:

If Earth were to stop spinning on its axis, gradually the oceans would migrate towards the poles from the equator.  The atmosphere would still be in motion with the Earth's original 1100 mile per hour rotation speed at the equator. And this means, rocks, topsoil, trees, buildings,  and so on, would be swept away into the atmosphere.

Hope that helped <3

Answer:

B

Explanation:

Answer:

Soil will loosen due to vast networks of underground tunnels.

Answer:

Yellow

Explanation:

It is for sure not red

Blackbody radiation graph for Star A, which has a surface temperature of 5,700 degrees Celsius. At the point where the curve peaks, the wavelength is approximately 550 nanometers, in the middle of the visible light spectrum. The color will Star A be Yellow.

A waveform signal that is carried in space or down a wire has a wavelength, which is the separation between two identical places (adjacent crests) in the consecutive cycles. This length is typically defined in wireless systems in meters (m), centimeters (cm), or millimeters (mm).

A blackbody's spectrum is continuous (it emits some light at all wavelengths), and it peaks at a particular wavelength. For hotter objects, the peak of the blackbody curve shifts to shorter wavelengths in a spectrum. If you conceive of a blackbody in terms of visible light, the hotter it is, the bluer its peak emission wavelength will be. For instance, the temperature of the sun is about 5800 Kelvin. With this temperature, a blackbody's peak is roughly 500 nanometers away, or the wavelength of yellow.

The color will Star A be Yellow.

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It is true that the sample of plasma has more energy than the sample of gas. Option D is correct.

Gas is a sample of matter that adopts the shape of the container in which it is housed and develops a uniform density inside the container.

Liquid molecules are more energetic than solid molecules. Further heating will cause the molecules to move so quickly that they won't stick together at all. The gas molecules have the highest energy content.

Although oxygen is a gas that you breathe in, it may also exist in all four other states. The mass of the four oxygen samples is the same, yet they are all in distinct states.

Regarding the relative levels of thermal energy in the samples, statement B is accurate. Regarding the proportional quantities of thermal energy in the samples,

It is true that the sample of plasma has more energy than the sample of gas.

Hence, option D is correct.

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Answer:

24000 kg·m/s

Explanation:

Momentum is Mass x Velocity, so 1200 kg time 20 m/s =  24000 kg-ms/s

The momentum of the car is  24000 Kg•m/s

Momentum is defined as the product of mass and velocity. Mathematically, it can be expressed as:

Momentum = mass × velocity

With the above formula, we can obtain the momentum of the car as follow:

Momentum = mass × velocity

Momentum = 1200 × 20

Momentum of car = 24000 Kg•m/s

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Answer:

θ = 720º

Explanation:

Let's use the rotational kinematics relations for this exercise, remember that all angles must be given in radians.

Let's reduce the magnitudes to the SI system

         θ = 2 rev (2π rad / 1 rev) = 4π rad

we assume that the pulley has a radius r

linear and angular coordinates are related

         y = θ r

let's calculate

         y = 4π r

to convert the distance in angles to radians

        θ = y / r

        θ = 4π

let's change this angle to degrees

        π rad = 180º

         θ = 4π rad (180º /π rad)

         θ = 720º

Answer:

3 miles per hour

Explanation:

15 divided by 5 = 3

Answer:

The ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is 2.35 x 10⁻⁶

Explanation:

Given;

mass of the honeybee, m = 0.1 g

charge acquired by the honeybee, Q = 23pC = 23 x 10⁻¹² C

the electric field near the earth's surface, E = 100 N/C

The magnitude of the electric force on the bee is given by;

F = QE

F = (23 x 10⁻¹²)(100) = 23 x 10⁻¹⁰ N

The weight of the bee is given by;

W = mg

W = 0.1 x 10⁻³ x 9.8

W = 9.8 x 10⁻⁴ N

The the ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is given by;

[tex]\frac{F}{W} = \frac{23*10^{-10}}{9.8*10^{-4}} = 2.35 *10^{-6}[/tex]

Therefore, the ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is 2.35 x 10⁻⁶

Answer:

B. Ecosystem B, because its high species diversity could have resulted from increased competition among its members.

Explanation:

This is because, in the  ecosystem with varying level of biodiversity, Ecosystem B has medium level of species diversity found in them with High medium level of habitat diversity which causes increasing competitions among them.

Answer:

The answer is D

Explanation:

Because it is big and can fit the truck.

Answer:

Three different types of levers exist, depending on where the input force, fulcrum, and load are. A class 1 lever has the fulcrum between the input force and load. A class 2 lever has the load between the fulcrum and input force. A class 3 lever is a lever that has the input force in between the fulcrum and the load.

Explanation:

Answer:

A that the answer I think I'm not tryna do u bad thi

Explanation:

but try it

Answer:

A) q = 7.03 × 10^(-13) C

n ≈ 4388265 electrons

B) q = 1757.8 × 10^(-16) C

n ≈ 1097253 electrons

Explanation:

We are given;

Mass; m = 1.5 × 10^(−8)g = 1.5 × 10^(-11) kg

Speed; v = 50 m/s

Electric Field; E = 8 × 10⁴ N/C

Distance between plates; s = 2 cm = 0.02 m

A) Now, speed = distance/time

So, time(t) = distance/speed = 0.02/50 = 0.0004 s

From Newton's first law of motion, we know that;

d = ut + ½at²

But u is initial velocity and in this case it's zero.

But we are told that a drop is to be deflected by 0.30 mm. So, d = 0.3 × 10^(-3) m

Thus;

0.3 × 10^(-3) = 0 + ½a(0.0004)²

a = 3750 m/s²

Now, we know that force in motion normally can be expressed as;

F = ma

But in electric field, it's;

F = qE

Thus;

qE = ma

So, charge is; q = ma/E

Plugging in the relevant values;

q = (1.5 × 10^(−11) × 3750)/(8 × 10⁴)

q = 7.03 × 10^(-13) C

Now, number of electrons is given by the formula;

n = q/e

Where e is charge on electron with a value of 1.602 × 10^(-19) C/electron

So; n = (7.03 × 10^(-13))/(1.602 × 10^(-19))

n ≈ 4388265 electrons

B) We are told speed is now 25 m/s.

Thus;

time(t) = distance/speed = 0.02/25 = 0.0008 s

From Newton's first law of motion, we know that;

d = ut + ½at²

But u is initial velocity and in this case it's zero.

d remains 0.3 × 10^(-3) m

Thus;

0.3 × 10^(-3) = 0 + ½a(0.0008)²

a = 937.5 m/s²

Now, we know that force in motion normally can be expressed as;

F = ma

But in electric field, it's;

F = qE

Thus;

qE = ma

So, charge is; q = ma/E

Plugging in the relevant values;

q = (1.5 × 10^(−11) × 937.5)/(8 × 10⁴)

q = 1757.8 × 10^(-16) C

Now, number of electrons is given by the formula;

n = q/e

Where e is charge on electron with a value of 1.602 × 10^(-19) C/electron

So; n = (1757.8 × 10^(-16) )/(1.602 × 10^(-19))

n ≈ 1097253 electrons

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

[tex]E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}[/tex]

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

[tex]r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm[/tex]

The magnitude of the electric field is now given as;

[tex]E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m[/tex]

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Answer:

F

Explanation: Its common sence

Answer:

It’s false

Explanation:

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