An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's altitude is 532 m, the pilot releases a package.
(a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
(b) Relative to the ground, determine the angle of the velocity vector of the package just before impact. (a) Number Units (b) Number Units

Answer:

F

Explanation: Its common sence

Answer:

It’s false

Explanation:

2020 (WORST YEAR EVER) on Edge ;-)

Answer:

B

Explanation:

Answer:

Soil will loosen due to vast networks of underground tunnels.

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

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

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:

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)

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:

0.65 m

Explanation:

See attachment for calculation

Answer:

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

Explanation:

but try it

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:

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:

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:

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:

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

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

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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Answer:

a

[tex]P =0.1094 \ m[/tex] past the goal post

Yes the Flyers won the match

Explanation:

Generally the distance covered by the pluck during the last 1.25 seconds is mathematically represented as

     [tex]D = ut + \frac{1}{2} * a * t^2[/tex]

=>  [tex]D =30  * 1.25 + \frac{1}{2} * (-0.5) * (1.25)^2[/tex]

=>   [tex]D =37.1094 \ m  [/tex]

Generally the position of the puck when the game clock reaches zero is mathematically represented as

     [tex]P = 37.1094 -37[/tex]

       [tex]P =0.1094 \ m[/tex] past the goal post

Given that the Bruins where one point down and Flyers scored another goal  it means that the flyer are now two point up hence they won the match

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:

It is gaining potential energy

Answer: it could also be gravitational potential energy

This is because as it rises in the air it will gain more energy to fall back down

Answer:

27%

Explanation:

15.999 divided by 58.32 = .27433128

Move the decimal place over 2 places.

27%

Make use of percentages. Given that the molecular mass of magnesium hydroxide (MOH.) is 58.32 amu and the atomic mass of an oxygen atom is 15.999 amu, oxygen accounts for 27% of this compound.

Chemical substances known as molecular compounds assume the shape of distinct molecules. Examples include common compounds like carbon dioxide (CO2) and water (H2O) (Figure 3.1.1 ).

These substances differ significantly from ionic substances like sodium chloride (NaCl). When metal atoms give up one or more of their electrons to non-metal atoms, ionic compounds are created. The cations and anions that result

Therefore,

58.32 / 15.999 (27.3%) =.27433128

Add two places to the decimal place.

27%

The molar mass, which is stated in g/mol, is defined as the mass of a specific material divided by the amount of a substance. In light of this, the molar mass is average. Amu, short for atomic mass units, is the unit used to quantify molecular mass.

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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

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:

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:

The answer is D

Explanation:

Because it is big and can fit the truck.

Answer:

1233 N

Explanation:

force=mass×accerelation

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:

3 miles per hour

Explanation:

15 divided by 5 = 3