A hummingbird gains a significant electric charge while flying and feeding. This has consequences: When a charged bird approaches a flower (Figure 1), the stamens of the flower bend toward the bird, even though the stamans are uncharged. Part A Explain how this happens Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Reset Help toward The charge of the hummingbird ____ the stamen polarizes The charged opposite to the bird's charge particles of the stamen move ___ the bird, and like charged particles move ____the bird Therefore, the net electric force on the stamen is directed ____the hummingbird Submit Request Answer

Therefore, the magnitude of the net force on the mower is 134.9 N.

Force is an influence that can cause an object to undergo a change in motion or acceleration. It is a vector quantity, which means it has both magnitude (size or strength) and direction. Force can be measured in units such as newtons (N) or pounds (lbs). It can result from a variety of sources, including gravity, electromagnetic fields, and contact between objects.

Here,

First, we need to resolve the pushing force into horizontal and vertical components.

F_horizontal = 264.8 N * cos(55.0°) = 152.9 N

F_vertical = 264.8 N * sin(55.0°) = 210.7 N

Since the net external force on the mower is horizontal to the right and the force of friction opposes the motion, we can write:

Net force = F_horizontal - f

where f is the force of friction.

Substituting the given values, we get:

Net force = 152.9 N - 18.0 N = 134.9 N

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The diver's highest point above the board is 0.60 meter, her feet are in the air for 0.684 second, and her velocity when her feet hit the water is 3.43 m/s downward.

To find the diver's highest point, we can use the fact that at the highest point, her velocity will be zero. We can use the equation:

v² = u² + 2as

where u is the initial velocity, v is the final velocity (which is zero at the highest point), a is the acceleration due to gravity (g = 9.81 m/s²), and s is the distance traveled. Solving for s, we get:

s = (v² - u²) / 2a

s = (0 - (3.35 m/s)²) / (2 × -9.81 m/s²)

s = 0.60 m

So the diver's highest point is 0.60 m above the diving board.

To find how long the diver's feet are in the air, we can use the fact that the time of flight (i.e. the time spent in the air) is twice the time it takes to reach the highest point. We can use the equation:

v = u + at

where t is the time, and v and u are the final and initial velocities, respectively. Solving for t, we get:

t = (v - u) / a

At the highest point, the velocity is zero, we can use u = 3.35 m/s and v = 0:

t = (0 - 3.35 m/s) / -9.81 m/s² = 0.342 second

So the time of flight is twice this value, or 0.342 × 2 = 0.684 second.

To find the diver's velocity when her feet hit the water, we can use the same equation as in part (b), but with u = 0 and v as the velocity we want to find. We can also use the fact that the distance traveled from the highest point to the water is the same as the distance traveled from the takeoff point to the highest point (0.60 m). So we have:

s = (v² - u²) / 2a

0.60 m = (v² - 0) / (2 × -9.81 m/s²)

v = -√(2 × -9.81 m/s² × 0.60 m) = -3.43 m/s

The negative sign indicates that the velocity is downward, which makes sense since the diver is falling feet first into the water. So the diver's velocity when her feet hit the water is 3.43 m/s downward.

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a.) The acceleration of the cart can be calculated using the formula g = 9.8 m/s², where g is the acceleration due to gravity. a = F/m = m × g = 4.40 kg × 9.8 m/s² = 43.12 m/s²

b.) The time it takes the cart to travel from the starting point to the ending point can be calculated using the kinematic equation:

d = v_0t + 1/2at²,

366 m - 53 m = 313 m

313 = 0t + 1/2(43.12)t² = 1/2(43.12)t²

313 = 21.56t²,  t² = 313 / 21.56

t² = 14.52, t = √14.52 = 3.8 sec.

c.) The acceleration of the cylinder can be calculated using the formula a = g - (v²)/Rg

d.) The time it takes the cylinder to travel from the top of the ramp to the bottom can be calculated using the kinematic equation:

d = v_0t + 1/2at²,

What is acceleration?

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The required frequency of the sound when distance between the two speakers is given is calculated to be 21.25 kHz.

Sound waves are travelling waves and they can be modelled as

A(r, t) = A₀(r)sin(kr - ωt + Ф₀)

Where,

A₀ is the initial amplitude of the wave

r is the distance

ωt is the frequency

Ф₀ is the initial phase shift

First we need to find out the phase difference (ΔФ) between two waves at different distances.

ΔФ = 2πΔr/λ + ΔФ₀

When you stand centred between the two waves you hear maximum intensity of sound so the the two waves must be in phase

ΔФ = 2πΔr/λ + 0

λ = 2πΔr/ΔФ

The distance when listening in front of the speakers is given by

Δr = r₂ - r₁

r₁ = 6.0 mm = 0.006 m

r₂ = √(0.013²+0.006²) = √(0.000169 + 0.000036) = 0.014 m

Δr = r₂ - r₁ = 0.014 - 0.006 = 0.008 m

λ = 2π × 0.008/ΔФ

The phase difference ΔФ = π

λ = 2π × 0.008/π

λ = 0.016 m

As we know the relation between frequency and wavelength is given by

f = c/λ

Where,

c = 340 m/s is the speed of light

f = 340/0.016

f = 21250 Hz

f = 21.25 kHz

Thus, the frequency of the sound is calculated to be 21.25 kHz.

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This assertion is true: There is no electric field inside a hollow charged conductor.

Any sort of charge causes an electric field to be associated to a location in space. The strength and direction of the electric field are expressed by the value of E, also referred to as the electric field strength, electric field intensity, or simply the electric field.

The electric field is the region of space around an electrically charged particle or object where the charge body feels force. Examples: -Electric fields are created by charges and their configurations, such as capacitors and battery cells.

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The pressure of the gas is 0.018 atm if the gas obeys the ideal gas law where a container with volume 1.69 l is initially evacuated.

What is the reason for this?

We know that;

√ 3RT/M

For N2;

vrms =  192 m/s

R = 8.314 J/K.mol

T = ?

M = 28 g/mol or 0.028 Kg/mol

So;

T =  vrms² M/ 3R

Temperature, T =  192² * 0.028 / 3* 8.314

T = 41.3K

From PV = nRT

n = 0.246 g/28 g/mol which can lead to 0.0087 moles

P = nRT/V

P =  0.0087 moles × 0.082 atmLK-1mol-1  ×  41.3K/1.61 L

P = 0.018 atm

Therefore, the pressure of gas is 0.018 atm.

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a container with volume 1.69 l is initially evacuated. then it is filled with 0.236 g of n2, the pressure is P = 1652 Pa

From the inquiry we are informed that

   The volume of the compartment is V = 1.83 L = [tex]183 * 10^{-3} m^{3}[/tex]

    The mass of [tex]N_{2}[/tex] is [tex]m_{n} = 0.246 g = 0.246 * 10^{-3} kg[/tex]

   The root-mean-square volume is v = 192 m/s

The root - mean square velocity is mathematically addressed as

[tex]v = \sqrt{3RT}/Mn[/tex]

Presently the ideal gas regulation is mathematically addressed as

 PV = nRT

=> RT = PV/n

Where n is the quantity of moles which is mathematically addressed as

n= mn/M

Where M is the molar mass of [tex]N_{2}[/tex]

So

             RT = [tex]PV /M_{n}[/tex]

=>         [tex]v = \sqrt{3^{ p* v*Mn/Mn } } /Mn[/tex]

=>         [tex]V = \sqrt{3 * P * V} / m_{n}[/tex]

=>         [tex]P = V^{2} * m_{n[/tex] / 3 * V

substituting values

   =>     [tex]P = (192)^{2} * 0.246 * 10^{-3} / 3 * 1.83 * 10^{-3}[/tex]

  =>      P  = 1652  Pa

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Answer:

Explanation:

The work done (W) by a force (F) over a distance (d) can be calculated using the equation:

W = F * d

In this case, the work done by the tractor is 11,100 N-m and the force applied is 257.9 N. To find the distance the tractor pulled the plow, we need to rearrange this equation to solve for d:

d = W / F

Plugging in the values:

d = 11,100 N-m / 257.9 N = 43.01 meters

So, the tractor pulled the plow a distance of 43.01 meters.

The length at which the restoring force is 2F is given by the following expression:

[tex]x = (-2F + F / (13 cm - 10 cm) * 10 cm) / (F / (13 cm - 10 cm))[/tex]

The restoring force of a spring is proportional to its displacement from its unstretched length, according to Hooke's law:

F = -kx

where F is the restoring force, k is the spring constant, and x is the displacement from the unstretched length.

If we know the restoring force F when the spring is stretched to a length of 13 cm, we can find the spring constant k:

[tex]F = -k * (13 cm - 10 cm)\\k = F / (13 cm - 10 cm)[/tex]

Next, we can find the displacement that gives a restoring force of 3F:

[tex]3F = -k * (x - 10 cm)\\3F = -k * x + k * 10 cm\\-3F = k * x - k * 10 cm\\-3F + k * 10 cm = k * x\\x = (-3F + k * 10 cm) / k[/tex]

Substituting the value of k, we get:

[tex]x = (-3F + F / (13 cm - 10 cm) * 10 cm) / (F / (13 cm - 10 cm))[/tex]

So, the length of the spring for which its restoring force is 3F is given by the above expression.

b. To find the length at which the restoring force is 2F, we can use the same method as above and replace 3F with 2F:

[tex]2F = -k * (x - 10 cm)x = (-2F + k * 10 cm) / k[/tex]

Substituting the value of k, we get:

[tex]x = (-2F + F / (13 cm - 10 cm) * 10 cm) / (F / (13 cm - 10 cm))[/tex]

So, the length at which the restoring force is 2F is given by the above expression.

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To lessen electrostatic attraction between them, the extra charges will be dispersed equally on the enormous conducting plates as widely as feasible.

a. When the battery is connected to the wires, the negatively charged electrons on the wire attached to the negative terminal of the battery will reject one another, driving them to spread as much as possible down the wire.

Positively charged holes (holes left over when electrons are missing) behave similarly to negatively charged holes in that they repel one another and spread out as much as possible down a wire connected to a positive terminal of a battery.

b. In order to prevent the battery from pushing more charge onto the plates, the extra charges on the plates will generate an electric field between them.

The back EMF, sometimes referred to as the counter EMF, grows as the charge on the plates rises. The back EMF will eventually equal the battery's EMF, at which time the battery will be unable to push any more charge onto the plates.

c. Yes, as the battery charges the plates, there will be a current going through the cables. The current is the passage of electrons from the battery's negative terminal onto the negative plate through the wire.

d. Because it shows how well the two plates can hold an electrical charge, much like a capacitor does, the amount C is known as capacitance.

Therefore, For a given voltage differential between the plates, the capacitance increases the amount of charge that may be stored.

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The transition from n=2 to n=1 emits the longest wavelength.

The energy of a photon is directly proportional to its frequency and inversely proportional to its wavelength. A transition of an electron from an excited state to the ground state (n=1) results in the emission of a photon with the longest wavelength in the Lyman series.

The Lyman series corresponds to electron transitions to the n=1 level and includes ultraviolet wavelengths. The transition with the longest wavelength in the Lyman series is from n=2 to n=1, and it corresponds to the emission of a photon with a wavelength of 121.6 nm.

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--The complete question is, When an electron in excited energy level drops to a lower energy level a photon is emitted, if the electron is dropping to n=1 which transition will emit the longest wavelength?--

Suppose an object experiences a constant force of magnitude F. In that case, it accelerates according to Newton's second law of motion, which states that F = ma, where m is the mass of the object and a is its acceleration.

Given that the object experiences a constant force that produces an acceleration of 8 m/s^2, we can write:

F = ma = m × 8

What will the acceleration of this object be if the force is halved?

If the force is halved, the new force is F/2. Using Newton's second law, we can find the new acceleration as follows:

F/2 = ma_new

ma_new = (F/2) / m

ma_new = F/2m

ma_new = (m × 8) / 2m

ma_new = 4 m/s^2

Therefore, the object's acceleration will be 4 m/s^2 if the force is halved.

What will the acceleration of this object be if the mass is halved?

If the mass is halved, the new mass is m/2. Using Newton's second law, we can find the new acceleration as follows:

F = (m/2) × a_new

a_new = F / (m/2)

a_new = 2F/m

a_new = 2(m × 8)/m

a_new = 16 m/s^2

Therefore, the acceleration of the object will be 16 m/s^2 if the mass is halved.

If the force is halved and the mass is halved, the new force is F/2 and the new mass is m/2. Using Newton's second law, we can find the new acceleration as follows:

F/2 = (m/2) × a_new

a_new = (F/2) / (m/2)

a_new = F/m

a_new = (m × 8) / m

a_new = 8 m/s^2

Therefore, the object's acceleration will be 8 m/s^2 if the force is halved and the mass is halved.

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In physics and engineering, stress refers to the internal force per unit area that a material experiences when subjected to an external force. When a force is applied to a material, the material undergoes deformation, or a change in shape. The stress describes the intensity of the internal forces that arise due to the deformation of the material.

There are different types of stress, such as tensile stress (stretching), compressive stress (squeezing), shear stress (sliding), and torsional stress (twisting). The critical stress for a material depends on the type and magnitude of the force applied, as well as the material properties such as its ultimate strength and stiffness. When the stress exceeds the critical value, the material can undergo plastic deformation or fracture.

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The change in velocity in y direction, if the runner accelerates at 0.350 m/s² at 52° angle is 2.83 m/s.

Acceleration of an object is the rate of change in velocity. It is a vector quantity having both direction and magnitude.

a = Δv/t

The initial velocity u = 2.88 m/s

a = 0.350 m/s²

the time t = u/a = 2.88 /0.350 = 8.2 s.

vy = uy + ay t

uy = u sin θ = 2.88 sin 52 =2.84 m/s

ay = 0.350 sin 52 = 0.345 m/s²

then vy = 2.84 m/s + (0.345 m/s²) 8.22 s = 5.67 m/s

then change in velocity in the y-direction is calculated as:

Δy = vy - uy

    = 5.67 m/s  - 2.84 m/s = 2.83 m/s

Therefore, the change in velocity in y-direction Δy for the runner is 2.83 m/s.

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The normal component of acceleration may be negative in the motion of curved particles.

The normal component of acceleration may be negative in the motion of curved particles. The component of acceleration that is normal to the motion path's tangent and pointed in the direction of the path's centre of curvature is known as the normal component of acceleration. The normal component of acceleration will be negative if the particle is going along a curved path and its speed is decreasing since it is directed in the opposite direction from the direction the particle is moving. Depending on the particle's motion and the direction of the tangent to the path at any given location, the tangential component of velocity and acceleration may potentially be negative.

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A force that is inversely proportional to the distance of the object from equilibrium is consistent with simple harmonic motion.

Mathematically, the restoring force can be expressed as [tex]F = -kx,[/tex]

where F is the restoring force, x is the displacement from equilibrium, and k is the spring constant, which determines the strength of the restoring force. Negative sign indicates that the restoring force is opposite in direction to the displacement. If we rearrange this equation, we get x = -(1/k) * F, which shows that the displacement is inversely proportional to the force. Therefore, a force that is inversely proportional to the distance of the object from equilibrium is consistent with simple harmonic motion.

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A) The speed of the river is:  3.2 km/h

B) The canoe speed in still water for the same paddling effort is 6.4 km/h.

Speed is a term used to describe how rapidly something moves or a process takes place. It is a scalar quantity that represents the speed at which an item moves or changes position with respect to time. By dividing the distance traveled by the time needed to cover that distance, speed may be computed. Units like meters per second, miles per hour, or kilometers per hour are frequently used to measure it. Numerous industries, such as science, sports, transportation, and telecommunications, depending on speed. It is essential in determining the effectiveness and functionality of many systems, including those in cars, machinery, and computer networks.

distance = rate x time

to set up two equations, one for each leg of the journey:

Upstream leg: distance = (c - v) x 2.0 km + 1.0 km

Downstream leg: distance = (c + v) x t

where t is the time it takes to catch up with the bottle.

We know that the distance traveled on both legs is the same because the students end up where they started. So we can set the two equations equal to each other and solve for v:

(c - v) x 2.0 km + 1.0 km = (c + v) x t

Expanding and simplifying:

2c - 2v + 1 = ct + vt

2c + 1 = (c + v)t

t = (2c + 1)/(c + v)

Now we need to find t. We know that the distance the bottle traveled downstream is 4.7 km, and the time it took to get there is t:

4.7 km = v x t

t = 4.7/v

Setting the two expressions for t equal to each other and solving for v:

(2c + 1)/(c + v) = 4.7/v

2cv + v = 4.7(2c + 1)

2cv + v = 9.4c + 4.7

v(2c + 1) = 9.4c + 4.7

v = (9.4c + 4.7)/(2c + 1)

This is our formula for the speed of the river in terms of the speed of the canoe in still water. Now we can solve for c by setting the canoe speed in still water equal to the relative speed upstream:

c - v = (1.0 km)/(2.0 h) = 0.5 km/h

c - (9.4c + 4.7)/(2c + 1) = 0.5 km/h

Simplifying and solving for c:

c = 5.7 km/h

So the speed of the river is:

v = (9.4c + 4.7)/(2c + 1) = 3.2 km/h

To find the canoe speed in still water for the same paddling effort, we can use the formula:

paddling effort = 2v = 2(c - v)

Setting v = 3.2 km/h and solving for c:

2(3.2) = 2(c - 3.2)

6.4 = 2c - 6.4

2c = 12.8

c = 6.4 km/h

So the canoe speed in still water for the same paddling effort is 6.4 km/h.

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The velocity distribution in the fluid can be found using the Navier-Stokes equation, which states that the net force on a fluid element is equal to its mass times its acceleration.

(a) Velocity distribution and torques on two coaxial cylinders:

The fluid is incompressible, so the continuity equation can be used to relate the fluid velocities at different radii. For the coaxial cylinders, the velocity distribution can be found by assuming a linear velocity profile between the two surfaces, where the velocity at the inner surface is ai * kr and the velocity at the outer surface is a * r. Therefore, the velocity profile is given by:

v(r) = (a - ai) / (r - kr) * (r - kr) + ai * kr

The torque required to maintain the motion of the cylinders can be found using the formula:

T = I * α

where T is the torque, I is the moment of inertia, and α is the angular acceleration. For each cylinder, the moment of inertia is given by:

I = (1/2) * m * R²

where m is the mass of the cylinder and R is its radius. The angular acceleration is related to the angular velocity by:

alpha = (a - ai) / (r - kr)

Therefore, the torque on the inner cylinder is:

Ti = (1/2) * m * kr² * (a - ai) / (r - kr)

and the torque on the outer cylinder is:

To = (1/2) * m * r² * (a - ai) / (r - kr)

(b) Velocity distribution and torques on two concentric spheres:

The velocity distribution and torques on two concentric spheres can be found in a similar way to the coaxial cylinders. Assuming a linear velocity profile between the two spheres, where the velocity at the inner sphere is ai * kr and the velocity at the outer sphere is a * r, the velocity profile is given by:

v(r) = (a - ai) / (r - kr) * (r - kr) + ai * kr

The torque required to maintain the motion of the spheres can be found using the same formula as for the cylinders, with the moment of inertia for each sphere given by:

I = (2/5) * m * R²

where m is the mass of the sphere and R is its radius. Therefore, the torque on the inner sphere is:

Ti = (2/5) * m * kr² * (a - ai) / (r - kr)

and the torque on the outer sphere is:

To = (2/5) * m * r² * (a - ai) / (r - kr)

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The required electric potential energy when charge and diameter of the copper plate are given is calculated to be -432 J/c.

The diameter of the copper plate is given as 15 cm = 0.15 m.

Radius r = diameter d/2 = 0.15/2 = 0.075 c

The charge over the copper plate is given as -3.6 nc = -3.6 × 10⁻⁹ c.

The expression of electric charged plate distributed over the surface is given by,

V = k (q/r)  

where,

V is the electric potential energy

q is the charge over the plate

r is the radius of the plate

k is coulomb's constant (9 × 10⁹ N)

Putting the values into the equation above gives us,

V = k (q/r) = [9 × 10⁹× (-3.6 × 10⁻⁹)]/0.075 = -432 J/c

Thus, the electric potential energy is calculated to be -432 J/c.

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The formulated sentences for each will be:

A: If length is 5cm, width 2cm, height 4cm, volume:  40 cubic centimeters.

B: If the length is 3cm, width 4cm, height 2cm, then Volume is: 24 cubic centimeters.

C:  If the length is 6cm, width 2cm, height 3cm, the  volume is 36 cubic centimeters.

D:  If the length is 3cm, width 2cm, height 4cm, the  volume 24 cubic centimeters.

To calculate the volume of a rectangular prism using multiplication, you need to multiply the length, width, and height of the prism. The multiplication sentence for this is:

Volume = length (m) x width (m) x height (m)

For a. The multiplication sentence for calculating the volume of rectangular prism C is:

b. The multiplication sentence for calculating the volume of rectangular prism B is:

c. The multiplication sentence for calculating the volume of rectangular prism C is:

Lastly, for d. The multiplication sentence for calculating the volume of rectangular prism D is:

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The resultant velocity of two significant figures is 192 km/h.

To calculate the resultant of the pair of velocities, we need to use the Pythagorean theorem.

The Pythagorean theorem states that for any right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

In this case, the hypotenuse is the resultant velocity, and the other two sides are the given velocities.

So, the resultant velocity is given by:

Resultant velocity = [tex]\sqrt{((120 km/h)^2 + (72 km/h)^2)[/tex]

Resultant velocity = [tex]\sqrt{(14400 km^2/h^2 + 5184 km^2/h^2)[/tex]

Resultant velocity = [tex]\sqrt{(19584 km^2/h^2)[/tex]

Resultant velocity = 140.01 km/h

To two significant figures, the resultant velocity is 140 km/h.

If both of the velocities are directed north, then the resultant velocity is simply the sum of the two velocities:

Resultant velocity = 120 km/h + 72 km/h

=> 192 km/h
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The rock's velocity as it hits the bottom of the hole is approximately 27.5 m/s and the rock is in the air for approximately 8.98 seconds.

a) The final velocity of the rock as it hits the bottom of the hole can be found using the kinematic equation vf^2 = vi^2 + 2ad, where vf is the final velocity, vi is the initial velocity, a is the acceleration due to gravity (-9.8 m/s^2), and d is the distance the rock falls (14 m). We know that vi = 22 m/s and d = -14 m (since the rock falls downward), so:

vf^2 = (22 m/s)^2 + 2(-9.8 m/s^2)(-14 m)

vf^2 = 484 + 274.4

vf^2 = 758.4

vf = sqrt(758.4)

vf ≈ 27.5 m/s

Therefore, the rock's velocity as it hits the bottom of the hole is approximately 27.5 m/s.

b) The time the rock is in the air can be found using the kinematic equation

d = vit + (1/2)at^2

where d is the distance the rock travels (which is equal to the height it was thrown from), vi is the initial velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time the rock is in the air. We know that d = vi*t + (1/2)at^2 and that d = 0 (since the rock returns to its starting height), so:

0 = (22 m/s)t + (1/2)(-9.8 m/s^2)t^2

0 = 22t - 4.9t^2

t(4.9t - 22) = 0

Therefore, t = 0 s or t = 4.49 s (rounded to three significant figures). Since the rock was released and then caught, we are interested in the time it takes for the rock to go up and then come back down, so the total time the rock is in the air is twice the time it takes for it to go up:

t_total = 2*t

t_total ≈ 8.98 s

Therefore, the rock is in the air for approximately 8.98 seconds.

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Displacement is expressed magnitude and a direction in m, km, ft etc. Among the given options, the one indicting a displacement is 40 km southwest.

Displacement is a physical quantity measuring how far an object is travelling and in which direction from the initial position. The displacement is a vector quantity having both magnitude and direction.

Displacement can be expressed in different units such as m, km, ft etc. Here, 32 ft/s²  and 9.8 m/s² are acceleration. 120 m/s is velocity. Then, 186000 ml is expressing volume of a substance.

Therefore, among the given options, 40 km southwest is indicating the displacement of an object with its magnitude and direction. 40 Km is magnitude and southwest is the direction.

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Protons are located in the nucleus of an atom with a positive charge, and they have a mass of approximately 1 atomic mass unit (amu) and every atom is made up of the electron, proton, and neutron.

A proton's mass is about 1,836 times greater than that of an electron, and these protons are affected by the strong nuclear force and they can be changed into neutrons through a process called beta decay.

Hence, protons are located in the nucleus of an atom with a positive charge, and they have a mass of approximately 1 atomic mass unit (amu) and every atom is made up of the electron, proton, and neutron.

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The question is incomplete, the complete and correct question is below,

Describe protons Location: Charge: Mass: ?

The only force operating on the ball once it has left its racquet is gravity, which the earth exerts on everything in its immediate vicinity. The ball is accelerated vertically by gravity at a velocity of around 9.8 m/s. ^2,

Hit into their deuce back for the opening point while standing on the right side of the center line; for the second part, serve from of the left side toward what is identified as the advantages service box. Until the game is won, the same player serves; after that, the other player takes over as the server.

In the middle ages, tennis score were shown on two watch faces that ranged from 0 to 60. The pointer changed position on each score, going from Zero to 15, 30, 45, and 60 for a victory.

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i) The period of oscillation of the vertical spring-mass system is 0.643 seconds.

ii) There are two quiet spots along the line segment connecting the speakers.

Over time, oscillation is a form of repeating motion. It describes how an object or system moves back and forth around a fixed point, usually an equilibrium center. Depending on the characteristics of the system, the oscillation may be periodic or non-periodic, and its frequency and amplitude may change. A pendulum's motion, the sound waves produced by a guitar string, and the alternating current in electrical circuits are all examples of oscillation. Oscillation is the repeating or periodic change of a quantity around a central value or between two or more states, often in time. An alternating current and a swinging pendulum are two common examples of oscillation. Numerous physical and biological systems depend on oscillations in one way or another.

(i) The period of oscillation of a horizontal spring-mass system is given by:

T = 2π√(m/k)

where m is the mass of the object attached to the spring, and k is the spring constant.

Using the values given in the problem, we can calculate the period T:

T = 2π√(m/k) = 2π√(1.0 kg / 78 N/m) = 0.571 s

Now, when the same spring-mass system oscillates vertically, the period of oscillation is given by:

T = 2π√(m/k_eff)

where k_eff is the effective spring constant, which takes into account the weight of the mass and the spring.

k_eff = k - mg

where g is the acceleration due to gravity. Substituting the values given in the problem, we get:

k_eff = 78 N/m - (1.0 kg) × (10.0 m/s^2) = 68 N/m

Using the above value of k_eff, we can calculate the period of oscillation T:

T = 2π√(m/k_eff) = 2π√(1.0 kg / 68 N/m) = 0.643 s

(ii) The distance between the two speakers is 7.0 m, which means that the halfway point (where the sound waves from the two speakers would be perfectly out of phase) is located at a distance of 3.5 m from each speaker.

Let's start by considering the point directly in front of one of the speakers. At this point, the distance traveled by the sound wave from the first speaker is simply the distance from that speaker to the point or 3.5 m. The distance traveled by the wave from the second speaker is the distance from that speaker to the point, plus the extra distance of 1.2 m due to the phase difference. Using the Pythagorean theorem, we can calculate this distance as:

sqrt((7.0/2)^2 + (1.2)^2) ≈ 3.65 m

This means that the first quiet spot is located 3.65 m in front of the second speaker.

Similarly, we can find the second quiet spot by considering the point directly in front of the second speaker. At this point, the distance traveled by the wave from the second speaker is simply 3.5 m, while the distance traveled by the wave from the first speaker is 7.0 m - the distance from the point to the second speaker, minus the extra distance of 1.2 m. Using the Pythagorean theorem again, we get:

sqrt((7.0/2)^2 + (1.2)^2) ≈ 3.65 m

This means that the second quiet spot is also located 3.65 m in front of the second speaker, but on the opposite side of the line connecting the speakers.

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Answer:

Below

Explanation:

Eforce =  C q1 q2 / r^2     C = coulomb's constant

            = (9 x 10^9)  (2.6x 10^-6)(3.8x 10^-6) / (.075)^2 = 15808 x 10^-3

            = 15.8 N     (attractive force because the forces are opposite charge)

Escherichia coli use long, whip-like structures called flagella to propel themselves.

Escherichia coli, also known as E. coli, is a Gram-negative, facultative anaerobic, rod-shaped, coliform bacterium of the genus Escherichia that is commonly found in the lower intestine of warm-blooded organisms

Given is to find what two structures help E.coli move.

Escherichia coli use long, whip-like structures called flagella to propel themselves. Motors in the cell's wall spin the flagella into bundles that rotate counter-clockwise, creating a twist that causes the bacterium to rotate clockwise, or towards the right when viewed from above.

Therefore, Escherichia coli use long, whip-like structures called flagella to propel themselves.

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Neglecting air resistance the ball will be 2.3 s in flight and will fall at a distance of 69 m with velocity 22.2 m/s

a) To calculate the time the ball is in flight, we can use the equation:

Time = 2Velocitysin(angle) / acceleration due to gravity

Where velocity = 30 m/s, angle = 40° and acceleration due to gravity = 9.8 m/s²

Therefore, time = 230 m/s sin(40°) / 9.8 m/s² = 2.3 s

b) To calculate the distance from the base of the cliff, we can use the equation:

Distance = Velocity × Time

Where velocity = 30 m/s and time = 2.3 s

Therefore, distance = 30 m/s × 2.3 s = 69 m

c) The ball's speed just before it hits the ground can be calculated using the equation:

Final Velocity = Initial Velocity × cos(angle) - acceleration due to gravity × Time

Where initial velocity = 30 m/s, angle = 40°, acceleration due to gravity = 9.8 m/s² and time = 2.3 s

Therefore, final velocity = 30 m/s × cos(40°) - 9.8 m/s² × 2.3 s = 22.2 m/s

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In the sentences for the telescope in the right-hand column from the first-hand column, the relevant blank has been filled.

Describe the telescope.

A telescope is the device used to show an enlarged view of a far-off object.

There are different types of telescopes, and each is used for a particular purpose.

The blank that should be filled in correctly from the first column is as follows:

1. Shorter (bluer) wavelengths of light have better angular resolution than longer (redder) wavelengths of light for the Hubble Space Telescope.

2. Huge reflecting telescopes are employed by the numerous research observatories on Mauna Kea.

3. The spectrograph, which divides light into its various colors, enables astronomers to ascertain the stellar composition as well as a number of other stellar characteristics.

4. The identical 10-m Keck

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The combined mass of the six blocks of steel on the Moon would be 10.0kg. This is because the gravitational force on the Moon is only one sixth of the gravitational force on Earth, meaning that the mass of the blocks would be correspondingly reduced. On the Moon, the blocks of steel would weigh only 1.67 kg each, for a total mass of 10.0kg.

The effect of the reduced gravity on the blocks of steel is due to the inverse square law of gravitation. This law states that the force of gravity between two objects is inversely proportional to the square of the distance between them. Since the Moon is much farther away from the Earth than the blocks of steel were on Earth, the gravitational force on the blocks is much weaker. The reduced gravitational force on the Moon means that the six blocks of steel have a collective mass of 10.0kg. This mass is significantly lower than their combined mass of 60 kg on Earth, which is due to the inverse square law of gravitation.

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The six steel blocks on the Moon would weigh a total of 10.0kg. This is due to the fact that the gravitational force on the Moon is only one sixth that on Earth, which implies that the mass of the blocks would be lowered accordingly.

The steel blocks would only weigh 1.67 kg each, for a total mass of 10.0 kg, on the Moon.

The inverse square law of gravitation is what causes the steel blocks to be affected by the decreasing gravity. According to this rule, the force of gravity is inversely proportional to the square of the distance between two objects. The gravitational pull on the steel blocks is much weaker on the Moon since it is much further away from the Earth than the steel blocks were on Earth. The six steel blocks weigh a total of 10.0 kg due to the Moon's lower gravitational pull. The inverse square law of gravitation explains why this mass is far smaller than their total mass of 60 kg on Earth.

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