Residents of a community currently pay a fee for recreational services in their neighborhood, such as tennis courts, running trails, a pool. To assess whether there is support for raising a fee for the upcoming summer, a random sample of 50 residents is contracted and asked this question: “Would you support a $10 increase in the recreational fee to support programs for the health and well-being of the children in the community?” Of the 50 residents contacted, 42 approved of the fee increase. Which statement describes a likely source of bias present in survey results?

Answer:

Yes, single-use batteries are now made of common metals deemed non-hazardous by the federal government and can be disposed of in your regular trash in all states except California, where it is illegal to throw away all types of batteries.

I believe the answer is 16.

 Have a blessed day!

Explanation:

Newton’s second law is a quantitative description of the changes that a force can produce on the motion of a body. It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.

For the magnitude in N.m of the torque exerted by the fish about an axis perpendicular to the page and passing through the angler's hand is mathematically given as

Z = 360.33Nm

Generally, the equation for the Torque is mathematically given as

Z = force x perpendicular distance

Z = Tcos30° x L

Z = 166(0.886) x 2.45

Z = 360.33Nm

In conclusion s the magnitude in N.m of the torque exerted by the fish about an axis perpendicular to the page and passing through the angler's hand

Z = 360.33Nm

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Answer:

Explanation:

The position between point charge q to x = -8.0 m and y = 1.5 m is:

[tex]r=\sqrt{x^{2}+y^{2}}=\sqrt{(-8)^{2}+(1.5)^{2}}=\sqrt{66.25}\approx 8.13[/tex] meter

Then the magnitude of the electric field E is:

[tex]E=\frac{1}{4\pi\epsilon_{o}} \frac{q}{r^{2}}=(9\times 10^{9}) \frac{6.0\times 10^{-9}}{66.25}\approx 0.81 N/C[/tex]

For the vector of E:

[tex]\tan\theta=\frac{1.5}{-8}=-0.1875 \rightarrow \theta = -10.61^{0}[/tex]

[tex]E_{x}=E\cos (-10.16)=(0.81)(0.984)=0.79704 N/C[/tex]

[tex]E_{y}=E\sin\theta = -0.81(0.184)=-0.14904 N/C[/tex]

[tex]\vec{E}=0.79704\hat{i}-0.14904\hat{j}[/tex] in N/C

Answer:

After 1 decay it has .9375 of its value left  

.9375^n = 1/2

What is the value of n (number of decays)

n log .9375 = -.693

n =.693 / .0645 = 10.7

After 10.7 periods of 50 sec it will have 1/2 of its original nuclei left

1/2 life = 10.7 * 50 sec = 537 sec = 8.95 min

Answer:

4

Explanation:

We know that :

Solving

Answer:

Explanation:

You can use that [tex]v = v_{o}+gt[/tex] or [tex]v^{2}=v_{o}^{2}+2gh=(25)^{2}+(2)(10)(381)=8245 \rightarrow v \approx 90.8[/tex] m/s.

Answer:

234 90 +energy

Explanation:

because alpha decrease by 4 2helium

Answer:

very fast

Explanation:

Answer:

it is Newton's third law of motion?

Hope This Helps

[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]

Gravitational Potential of an object with mass m, and height of h metre is :

[tex]\qquad \sf  \dashrightarrow \:mgh[/tex]

Now, if the mass of object is 2100 kg, height of 18 m is placed in there, then it's gravitational potential energy is ~

[tex]\qquad \sf  \dashrightarrow \:2100 \times 10 \times 18[/tex]

[tex]\qquad \sf  \dashrightarrow \:378000 \: \: joules[/tex]

[tex]\qquad \sf  \dashrightarrow \:3.78 \times 10 {}^{5} \: \: joules[/tex]

Hi there!

We can begin by identifying key characteristics of both graphs.

Graph A.

Looking at the graph, we can see that the maximum distance (amplitude) is 10 cm (0.1 m). Additionally, its period (T) is 2 seconds (one full cycle).

We also know that:
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]

We can use this equation to compare with the other graph. Notice how the period does NOT depend on how far the spring is stretched. We can eliminate choice A for this reason.

Graph B.

The amplitude is 20 cm (0.2 m), and each period is 4 seconds.

We can now eliminate choice B because the springs are identical, so their spring constants are equal. Distance stretched has no impact on the spring constant.

For the other choices, we must look at forces and work.

Recall that:
Spring potential energy = [tex]\frac{1}{2}kx^2[/tex]

Kinetic energy = [tex]\frac{1}{2}mv^2[/tex]

Using the work-energy theorem:

[tex]\frac{1}{2}kx^2 = \frac{1}{2}mv^2[/tex]

Even if the mass of Block B is greater, its displacement is larger than that of Block A. Since displacement is squared in this equation, it would have a greater effect on the speed. Thus, Choice C is incorrect.

Using Hooke's Law:
[tex]\Sigma F = -kx\\\\ma = -kx\\[/tex]

[tex]a = \frac{-kx}{m}[/tex]

If the mass is greater, the acceleration will be smaller. Choice D is correct.

The vertical component of the net force that is moving the rocket away from Earth is determined as 171,010.1 N.

The vertical component of the net force that is moving the rocket away from Earth is determined as follows;

Fy = Fsinθ

where;

Fy = 500,000 x sin(20)

Fy = 171,010.1 N

Thus, the vertical component of the net force that is moving the rocket away from Earth is determined as 171,010.1 N.

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Answer:

250,000

Explanation:

i. The electron density of maganese is 2.44 × 10²⁸ electrons/m³

ii. The current that flows through a cylindrical manganese wire of volume 27 cm3 is 21.11 kA

To solve the question, we need to know what electron density is.

This is the number of free electrons per unit volume of material.

To find electron density, we need to find the atom density which is the number of atoms per unit volume, n.

So, the atom density of manganese is n = Nρ/A where

So, n = Nρ/A

n = 6.022 × 10²³ atoms/mol × 7430 kg/m³/0.05494 kg/mol

n = 44743.46 × 10²³ atoms/mol × kg/m³/0.05494 kg/mol

n = 81440.59 × 10²³ atoms/m³

n = 8.144059 × 10²⁷ atoms/m³

n ≅ 8.14 × 10²⁷ atoms/m³

The electron density of maganese is 2.44 × 10²⁸ electrons/m³

So, the electron density of manganese n' = n" × n where

So, n' = n" × n

= 3 electrons per atom × 8.14 × 10²⁷ atoms/m³

= 24.42 × 10²⁷ electrons/m³

= 2.442 × 10²⁸ electrons/m³

≅ 2.44 × 10²⁸ electrons/m³

So, the electron density of maganese is 2.44 × 10²⁸ electrons/m³

The current that flows through a cylindrical manganese wire of volume 27 cm3 is 21.11 kA

The current flowing through the wire is given by i = n'eV/t where

So, substituting the values of the variables into the equation, we have

i = n'eV/t

i = 2.44 × 10²⁸ electrons/m³ × 1.602 × 10⁻¹⁹ C × 27 × 10⁻⁶ m³/5 s

i = 105.54 × 10³ C/5 s

i = 21.11 × 10³ C/s

i = 21.11 × 10³ A

i = 21.11 kA

So, the current that flows through a cylindrical manganese wire of volume 27 cm3 is 21.11 kA

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The period (T) of the right pendulum for small displacements in s is equal to 1.49 seconds.

Mathematically, the time taken by this pendulum to undergo a small displacement is given by this formula:

[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex]

Where:

Scientific data:

Acceleration due to gravity (g) = 9.8 m/s².

Assuming a randomized variable for the length of the rope to be 1.2 meter. Also, we know that the right pendulum has a radius of 1/2L.

1/2L = L/2

L/2 = 1.1/2

L/2 = 0.55 meter.

Substituting the parameters into the formula, we have;

[tex]T=2\times 3.142 \sqrt{\frac{0.55}{9.8} }\\\\T=6.284 \times \sqrt{0.0561} \\\\T=6.284 \times 0.2369[/tex]

T = 1.49 seconds.

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Answer:

our acceleration is 0 because there is a constant speed..

a = final speed - initial speed / time

final speed = initial speed ( since speed is constant)

therefore acceleration is 0

Answer:

0.1111 W/m²

Explanation:

If all other parameters are constant, sound intensity is inversely proportional to the square of the distance of the sound. That is,

I ∝ (1/r²)

I = k/r²

Since k can be the constant of proportionality. k = Ir²

We can write this relation as

I₁ × r₁² = I₂ × r₂²

I₁ = 0.25 W/m²

r₁ = 16 m

I₂ = ?

r₂ = 24 m

0.25 × 16² = I₂ × 24²

I₂ = (0.25 × 16²)/24²

I₂ = 0.1111 W/m²

Answer:

If the Sun was on the left side of the diagram it would be points A and C

Explanation:

Spring tides are when the Sun, Earth, and Moon are aligned together. Neap tides occur when when the Sun, Earth, and Moon are forming a right angle.

Answer:

C and d mark mwe braliest

Explanation:

Answer:

0.568

Explanation:

velocity = -9.8 (t) + Vo

Vo= 0 so v= 9.8(t)

Momentum = mass x velocity

v * 0.14 kg = 0.78 kg m/s

v = 5.571 m/sec

v= 9.8(t) , so substituting v we get , 5.571 = 9.8 * t

t = 5.571/9.8 = 0.568 seconds

The torque experienced by the root of the tooth about point A when the steel band exerts the given force is 0.713 Nm.

The torque experienced by the root of the tooth about point A is calculated as follows;

τ = rFsinθ

where;

τ = (0.012) x (80) x (sin 48)

τ = 0.713 Nm

Thus, the torque experienced by the root of the tooth about point A when the steel band exerts the given force is 0.713 Nm.

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The speed of the wave with the given frequency and wavelength is 1.2m/s.

Given the data in the question;

Wavelength is  the distance over which the shapes of waves are repeated. It is the spatial period of a periodic wave.

It is expressed as;

[tex]\lambda = \frac{v}{f}[/tex]

Where v is velocity/speed and f is frequency.

Now, we can easily get the speed of the wave by substituting our given values into the expression above.

[tex]\lambda = \frac{v}{f} \\\\0.4m = \frac{v}{3s^{-1}}\\ \\v = 0.4m * 3s^{-1}\\\\v = 1.2ms^{-1}\\\\v = 1.2m/s[/tex]

Therefore, the speed of the wave with the given frequency and wavelength is 1.2m/s.

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Question :-

Answer :-

Explanation :-

As per the provided information in the given question, The Force is given as 10 Newton . The Distance is given as 30 cm [ 0.3 m ] . And, we have been asked to calculate the Amount of Force .

For calculating the Force , we will use the Formula :-

[tex] \bigstar \: \: \: \boxed{ \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} \: } [/tex]

Therefore , by Substituting the given values in the above Formula :-

[tex] \dag \: \: \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} [/tex]

[tex] \longmapsto \: \: \: \sf {Moment \: of \: Force \: = \: 10 \: \times \: 0.3} [/tex]

[tex] \longmapsto \: \: \: \textbf {\textsf {Moment \: of \: Force \: = \: 3}} [/tex]

Hence :-

[tex] \underline {\rule {185pt}{4pt}} [/tex]

Answer:

Explanation:

You can use [tex]\Delta E=\frac{\hbar c}{\lambda} \rightarrow \lambda =\frac{\hbar c}{E} = \frac{1240 eV.nm}{0.0141 eV} \approx 87943.26[/tex] in nm.

Here the hc = 1240 eV.nm in eV and nanometer unit.

The relative velocity of the ball with respect to the truck is same as that of the truck with respect to the ball at equal time.

The relative velocity of an object is the velocity of the object observed with respect to rest frame of another object.

The distance traveled by each object is determined from the prouduct of velocity and time of motion.

d = vt

where;

Thus, if the ball moved the same distance when it was on the flat bed truck, then the relative velocity of the ball with respect to the truck is same as that of the truck with respect to the ball at equal time.

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