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15. A puck is set in motion across a frozen pond. Friction and air resistance may be neglected. If you see the puck change its direction but not its speed, then the
force on the puck is
OA Non-zero, and equal to the product of its mass times its weight
OB. Non-zero, and equal to its weight
OC. Zero
OD. Non-zero, and depends on the puck's new direction
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The 3rd  overtone or the third harmonic given that the fundamental frequency is 330Hz is 990Hz

Hamonics defined as the number gotten by multiplying the nth overtone by the fundamental frequecency.

In summary, the expression for the nth Hamonic or overtone is given as

Fn = Fundamental Frequency*N

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Answer:

when a medium begins to vibrate

Answer:

The giraffe is the tallest of all mammals. It reaches an overall height of 18 ft (5.5 m) or more. The legs and neck are extremely long. The giraffe has a short body, a tufted tail, a short mane, and short skin-covered horns

Answer:

The quantity and quality of sleep both impact memory.

Explanation:

Explanation:

i dont to i want to copy and paste on my printer

[tex]~~~~~\dfrac{T_1}{T_2}= \dfrac{1.22}{0.54}\\\\\\\implies \dfrac{2\pi \sqrt{\dfrac{L_1}{g}}}{2\pi \sqrt\dfrac{L_2}g}}= 2.259\\\\\\\implies \dfrac{\sqrt{L_1}}{\sqrt g} \times \dfrac{\sqrt g}{\sqrt{L_2}} = 2.259\\\\\\\implies \sqrt{\dfrac{L_1}{L_2}} = 2.259\\\\\\\implies \dfrac{L_1}{L_2} = 5.104\\\\\\\implies L_1:L_2 = 5.104[/tex]

Answer:

I have no clue I'm just trying to get points

Explanation:

:) sorry

Answer:

anything that contains atoms is the correct answer of given statement

We need frequency:-

[tex]\\ \rm\Rrightarrow \nu=\dfrac{c}{\lambda}[/tex]

[tex]\\ \rm\Rrightarrow \nu=\dfrac{3\times 10^8m/s}{590\times 10^{-9}m}[/tex]

[tex]\\ \rm\Rrightarrow \nu=0.0051\times 10^{17}Hz[/tex]

[tex]\\ \rm\Rrightarrow \nu=5.1\times 10^{14}Hz[/tex]

Answer:

See below

Explanation:

rho = R A/l     R = resistance   A = cross sectional area l = length

Answer:

The water cycle shows the continuous movement of water within the Earth and atmosphere. It is a complex system that includes many different processes. Liquid water evaporates into water vapor, condenses to form clouds, and precipitates back to earth in the form of rain and snow.

Explanation:

Hi there!

In order for a block to begin sliding, the force due to STATIC friction must be overcome.

In this instance, the following forces are acting on the block ALONG the axis of the incline.

Force due to gravity:

On an incline, the component of the force due to gravity contributing to the object's downward movement is equivalent to the horizontal (sine) component.

[tex]F_g = Mgsin\theta[/tex]

Force due to static friction:
The force due to friction is equivalent to the normal force multiplied by the coefficient of friction.

The normal force is the cosine component (perpendicular to the incline), so:
[tex]N = Mgcos\theta\\\\F_s = \mu_sMgcos\theta[/tex]

To find the minimum angle for the block to begin sliding, we can set the two forces equal to 0. They work in opposite directions (let down the incline be negative and up the incline be positive).

[tex]\Sigma F = F_s - F_g\\\\0 = F_s - F_g\\\\0 = \mu_sMgcos\theta - Mgsin\theta\\\\Mgsin\theta = \mu_sMgcos\theta[/tex]

Cancel out 'Mg' and rearrange to solve for theta.

[tex]sin\theta = \mu_scos\theta\\\\tan\theta = \mu_s\\\\\theta = tan^{-1}(\mu_s) = tan^{-1}(.45) = \boxed{24.228^o}[/tex]

Answer: The math behind this is quite simple. If you double the force, you double the acceleration, but if you double the mass, you cut the acceleration in half.

Explanation:

Answer:

Digital logic is the basis of computing and many other electronic devices as well as control systems found in this continually advancing digital world.

(too short, sorry)

Answer:

C

Explanation:

Generally, the speed of light slows down when passing through a medium that is not a vacuum. This is not always the case, but I will be ignoring the rare/exotic exceptions. Light has a harder time traveling through solids and liquids than it does with gases.

Answer:

its c

Explanation:

Answer:

galaxies were moving away from us

Explanation:

i got 100% on my quiz

Answer~

Both life science and physical science helps to understand the natural habitat of the earth. Life science deals with the living thing; thus, it is a study of the organic world. On the other hand, physical science deals with nonliving things; thus, it is a study of the inorganic world.

Explanation:

The sound mixer will need to increase the amplitude of the sound wave produced by the singer which will increase the loudness of the sound.

The amplitude of a sound wave is the maximum vertical displacement of the sound wave.

The sound mixer will need to increase the amplitude of the sound wave produced by the singer.

The increase in the amplitude of the sound wave produced by the lower tune singer will result in increased loudness of the sound.

Thus, the sound mixer will need to increase the amplitude of the sound wave produced by the singer which will increase the loudness of the sound.

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From the activity values and the decay constant, the mass of of Strontium in the sample is:

[tex]1.62 × 10^{-7}g[/tex]

The decay constant of an element is the probability of decay of a nucleus per unit time.

{λ = ln 2 / t1/2 

where;

t1/2 is the half-life of the isotope.

The half-life is converted to seconds since the decay constant is asked in per seconds.

[tex]28.8 years = 28.8 × 3.156 × 10^{7} = 908928000 s \\ [/tex]

Hence;

[tex]λ = \frac{ln2}{90892800s} = 7.626 s^{-1}[/tex]

The activity of the element, A, the decay constant, λ and the number of nuclei, N are related as follows:

[tex]N = \frac{8.25 ×10^{5}}{7.626×10^{-10}} = 1.082 × 10^{15} [/tex]              

Molar mass of Strontium-90 is 90 g.

1 mole of Strontium-90 contains 6.02×10^23 nuclei.

The mass, m of Strontium in the sample is calculated:

[tex]m = 1.082 × 10^{15} × \frac{90 g}{6.02 × 10^{23}} = 1.62 × 10^{-7}g \\ [/tex]

Therefore, the mass of of Strontium in the sample is:

[tex]1.62 × 10^{-7} \: g[/tex]

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Sebkhet Arina

Explanation:

sorry if it's wrong need more information

Answer:

According to the question,

Initial velocity of car (at t=0) = 10 m/s

Final velocity of car (after 15 sec) = 40m/s

Time taken = 15 sec.

Now use, first equation of motion viz.

Explanation:

v = u + at.

40 = 10 + a *15

30 = a*15

a = 30/15

a = 2 m/sec square.

B. False, the surface area of the cone and the cylinder is not the same.

The area or region that an object’s surface occupies is known as its surface area. Volume, on the other hand, refers to how much room an object has. Geometry has numerous shapes and dimensions, including spheres, cubes, cuboids, cones, cylinders, etc. Each form has its own volume and surface area.

If the slant height of the cone is 2r, then we can use the Pythagorean theorem to find the radius of the cone:

l² = (2r)² - r²

l² = 3r²

l = r√3

Using this value for the radius of the cone, the formula for its surface area becomes:

A(cone) = πrl + πr²

A(cone) = π(r)(r√3) + πr²

A(cone) = πr²(1 + √3)

For the cylinder, we can use the given height and radius to find its surface area:

A(cylinder) = 2πrl + 2πr²

A(cylinder) = 2π(r)(r) + 2π(r²)

A(cylinder) = 2πr²(1 + r)

Comparing the two surface area formulas, we can see that they are not equal in general, since each figure's coefficient in front of the second term (πr²) is different.

Therefore, we cannot conclude that the two figures have the same surface area based on the information.

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Answer:

see below

Explanation:

Kind of like when yo open a warm soda .....excess gas is released

Answer:

the facial-feedback hypothesis

Explanation:

For an airplane of mass 270,000 kg , the magnitude of the force exerted is mathematically given as

N2=1584431.13N

N1=1061568.87N

Generally, the equation for the Force balance is mathematically given as

N1+N2=270000*g

Therefore

N2*10=N1*15

N1=10N2/15

N1=0.67N2

Hence

0.67N2+N2=270000*g

N2=1584431.13N

N1=1061568.87N

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Hi there!

For an object on an incline with friction being pulled, the following forces are present.

The force due to friction opposes the force due to gravity which would cause the object to slide down. (The force due to friction acts up the incline). Additionally, the force due to the rope is also upward.

Let up the incline be positive, and down the incline be negative.

Doing a summation of forces:
[tex]\Sigma F = F_T + F_f - F_g[/tex]

For the crate to be moving at a constant velocity, there is NO net force acting on the crate, so:
[tex]0 = F_T + F_f - F_g[/tex]

Now, we can express each force as an equation.

Force due to tension:

Force due to gravity:

This is expressed as:
[tex]F_g = Mgsin\theta[/tex]

Force due to friction:

[tex]N = Mgcos\theta[/tex]

[tex]F_f = \mu Mgcos\theta[/tex]

Now, plug these expressions into the above equation.

[tex]0 = F_T + \mu Mgcos\theta - Mgsin\theta\\\\Mgsin\theta - \mu Mgcos\theta = F_T[/tex]

Mg = 245 N (weight). Plug in all values:
[tex]245sin(20) - 0.270(245)cos(20) = F_T\\\\F_T = \boxed{21.634 N}[/tex]

The normal force is equivalent to the vertical component (PERPENDICULAR TO THE INCLINE) of the weight (cosine), so:

[tex]N = Mgcos\theta\\\\N = 245cos(20) = \boxed{230.225 N}[/tex]

To find the tension in the rope needed to make the crate move at a constant velocity, we need to consider the forces acting on the crate.

1. Tension in the rope (T): This force is applied upwards at an angle of 20 degrees above the horizontal.

2. Weight of the crate (W): This force acts downward vertically and is equal to 245 N.

3. Normal force (N): This force is exerted by the floor on the crate and acts perpendicular to the surface of the floor.

4. Kinetic friction force (f_k): This force opposes the motion of the crate and acts parallel to the surface of the floor. The magnitude of kinetic friction is given by: f_k = μ_k * N, where μ_k is the coefficient of kinetic friction (0.270 in this case).

Since the crate is moving at a constant velocity, the net force on the crate must be zero. In other words, the forces pulling the crate forward (T and the horizontal component of the weight) must balance the forces opposing its motion (kinetic friction). The vertical forces (the vertical component of the weight and the normal force) must also balance.

Let's proceed with the calculations:

Horizontal Forces:

T * cos(20°) - f_k = 0

T * sin(20°) + N - W = 0

We can now solve for T and N:

From the horizontal forces equation:

T * cos(20°) = f_k

vertical force:

T = f_k / cos(20°)

T = (0.270 * N) / cos(20°)

From the vertical forces equation:

T * sin(20°) + N = W

T * sin(20°) + N = 245 N

Now, substitute the expression for T from the horizontal forces equation:

(0.270 * N) / cos(20°) * sin(20°) + N = 245 N

Now, solve for N:

N * (0.270 * tan(20°) + 1) = 245 N

N = 245 N / (0.270 * tan(20°) + 1)

Now, plug in the values and calculate N:

N = 245 N / (0.270 * tan(20°) + 1)

N ≈ 208.75 N

Now, we can calculate the tension in the rope (T):

T = (0.270 * N) / cos(20°)

T ≈ (0.270 * 208.75 N) / cos(20°)

T ≈ 67.24 N

So, the tension in the rope needed to make the crate move at a constant velocity is approximately 67.24 N, and the normal force exerted by the floor on the crate is approximately 208.75 N.

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Answer:

7.07 radians per second

Let's see

[tex]\\ \rm\Rrightarrow x=Asin(\omega t)\dots(1)[/tex]

Now we know the formula of acceleration

[tex]\\ \rm\Rrightarrow \alpha=-A\omega^2sin(\omega t)[/tex]

[tex]\\ \rm\Rrightarrow \alpha=-Asin(\omega t)\times \omega^2[/tex]

[tex]\\ \rm\Rrightarrow \alpha=-x\omega^2[/tex]

Or

[tex]\\ \rm\Rrightarrow \alpha=-\omega^2x[/tex]

Given that the position x of a particle along X-axis varies with time t by the equation:

[tex]{:\implies \quad \sf x=A\sin (\omega t)}[/tex]

As it defines the position, so x is just displacement here, and we need to find the acceleration first for telling with what if varies, so by definition, the second differential coefficient of displacement is acceleration, so differentiating both sides w.r.t.x of the above equation in accordance with chain rule we have:

[tex]{:\implies \quad \sf \dfrac{dx}{dt}=A\cos (\omega t)\cdot \omega \cdot \dfrac{dt}{dt}}[/tex]

[tex]{:\implies \quad \sf \dfrac{dx}{dt}=A\omega \cos (\omega t)}[/tex]

Differentiating both sides w.r.t.x by chain rule again to get the 2nd order derivative

[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-A\omega \sin (\omega t)\cdot \omega \dfrac{dt}{dt}}[/tex]

[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-A\omega^{2}\sin (\omega t)}[/tex]

Re-write as :

[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-\omega^{2}\{A\sin (\omega t)\}}[/tex]

Can be further written as

[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}=-x\omega^{2}}[/tex]

This is the Required answer

If they ask you the differential equation for the acceleration of a wave (as the given equation was general equation of a wave), you can simply write:

[tex]{:\implies \quad \sf \dfrac{d^{2}x}{dt^{2}}+x\omega^{2}=0}[/tex]