13. What happens to the number of times you must rotate a screw in order to drive it into
an object when you increase the number of threads on the screw?
a. You must rotate the screw more times and apply a stronger force
b. You must rotate the screw more times but can apply a smaller force
c. You will rotate the screw less times but must apply a stronger force
d. You will rotate the screw less times and can apply a smaller force
14. What do you
hvor

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:

Answer:

Take north as y and west as -x.

distance=√(x^2+y^2)

displacement= tan theta(tan^-1) (y/x)

answers should be, (9.85m, 66.04°)

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: 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:

[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]

Sebkhet Arina

Explanation:

sorry if it's wrong need more information

Answer:

3 14/3

Explanation:

you put a 1 under the 7 then times it with 3 2/3 and you get 3 14/3

Answer:

anything that contains atoms is the correct answer of given statement

Answer:

See below

Explanation:

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

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:

7.07 radians per second

Answer:

the facial-feedback hypothesis

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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Explanation:

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

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:

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:

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)

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]

When the spring is replaced with a spring with double of initial spring constant k2 = 2k₁, the new period, T2, is [tex]\sqrt{\frac{1}{2} } \ T_1[/tex].

The period of the oscillation of the mass or cart on the spring is given by the following formula;

[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\[/tex]

at a costant mass;

[tex]T_1\sqrt{k_1} = T_2\sqrt{k_2} \\\\T_2 = \frac{T_1\sqrt{k_1} }{\sqrt{k_2} }[/tex]

when spring constant is doubled, k2 = 2k1. the new period, T2 is determined as follows;

[tex]T_2 = \frac{T_1\sqrt{k_1} }{\sqrt{2k_1} } \\\\T_2 = \frac{T_1\sqrt{k_1} }{\sqrt{2} \times \sqrt{k_1} } \\\\T_2 = \frac{T_1}{\sqrt{2} }\\\\T_2 = \sqrt{\frac{1}{2} } \ T_1[/tex]

Thus, when the spring is replaced with a spring with double of initial spring constant k2 = 2k₁, the new period, T2, is [tex]\sqrt{\frac{1}{2} } \ T_1[/tex].

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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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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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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.

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

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:

It takes total 248 years for Pluto to orbit the sun.

Pluto will take 253 year to orbit the sun.

Earth's average distance to the Sun is about 93 million miles (150 million kilometers) from the Sun.

Pluto has a semi-major axis distance a=40 then T2=403, which gives T=253 years.

Therefore, Pluto will take 253 year to orbit the sun.

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