HELP HELP!


Which of the following surfaces would have the lowest amount of friction?

a. Glass
b. Sand
c. Wood
d. Ice

Answer:

Light traveling through a vacuum moves at exactly 299,792,458 meters (983,571,056 feet) per second. That's about 186,282 miles per second

Explanation:

Answer:

(neutral)

(protons)

(electrons)

Explanation:

Electrons have a negative charge (-) while protons have a positive (+) charge.

Atoms will usually be neutral, which means that there will be no charge.

For an atom to have a neutral charge, protons and electrons must "cancel" each other out. For this to happen, you need to have the same amount of each.

Positive will "cancel" out the negative.

Answer:

I will use (a / R) for alpha the angular acceleration

T R = I a / R     torque equals angular acceleration for cylinder

M g - T = M a    linear acceleration of center of mass

T = M (g - a) = I a / R^2 from first equation

If I = 1/2 M R^2 then M ( g - a) = M a / 2    from above

or g = 3 a / 2  and a = 2 g / 3

Also we have T = M (g - a) = M (g - 2 g / 3) = g / 3

Substitute I = M R^2 for the hollow cylinder

Looks like a = g/2 for hollow cylinder

Your answer is → Mass number

Answer:

The elasticity of the bungee cord reduces the gravitational forces applied on the body during bungee jumping. For example, if a 100-pound individual jumps from a building and encounters 900 pounds of deceleration force, they will feel 9 "G's" of force.

hopefully this'll help

have a nice day!!! :D

Answer:

I think the answer is option 3

Answer:

A solid air freshener loses mass and volume as a result of sublimation, where solid particles skip the liquid state and change directly from a solid to a gas. Sublimation requires that the particles in the solid state gain enough energy to immediately become a gas.

Explanation:

Hope this helps.

Answer:

magnets are magnetic :)

Explanation:

The definition of average acceleration allows to find the result for the average acceleration in the given time interval is:

          [tex]a_{average}= 1.5 \ \frac{m}{s^2}[/tex]

Instantaneous acceleration is defined as the derivative of velocity with respect to time.

           a =   [tex]\frac{dv}{dt}[/tex]

Where a is the acceleration, v the velocity and t the time.

They indicate that the speed of the car is given by the relation.

          v = α t + β t²

With α = 3 m / s and β = 0.1 m / s³

Let's make  the derivative.

           a = α + 2β t

Let's substitute

            a = 3 + 2 0.1 t

Average acceleration is the change in velocity in the time interval.  

          [tex]a_{average} = \frac{\Delta v}{\Delta t }[/tex]

Let's find the velocity at the indicated time.

For t = 5 s

         v₅ = 3 + 0.1 5²

         v₅ = 5.5 m / s

For t = 10 s

          v₁₀ = 3 + 0.1 10²

          v₁₀ = 13 m / s

Let's calculate the average acceleration.

           [tex]a_{average} = \frac{13 - 5.5 }{ 10 - 5 }\\[/tex]

           [tex]a_{average}= 1.5 \ m/s^2[/tex]

In conclusion using the definition of mean acceleration we can find the result for the mean acceleration in the given time interval is:

           [tex]a_{average} =[/tex]  1.5 m / s²

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Answer:

335 - 30 g

Explanation:

V = Vo + g*T = 335

Vo + 9.8*30 = 335

Vo =

Answer: Liquid - Lotion

Suspension - Semisolid

Capsule - Solid

Explanation:

Give Brainliest if correct :)

Answer:

When the frequency of the wave is increased, the pitch of the sound increases; that is, he sound becomes sharper or higher.

Explanation: Just completed on Edmentum

A period of a satellite is the time taken by the satellite to travel round a

body.

The comparison between the periods  [tex]T_B[/tex], and [tex]T_A[/tex] is [tex]\underline {T_B = \dfrac{\sqrt{2} }{4 } \cdot T_A}[/tex]

Reason:

The period, T, of a satellite is given as follows;

[tex]T = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \cdot M} }[/tex]

Volume of the planet A = [tex]\dfrac{4}{3} \cdot \pi \cdot r^3[/tex]

Mass of planet A, [tex]m_A[/tex] = [tex]\dfrac{4}{3} \cdot \pi \cdot r^3 \times \rho[/tex]

Volume of the planet B = [tex]\dfrac{4}{3} \cdot \pi \cdot (2 \cdot r)^3 = \dfrac{32}{3} \cdot \pi \cdot r^3[/tex]

Mass of planet B, [tex]m_B[/tex] = [tex]\dfrac{32}{3} \cdot \pi \cdot r^3 \times \rho[/tex]

Period of the satellite on planet A, [tex]T_A[/tex], is given as follows;

[tex]T_A = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \times \dfrac{4}{3} \cdot \pi \cdot r^3 \times \rho} } = 2 \cdot \pi \cdot \sqrt{\dfrac{1}{G \times \dfrac{4}{3} \cdot \pi \times \rho} }[/tex]

Period of the satellite on planet B, [tex]T_B[/tex], is given as follows;

[tex]T_B = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \times \dfrac{32}{3} \cdot \pi \cdot r^3 \times \rho} } = 2 \cdot \pi \cdot \sqrt{\dfrac{1}{G \times \dfrac{32}{3} \cdot \pi \times \rho} }[/tex]

Therefore, get;

[tex]\dfrac{T_A}{T_B} = \dfrac{ 2 \cdot \pi \cdot \sqrt{\dfrac{3}{G \times 4 \cdot \pi \times \rho} }}{ 2 \cdot \pi \cdot \sqrt{\dfrac{3}{G \times 32 \cdot \pi \times \rho} }} = \sqrt{\dfrac{32}{4} } = \sqrt{8} = 2 \cdot \sqrt{2}[/tex]

Therefore, [tex]T_A[/tex] = (2·√2)·[tex]T_B[/tex]

[tex]T_B = \dfrac{T_A}{2 \cdot \sqrt{2} } = \dfrac{\sqrt{2} \cdot T_A}{4 }[/tex]

The comparison between [tex]T_A[/tex] and  [tex]T_B[/tex] is therefore;

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The distance through which the rider at the end of the beam moves is;

L = 15.5 m

We are told that the beam on which the carnival ride is fixed is 12.6m in length.

Since the seat is at the end of the beam with the other end pivoted and the beam sweeps through an angle of 141°, then we can say that the radius of this arc formed by the swing is;

Radius; r = 12.6 m

Also, θ = 141°

The distance through which the driver moves will be the length of the arc formed by the beam at angle of 141°.

Formula for length of arc is given as;

L = 2πr(θ/360)

Plugging in the relevant values gives;

L = 2π × 12.6 × 141/360

L = 15.5 m

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(1) As the angle of the ramp is increased, the normal force decreases.

(2) As the angle of the ramp is increased, the parallel force increases.

(3) The angle at which the force down the plane was equal to the force of friction is zero degree.

(4) The net force that would cause acceleration is 47.33 N.

Let the angle of inclination of the ramp = θ

(1)

The normal force on an object on the ramp inclined to the ramp is calculated as follows;

[tex]F_n = mgcos (\theta)[/tex]

when θ is 0;

[tex]F_n = mgcos (0)\\\\F_n = mg[/tex]

when θ is 90;

[tex]F_n = mgcos(90)\\\\F_n = 0[/tex]

Thus, as the angle of the ramp is increased, the normal force decreases.

(2)

The parallel force on an object on the ramp inclined to the ramp is calculated as follows;

[tex]F_x = mgsin(\theta)\\\\[/tex]

when θ is 0;

[tex]F_x = mgsin(\theta)\\\\F_x = mgsin(0) \\\\F_x = 0[/tex]

when θ is 90;

[tex]F_x = mgsin(90)\\\\F_x = mg[/tex]

Thus, as the angle of the ramp is increased, the parallel force increases.

(3)

The force of friction is calculated as follows;

[tex]F_n = \mu F_n[/tex]

[tex]F_k = \mu mgcos(\theta)[/tex]

[tex]F_k = \mu mg cos(0)\\\\F_k = \mu mg[/tex]

Thus, the angle is zero degree

(4)

The net force that would cause acceleration is calculated as follows;

[tex]F_k = Fn\\\\F_k = mg cos(\theta)\\\\F_k = 5 \times 9.8 \times cos(15)\\\\F_k = 47.33 \ N[/tex]

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By Newton's second law, the net force on the crate is

∑ F = 300 N - f = (45 kg) (4.44 m/s²)

where f is the magnitude of friction. Solve for f :

300 N - f = (45 kg) (4.44 m/s²)

f = 300 N - (45 kg) (4.44 m/s²)

f = 100.2 N ≈ 100 N (D)

Given what we know, we can confirm that option D,  Electromagnetic waves with high frequency are more energetic than electromagnetic waves with low frequency is true.

High-frequency waves are synonymous with short wavelengths. This means that the waves are oscillating much quicker and therefore carry more kinetic energy within them. This is transformed and released as electromagnetic radiation, which is the reason why high-frequency waves are more energetic than low-frequency electromagnetic waves.

Therefore, we can confirm that the statement "Electromagnetic waves with high frequency are more energetic than electromagnetic waves with low frequency" is true.

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The answer to the question What shape is the graph produced by a force vs acceleration graph is A. Linear

Since Force, F = ma where m = mass and a = acceleration. For constant mass, F ∝ a. That is, F is directly proportional to acceleration, a.

Since this is a linear relationship, the graph of force vs acceleration will be linear.

The answer to the question What shape is the graph produced by a force vs acceleration graph is A. Linear

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Answer:

P = 2 * pi * (L / g)^1/2        period of simple pendulum

P = 6.28 * (2 / 9.8)^1/2 = 2.84 sec

Answer:   6.605

Explanation:

Answer:

true

Explanation:

Explanation:

[tex]V = IR \Rightarrow R = \dfrac{V}{I}[/tex]

Plugging in the numbers, we get

[tex]R=\dfrac{4.5\:\text{V}}{0.5\:\text{A}} = 9.0[/tex] ohms

Answer:

hola como estas hablas español

Explanation:

Answer:

I think D it's the correct answer!! Let me know if i'm wrong

Explanation:

Answer:

Explanation:

006

They are acting in opposite directions. Therefore the net force is found by subtraction. The sign is the same as the larger number.

Net Force = 99.6 - 52.8 = 46.8 N acting in the same direction as the 99.6  which is upward.

008

If the two forces act in the same direction, the net force is found by addition.

Net Force = 99.6 + 52.8  = 152.4 N downward.

Answer:

What is instillation

Explanation:

Answer:

the ball will go up 3s and down 3s

v=gt

where t=3s and g=9.8m/s^2

distance=v0(t)+(1/2)gt^2

where initial velocity (v0)=0

Explanation:

The speed required to throw a ball straight up and returns 6 seconds later would be 29.43 meters/seconds.

There are three equations of motion given by Newton,

v = u + at

S = ut + 1/2×a×t²

v² - u² = 2×a×s

As given in the problem we have to find the speed required to throw a ball straight up and it returns 6 seconds later,

S = ut + 1/2*a*t²

0 = u×6 + 0.5×(-9.81)×6²

0 = 6u - 176.8

6u = 176.8

u = 176.8/6

u = 29.43 meters / seconds

Thus, the speed required to throw a ball straight up and returns 6 seconds later would be 29.43 meters/seconds.

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Answer:

Explanation:

Use kinematic equation v² = u² + 2as

Rearrange for distance

s = (v² - u²) / 2a

Realize that at the top of its flight, the ball has zero velocity and gravity is acting downward in an assumed upward positive reference frame.

s = (0² - 29²) / (2(-9.8))

s = 42.90816...

s = 43 m

the expression for diffraction grating allows to find the results for the questions for the angular separation are:

i) The third order is Δθ = 0.203 rad.

ii) The first order with water is Δθ = 0.046 rad.

The diffraction grating is a system formed by a large number of equally spaced lines whose diffraction is given by the expression.

          d sin θ = m λ

Where d is the distance between two lines, θ is the angle of diffraction, the order of diffraction and λ is the wavelength.

i) Let's start by looking for the separation between two lines

Let's use a rule of direct proportions. If there are 300 lines in 1 mm, what distance is there between two lines.

         d = 1 lines (1 mm / 300 lines) = 3,333 10⁻³ mm

         d = 3.333 10⁻⁶ m

Let's find the angle of diffraction for the third order (m = 3) for each wavelength.

λ₁ = 400 nm = 400 10⁻⁹ m

         sin θ₁ = [tex]\frac{m \ \lambda }{d}[/tex]m λ/ d

         sin θ₁ = [tex]\frac{3 \ 400 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]  

         θ₁ = sin⁻¹ 0.3600

         θ₁ = 0.368 rad

λ₂ = 600 nm = 600 10⁻⁹ m

         sin θ₂ = [tex]\frac{3 \ 600 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]  

         θ₂ = sin⁻¹ 0.5401

         θ₂ = 0.571 rad

The angular separation is

         Δθ = θ₂ - θ₁

         Δθ = 0.571 - 0.368

         Δθ = 0.203 rad

ii) In this case, the separation between the network and the observation screen is filled with water.

When the rays leave the network they undergo a refraction process, for which they must comply with the relationship.

           [tex]n_i \ sin \theta_1 = n_r \ sin \theta_r[/tex]

The incident side is in the air, therefore its refractive index is n_i = 1 and when it passes into the water with refractive index n_r = 1.33.

Let's start looking for the incident angles for the first order of diffraction.

      m = 1

λ₁ = 400 nm

         θ₁ = sin⁻¹  [tex]\frac{1 \ 400 \ 10^{-9}}{3.33 \ 10^{-6}}[/tex]

         θ₁ = 0.120 rad

λ₂ = 600 nm

        θ₂ = sin⁻¹¹ [tex]\frac{1 \ 600 \ 10^{-9} }{3.33 \ 10^{-6}}[/tex]

        θ₂ = 0.181 rad

we use the equation of refraction.

         [tex]\theta_r[/tex]  = sin⁻¹ ([tex]\frac{n_i}{n_r} \ sin \ \theta_i[/tex] )

λ₁ = 400 nm  

       θ₁ = sin¹ ([tex]\frac{1 sin 0.120}{1.33}[/tex]

       θ₁ = 0.090 rad

λ₂ = 600 nm

        θ₂ =sin⁻¹  [tex]\frac{1 sin 0.181}{1.33}[/tex]

        θ₂ = 0.1358 rad

The angular separation is

          Δθ = 0.1358 - 0.090

          Δθ = 0.046 rad.

In conclusion using the relation for the diffraction grating we can find the results for the questions about angular separation are:

       i) The third order is Δθ = 0.203 rad.

      ii) The first order with water is Δθ = 0.046 rad.

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