Pls help me with this one :)) ​

This question involves the concepts of tension in a string and the wavelength of the wave in a string.

The wavelength of a sinusoidal wave will "increase by square power" on a string if the tension in the string is increased when the frequency is kept constant.

The speed of a wave on a string is given by the following formula:

[tex]v=\sqrt{\frac{T}{\mu}}[/tex]

where,

v = speed of wave = fλ

f = frequency of the wave

λ = wavelength of the wave

T = tension force

μ = linear mass density of the string

Therefore,

[tex]f\lambda=\sqrt{\frac{T}{\mu}}\\\\T = f^2\lambda^2\mu[/tex]

It is given that the frequency is kept constant. The linear mass density is also constant for a string. Therefore,

[tex]T=(constant)\lambda^2\\T\ \alpha\ \lambda^2[/tex]

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Hi there!

We know that:

Initial Total Mechanical Energy = Final Total Mechanical Energy

(Ei = Ef)

In this instance:

Ei = Gravitational Potential Energy

Ef = Kinetic Energy

In the absence of friction, ALL of the initial potential energy will be changed into kinetic energy at the bottom of the slide. Thus, the maximum kinetic energy of the child will be 1800 J.

Answer:

thx for the points

Explanation:

no need brainliest

Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.

By Newton's second law, the net vertical force is

• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0

where a is the acceleration of the wagon.

Solve for n (the magnitude of the normal force) :

n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N

Then

f = 0.500 (58.0 N) = 29.0 N

Meanwhile, the horizontal component of the applied force has magnitude

(80.0 N) cos(30.0°) ≈ 69.3 N

Now calculate the work done by either force.

• friction: -(29.0 N) (10.0 m) = -290. J

• pull: (69.3 N) (10.0 m) = 693 J

Answer:

Explanation:

I will ASSUME that the struck car was initially at rest. Would not have to be, but a different mass will result if it was

Conservation of momentum

14000(8.1) + m(0) = (14000 + m)(3.6)

14000(8.1) = 14000(3.6) + m(3.6)

m = 14000(8.1 - 3.6) / 3.6

m = 17500 kg

The mass of the second car is 17500 kg.

The law of conservation of momentum asserts that, unless an external force is applied, the total momentum of two or more bodies operating upon one another in an isolated system remains constant. As a result, momentum cannot be gained or lost.

Newton's third law of motion has a direct impact on the idea of momentum conservation.

mass of the first car = 14000 kg.

Initial speed of first car = 8.1 m/s.

Final speed of the two cars = 3.6 m/s.

From law of conservation of momentum; we can write:

Total initial momentum = total final momentum

14000 kg ×8.1 m/s + m×0 = (14000 + m) kg × (3.6 m/s)

14000(8.1) = 14000(3.6) + m(3.6)

m = 14000(8.1 - 3.6) / 3.6

m = 17500 kg

Hence, the mass of the second car is 17500 kg.

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Answer:

F net ∞ [tex]\frac{1}{\sqrt{t} }[/tex]

Explanation:

In pic

_________________

(hopet his helps can I pls have brainlist (crown)☺️)

5.08 m

Explanation:

The weight of the electron is being counteracted by the attractive electrostatic force exerted by the proton above it. We can write the force equation as follows:

[tex]m_eg = k_e\dfrac{Q_pQ_e}{r^2}[/tex]

where the Q's are the charges of the proton and electron, r is the distance between the particles, g is the acceleration due to gravity, [tex]m_e[/tex] is the mass of the electrons and [tex]k_e[/tex] is the Coulomb constant. So solving for r, we get

[tex]r^2 = k_e\dfrac{Q_pQ_e}{m_eg}[/tex]

Taking the square root of r^2, we then get the distance as

[tex]r = \sqrt{k_e\dfrac{Q_pQ_e}{m_eg}}[/tex]

The values are given as follows:

[tex]m_e = 9.11×10^{-31}\:\text{kg}[/tex]

[tex]g = 9.8\:\text{m/s}^2[/tex]

[tex]Q_p = Q_e = 1.60×10^{-19}\:\text{C}[/tex]

[tex]k_e = 8.99×10^9\:\text{N-m}^2\text{/C}^2[/tex]

Putting in all of these values in our equation for r,

[tex]r = \sqrt{\dfrac{(8.99×10^9\:\text{N-m}^2\text{/C}^2)(1.60×10^{-19}\:\text{C})^2}{(9.11×10^{-31}\:\text{kg})(9.8\:\text{m/s}^2)}}[/tex]

[tex]\:\:\:\:\:= 5.08\:\text{m}[/tex]

Answer:

less

Explanation:

heat travels from hotter to colder object. hotter object has more kinetic energy.

Answer:

i dont know need points

Explanation:

Answer:

600

Explanation:

GPE=mgh (mass*gravitational force*height)

m=10 kg

g=10

h=6m

10*10*6= 600 Joules

Answer:

No - It is connected in parallel instead of series

Explanation:

The answer is No, the reason behind the answer that the ammeter is connected in parallel which is wrong it must be connected in series.

An ammeter is a device for detecting electric current in amperes, either directly (DC) or alternating (AC). Due to the fact that a shunt running parallel to the meter carries the majority of the electricity at high current values, the digital multimeter can measure a wide range of current values. A circle with the capital A in it serves as the icon for an ammeter for circuit diagrams.

The moving coil of the electrodynamics ammeter rotates in the field created by the fixed coil. It measures alternating and direct current with 0.1 to 0.25 percentage points (by converting Ac power To DC power using a rectifier).

The ammeter is connected in series and, on the other hand, the voltmeter is connected in parallel.

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Answer:

The kinetic molecular theory of matter asserts that: Matter is made up of particles that are continually moving. All particles contain energy, however the energy fluctuates depending on the temperature the sample of matter is in. This in turn decides whether the material exists in the solid, liquid, or gaseous form.

Explanation:

Hope it helps:)

The galaxies are in constant motion so it can be inferred that their position will not be the same as a hundred million years ago.

A galaxy is a term to refer to the set of stars, gas clouds, planets, cosmic dust, dark matter, and energy gravitationally united in a more or less defined structure in the universe.

Galaxies are in constant motion because the elements that make them up are not static at any time; The galaxies rotate around the center of the galaxy. If they were still, the gravitational attraction would cause them to immediately fall towards the center of the galaxy: it is the same that would happen to the Earth and the other planets if they stopped rotating around the Sun, they would fall towards the Sun.

According to the above, it can be inferred that the galaxies after a while have moved from the place where they were last seen.

Note: This question is incomplete because there is some missing information. However, I can answer based on my general prior knowledge.

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60 inches

because if 30=50% then 60=100% so its 60 inches

The sum of the series allows to find the result for the total distance that the ball bounces is:

            total distance = 59.52 in

A series is a set of things or numbers related by a specific operation.

They indicate that the ball falls from an initial height y₀ = 30 in. and it bounces 50% of the height and the process is repeated until it stops, see attached.

Let's build a table to observe the sequence.

drop   height    rebound

   1        30          15

   2       15            7.5

   3        7.5         3.75

If we call the first term y₀  

The first bounce can be found.

                             y₁ = [tex]\frac{y_o}{2}[/tex]

The second bounces.

                             [tex]y_2 = \frac{y_1}{2} \\y_2 = \frac{y_o}{4}[/tex]

The third bounce.

                             [tex]y_3 = \frac{y_2}{2} \\y_3 = \frac{y_0}{ 8}[/tex]

By observing this table we can construct a series of the form

      Total distance = [tex]y_o \ ( 1 + \frac{1}{2} + \frac{1}{4}+ \frac{1}{8} + ... +\frac{1}{2n} )[/tex]

The sum of the serie has a result of

        sum  = 127/64 = 1,984

Let's calculate

     distance total  = 30 1,984

     Distance total = 59.52 cm

In conclusion, using the sum of the series we can find the result for the total distance that the ball bounces is:

            total distance = 59.52 in

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The maximum magnitude of the friction force exerted on the box by the ramp is 44.41 N.

The given parameters;

The normal force on the wooden box is calculated as follows;

[tex]F_n = mg \times cos(\theta)\\\\F_n = 10 \times 9.8 \times cos(25)\\\\F_n = 88.8 2 \ N[/tex]

The maximum magnitude of the friction force exerted on the box by the ramp is calculated as follows;

[tex]F_f = \mu \times F_n\\\\F_f = 0.5 \times 88.82 \\\\F_f = 44.41 \ N[/tex]

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Hello!

We can use the following angular kinematic equation to solve:

α = Δω/Δt, or (ωf-ωi)/t

Plug in the given values:

α = (25 - 10)/5 = 15/5 = 3 rad/sec²

The distance between the glass to have the given heat loss is 2.54 m.

The given parameters:

The area of the glass window is calculated as follows;

[tex]A = 0.6 \times 0.9\\\\A = 0.54 \ m^2[/tex]

The distance between the glass is calculated as follows;

[tex]Q = \frac{KA \Delta T}{\Delta x} \\\\\Delta x = \frac{kA \Delta T}{Q} \\\\\Delta x = \frac{0.8 \times 0.54 \times 24 }{4.09} \\\\\Delta x = 2.54 \ m[/tex]

Thus, the distance between the glass to have the given heat loss is 2.54 m.

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Answer:

8.01e-22

Explanation:

The energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.

The extension produced by one mole of lead atom is calculated by applying Hooke's law;

F = kx

mg = kx

x = mg/k

x = (0.207 x 9.8) / (20)

x = 0.101 m

The Energy of one quantum of energy for an atomic oscillator in a block of lead is calculated as follows;

E = ¹/₂kx²

E = ¹/₂ (20)(0.101)²

E = 0.102 J

Thus, the energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.

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The acceleration due to gravity on the given planet is 8.71 m/s².

The given parameters;

The acceleration due to gravity on the planet is calculated by applying following formula as follows;

[tex]T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2 \pi } = \sqrt{\frac{l}{g}} \\\\\frac{T^2}{4\pi ^2} = \frac{l}{g} \\\\g = \frac{4\pi^2 l}{T^2} \\\\g = \frac{4 \times (\pi)^2 \times 0.45}{1.428^2} \\\\g = 8.71 \ m/s^2[/tex]

Thus, the acceleration due to gravity on the given planet is 8.71 m/s².

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Answer:

Our eyes are most sensitive to the wavelengths corresponding to the yellow and green colors of the spectrum. Flashy signs and some fire engines are painted in a yellowish-green color to attract our attention.

Acceleration = (change in velocity ( final speed - starting speed))/ (time)

Acceleration = (18-30)/10.5

Acceleration = -12/10.5

Acceleration = -1.14 m/s^2

Distance = 30m/s x 10.5s + 1/2(1.14)(10.5)^2

Distance = 252.2 meters

The only force opposing the block's sliding as it slows down is friction with magnitude f . By Newton's second law, the net force in this direction is

∑ F = -f = ma = (4.00 kg) a

Assuming constant acceleration a , the acceleration applied by friction is such that

(1.5 m/s)² - (11 m/s)² = 2a (4.00 m)

Solve for the acceleration :

a = ((1.5 m/s)² - (11 m/s)²) / (8.00 m) ≈ -14.8 m/s²

Then the frictional force exerted a magnitude of

-f = (4.00 kg) (-14.8 m/s²)

f ≈ 59.4 N

and was directed opposite the block's motion.

Answer:

Ammonium perchlorate NH4ClO4

Ammonium Nitrate

Calcium Cyanamide

When detonated, the reaction products are all gases, such as water vapor, nitrogen gas, and oxides of nitrogen.

Hopefully this helps :)

Answer:

option B is the correct answer

Explanation:

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Answer: An ice hockey player picks up a trophy as he slides past it.

Two pool balls colliding precisely inelastically and rebounding off one another is an example. So, the correct option is A.

Inelastic collisions are those in which the total kinetic energy is lower after the impact than it was before. The stick is travelling quickly in the direction of the ball before to the contact. The stick comes to a stop following the accident. It transfers some of its kinetic energy to the cue ball, which rolls forward.

The type of collision mentioned in the given example is known as an inelastic collision is one in which the system's momentum is preserved but its kinetic energy is not. In a ballistic pendulum, an example of an inelastic collision. Dropped ball of clay is another illustration of an inelastic collision.

Thus, the correct option is A.

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Your question is incomplete, most probably the complete question is:

Which of the following statements describes a perfectly inelastic collision?

A. Two billiard balls bounce off of each other,

B. A car crashes into a tree and rebounds in the other direction,

C. A wad of chewing gum is thrown and sticks to a moving truck,

D. Two ice skaters hit each other and fall over in opposite directions,

The actual weight of the gas = apparent weight + weight.

The actual weight = [tex]W_{A}[/tex] + W

Given that a plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again.

If the apparent weight of the gas is the difference between these two masses, then let the apparent weight = [tex]W_{A}[/tex]

The gas is squeezed out of the bag to determine its volume by the displacement of water. Since

density = mass / volume

The density of water is 1000 kg/[tex]m^{2}[/tex]

we can get the mass of the gas by making m the subject of the formula.

W = mg

The actual weight of the gas = apparent weight + weight

That is,

The actual weight =  [tex]W_{A}[/tex] + W

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Let m₁ and m₂ be the masses of the two objects, and v₁ and v₂ their initial velocities. So

m₁ = 30.0 g = 0.0300 kg

m₂ = 13.0 g = 0.0130 kg

v₁ = + 20.5 cm/s = 0.205 m/s

v₂ = + 15.0 cm/s = 0.150 m/s

and we want to find v₁' and v₂', the final velocities of either object after their collision.

Momentum is conserved throughout the objects' collision, so that

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

where v₁' and v₂' are the first and second object's velocities after the collision.

Kinetic energy is also conserved, so that

1/2 m₁v₁² + 1/2 m₂v₂² = 1/2 m₁(v₁')² + 1/2 m₂(v₂')²

or

m₁v₁² + m₂v₂² = m₁(v₁')² + m₂(v₂')²

From the first equation (omitting units), we have

0.0300 • 0.205 + 0.0130 • 0.150 = 0.0300 v₁' + 0.0130 v₂'

0.0810 = 0.0300 v₁' + 0.0130 v₂'

81 = 30 v₁' + 13 v₂'

From the second equation,

0.0300 • 0.205² + 0.0130 • 0.150² = 0.0300 (v₁')² + 0.0130 (v₂')²

0.00155 ≈ 0.0300 (v₁')² + 0.0130 (v₂')²

1.55 ≈ 30 (v₁')² + 13 (v₂')²

Solving both equations simultaneously gives two solutions, one of which corresponds to the initial conditions. The other yields

v₁' ≈ + 0.172 m/s

and

v₂' ≈ + 0.227 m/s

Answer:

hi

Explanation:

i like saying hi

Answer:

I feel exited and happy I enjoy it with my friend