list three things that spectra can tell us about objects in space​

build a fence.....................

The percent difference between GPE and KE in the experiment illustrates that the higher the percentage, the more efficient the energy transfer.

It should be noted that energy transformation simply means the process of changing energy from one form to another form.

In this case, the percent difference between GPE and KE in the experiment illustrates that the higher the percentage, the more efficient the energy transfer.

Also, the mass and height should be adjusted in such a way that the design of the system leads to GPE and KE values that are as close as possible.

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The experiment's percent difference between GPE and KE shows that the efficiency of energy transfer increases with percentage.

Thus, It should be emphasized that the term "energy transformation" simply refers to the act of altering energy's form.

The experiment's percent difference between GPE and KE in this instance demonstrates that the more significant the percentage, the more effectively energy is transferred.

The experiment's percent difference between GPE and KE in this instance demonstrates that the more significant the percentage, the more effectively energy is transferred.

Thus, Energy is capable of changing its forms. For instance, your body stores chemical energy from the food you eat until you can use it as kinetic energy when working or playing.

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Answer:

4 J

Explanation:

Original KE

Mass increases x 2

Velocity decreases x 4

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

[tex]\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\[/tex]

[tex]\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\[/tex]

Substitute the given parameters and solve for the angular speed;

[tex]\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s[/tex]

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

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[tex] \huge \mathbb \red{HEY \: THERE ♡}[/tex]

[tex] \huge \mathbb\pink{HOPE \: IT \: HELPS}[/tex]

Answer:

from what i got it is B

Hope This Helps

Answer:

proton is positively charged changechar

Explanation:

[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]

Answer:

The circular turning of roads

Explanation:

please mark brainliest

Answer:

Explanation:

The circular turning of roads

Driving on Curves

Hi there!

We can use the angular equivalent of the following kinematic equation:

[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta[/tex]

ωf = final angular velocity (1.9 rad/s)

ωi = initial angular velocity (11.3 rad/s)

α = angular acceleration (? rad/s²)

θ = angular displacement (17.95 rad)

We can rearrange the equation to solve for angular acceleration.

[tex]\omega_f^2 - \omega_i^2 = 2\alpha \theta\\\\\alpha = \frac{\omega_f^2 - \omega_i^2}{2\theta}[/tex]

Plug in the given values and solve.

[tex]\alpha = \frac{1.9^2 - 11.3^2}{2(17.95)} = \boxed{3.456 \frac{rad}{s}}[/tex]

Answer:

Find the heavy processes of CO2 and methane.

Explanation:

Step #1 Find a very reliable energy source.

Step #2 Find the heavy processes of CO2 and methane like in many factories and places with high carbon emissions.

Step #3 Shut down each of the heavy processes even 10 per day would do a lot.

Step #4 Use the renewable energy source as a replacement for the attempts to generate energy using carbon emissions.

Answer:

flammable  liquids

Explanation:when it catches on fire your going to get burned.hope this helps.

Answer:

If this is for Ultimate Frisbee, then FALSE

Explanation:

According to the rules, "In order for the disc to go from one player to another, it must at some time be in the air. No player may walk, run, or take steps in possession of the disc. The momentum of the receiver, however, must be taken into consideration." (crec.unl.edu) This means that if the player is already moving, they may take a maximum of two steps to come to a stop. (www.csun.edu).

.

.

.

Hope that helps?

Answer:

Rahman was a little bit 5AM but I think he was just try and the guy was the one who just got a a??ze and you know how are you to me that you

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:

Answer:

True

Explanation:

Answer:

4

Explanation:

We know that :

Solving

Given that the magnitude of the forces is greater towards the left; the ball slows down as it moves to the right because the horizontal forces are unbalanced.

We have to know that force is a vector quanty. The direction in which the ball will move is  dependent on the direction and magnitude of the resultant force.

Now considering the statements, the pairs of  statements that are true are;

description: the ball slows down as it moves to the right because the horizontal forces are unbalanced, with the larger force acting to the left.

explanation: The ball speeds up as it moves downward because the vertical forces are unbalanced, with the larger force acting downward.

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The answers to your questions are as written below

A) capacitance of a parallel plate capacitor = 442.5 * 10⁻¹³ F

B) The charge on each plate is = 2655 * 10⁻¹³ C

C) The electric field between the plates is ; 2 * 10³ N/C

D) The estimated area of the plates is ; 1.695 * 10⁹ m²

Plate dimension = 30 cm * 5 cm

Air gap ( d )  = 3 mm = 3 * 10⁻³ m

First step : calculate the area of the plate

Area of plate = 30 * 10⁻²m  *  5 * 10⁻² m

                      = 150 * 10⁻⁴ m²

Next step : determine the capacitance

Capacitance = ε₀A / d

                     = ( 8.85 * 10⁻¹² *  150 * 10⁻⁴ ) / 3 * 10⁻³

                     = 442.5 * 10⁻¹³ F

charge on each plate can be calculated with the formula below

Q = CV

   =  442.5 * 10⁻¹³  * 6

   = 2655 * 10⁻¹³ C

Electric field between plates ( E ) = V / d

                                                       = 6 / 3 * 10⁻³

                                                       = 2 * 10³ N/C

Required capacitance = 5 F

Air gap ( d ) =  3 * 10⁻³ m

applying the equation below

C = ε₀A / d -----

5 = ( 8.85 * 10⁻¹²  * A ) / 3 * 10⁻³

Therefore A :

A = ( 5 * 3 * 10⁻³ ) /  8.85 * 10⁻¹²  

   = 15 * 10⁻³ / 8.85 * 10⁻¹²

   = 1.695 * 10⁹ m²

Hence we can conclude that the answers to your questions are as written above.

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Answer:

the distance between two consecutive crests or troughs

it has no energy because its being held still

it has motion energy because it will fall if let go

Answer:

who can apply for the award

Answer:

1. Two bright bands would appear on the screen in line with the slits.

2. Waves that make up the radiation collide with each other so that they add together or cancel each other out.

3. Waves that are in phase constructively interfere to create bright bands.

4. Waves that are out of phase destructively interfere to form dark bands.

5. The spacing between the bright fringes will decrease.

6. 581 nm

Explanation:

Hi there!

We can begin by using Coulomb's Law:

[tex]E = \frac{kq}{r^2}[/tex]

k = Coulomb's Constant (8.99 × 10⁹ Nm²/C²)

E = Electric field strength (N/C)
r = distance from point (m)

q = charge (C)

Since this is a continuous charge, we must use calculus.

We can express this as the following:
[tex]q = \lambda L[/tex]

λ = Linear charge density (C/m)

L = Length of rod (m)

Now, since this is an arc, L = s (arc length). Additionally, we must find the differential elements of each:
[tex]dq = \lambda ds\\\\dq = \lambda rd\theta[/tex]

Our new equation is:
[tex]dE = \frac{kdq}{r^2}\\\\dE = \frac{k\lambda rd\theta}{r^2}[/tex]

However, we will only take the cosine component of the electric field since the vertical components will cancel out. (Electric fields are a vector). Therefore:
[tex]dE = \frac{k\lambda rd\theta}{r^2}cos\theta\\\\dE = \frac{k\lambda}{r}cos\theta d\theta[/tex]

Integrate. For a semicircle, the bounds will be from -π/2 to π/2.

[tex]E = \frac{k\lambda}{r}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}} {cos\theta} \, d\theta\\\\E = \frac{k\lambda}{r}sin\theta\left \|{\frac{\pi}{2}} \atop {-\frac{\pi}{2}}} \right. \\\\E = \frac{k\lambda}{r}(1 - (-1)) = \frac{2k\lambda}{r}[/tex]

We need to solve for λ, which is Q/ L:
[tex]\lambda = \frac{3.10 \times 10^{-9} C}{\pi (0.05)} = 1.9735 \times 10^{-8} \frac{C}{m}[/tex]

Now, plug and solve for the electric field strength:
[tex]E = \frac{2(8.99\times 10^9)(1.9735\times 10^{-8})}{0.05} = \boxed{7096.783 \frac{N}{C}}[/tex]

**A diagram was not provided, but if the hemisphere's focus was to the right, the electric field would be to the right, and etcetera.

The linear mass density of the rope in the given motion of the wave is 0.15 kg/m.

The linear mass density of the rope in the given motion of the wave is determined by dividing the mass of the rope with the length of the entire rope.

The linear mass density of the rope is calculated as follows;

μ = m/L

μ = 0.6 kg / 4 m

μ = 0.15 kg/m

Thus, the linear mass density of the rope in the given motion of the wave is 0.15 kg/m.

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hi you need to stop it get some help okay hope that worked

The wavelength (in m) of the note middle "C" (frequency = 256 Hz) in the air at room temperature and pressure will be 1171875 m.

The distance between two successive troughs or crests is known as the wavelength. The peak of the wave is the highest point, while the trough is the lowest.

The relationship between the wave's wavelength, time, and speed is given as;

[tex]\rm \lambda = \frac{c}{v} \\\\ \rm \lambda = \frac{3 \times 10^8}{256} \\\\ \lambda=1171875 \ m[/tex]

Hence the wavelength (in m) of the note middle "C" (frequency = 256 Hz) in the air at room temperature and pressure will be 1171875 m.

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Based on law of conservation of angular momentum, the rotational speed of the star is equal to 6,000 rev/s.

Given the following data:

First of all, we would determine the initial angular speed of the neutron star as follows:

[tex]\omega_i = \frac{1 \;Rev}{10 \;days} \\\\\omega_i = \frac{1 \;Rev}{10 \times 24 \times 60 \times 60}\\\\\omega_i = 1.157 \times 10^{-6}\;rev/s[/tex]

Mathematically, the moment of inertia of a uniform solid sphere is given by this formula:

[tex]I=\frac{2}{5} mr^2[/tex]

Where:

In order to determine the rotational speed of this neutron star, we would apply the law of conservation of angular momentum:

[tex]L_1 = L_2\\\\I_1\omega_1 = I_2\omega_2\\\\\frac{2}{5} m_1r_1^2 \omega_1 = \frac{2}{5} m_2r_2^2\omega_2\\\\m_1r_1^2 \omega_1 = m_2r_2^2\omega_2\\\\\omega_2 =\frac{r_1^2 \omega_1}{r_2^2}[/tex]

Substituting the given parameters into the formula, we have;

[tex]\omega_2 = \frac{ (7 \times 10^5 )^2 \times 1.157 \times 10^{-6}}{ (10 )^2}\\\\\\\omega_2 = \frac{ (7 \times 10^5 )^2 \times 1.157 \times 10^{-6}}{ 10 0}\\\\\omega_2 =\frac{566,930}{100}[/tex]

Final angular speed = 5,669 ≈ 6,000 rev/s.

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Answer:

its counting by 4 the multi 4,8,12,16