are there really dead stars in the sky?​

h= height = 19 m

g.e = gravity on earth = 9.8 m/s^2

g.u = gravity on Uranus = 9.8 x (88.9/100) = 8.71 m/s^2

Apply :

h = vot + 1/2 gt^2

h= 1/2 g t^2

t = √(2h/g)

Earth:

t= √(2(19)/9.8) = 1.97 s

Uranus

t= √(2(19)/8.71) = 2.09 s

Tu - Te = 2.09 - 1.97 = 0.12 s

Answer : 0.12 s

Answer:

I KNOW THE ANSWER IT WILL COST 30$

Explanation:

The net force on particle q₂ is 1.05 x 10¹³ N, directed towards particle q1.

To find the net force on particle q₂, we need to calculate the force between q₁ and q₂, and the force between q₂ and q₃, and then add these two forces together.

The force between two point charges can be calculated using Coulomb's law, which states that

F = k × q₁× q₂ / r²

where F is the force, k is Coulomb's constant (9.0 x 10⁹ N m² / C²), q₁ and q₂ are the magnitudes of the charges, and r is the distance between the charges.

The force between q₁ and q₂ is

F₁ = k × q₁ × q₂ / r₁²

where r₁ = 0.10 m.

Substituting the values, we get

F₁ = 9.0 x 10⁹ N m² / C²  × 8.0 C  × 3.5 C / (0.10 m)² = 2.52 x 10¹³ N

The force between q₂ and q₃ is

F₂ = k  × q₂  × q₃ / r₂²

where r₂ = 0.15 m.

Substituting the values, we get

F₂ = 9.0 x 10⁹ N m² / C²  × 3.5 C  × (-3.5 C) / (0.15 m)² = -1.47 x 10¹³ N

Note that the negative sign indicates that the force is attractive, since q₂ and q₃ have opposite signs.

The net force on q₂ is the vector sum of F₁ and F₂

\(F_{net}\) = F₁ + F₂ = 2.52 x 10¹³ N - 1.47 x 10¹³ N = 1.05 x 10³ N

To know more about net force here

https://brainly.com/question/29261584

#SPJ1

Answer:

Gravity is a contact force on both the cat and the flap. - False

The push from the flap is a contact force on the cat. - True

The push from the cat is a contact force on the flap. - True

Explanation:

Gravity is a contact force on both the cat and the flap. - False

Gravity is a non-contact force that acts between objects with mass. It does not require direct contact between the objects.

The push from the flap is a contact force on the cat. - True

When the flap pushes against the cat, it exerts a contact force on the cat.

The push from the cat is a contact force on the flap. - True

When the cat pushes against the flap, it exerts a contact force on the flap.

The caps used with boomwhacker tubes alter the sound of the notes by changing the effective length of the air column inside the tube.

This modification affects the fundamental frequency produced when the tube is struck or tapped against a surface.The fundamental frequency determines the pitch of the sound we hear.

When a boomwhacker tube is struck without any cap, the air column inside vibrates at its natural frequency, producing a specific pitch. By adding caps to the tube, we effectively shorten its length, which increases the frequency of the vibrations and raises the pitch. Conversely, removing caps increases the effective length, lowering the frequency and lowering the pitch.

To visualize this, imagine a tube with a cap on one end. The cap acts as a barrier, preventing the air column from vibrating in that direction. Therefore, the effective length of the tube becomes the distance between the cap and the open end. This shorter length results in a higher pitch. Removing the cap increases the effective length, resulting in a lower pitch.By using different combinations of caps on the boomwhacker tubes, we can produce a range of pitches, creating a musical and colorful experience.

for such more questions on sound

https://brainly.com/question/13679347

#SPJ8

Answer:

when you lag in culture

Explanation:

it makes sense

Answer:

Well it could possibly be because of where you live, or if there is a strong power source or WiFi thing.

Explanation:

I’m sorry if my answers not the best I’m just doing it of my own experience from bad WiFi.

The velocity of the car A as observed from car B is (15i - 22.5j) m/s

The acceleration of car A as observed from car B is 4.5j m/s²

The velocity of car A and car B in m/s is equal:

VA = 54 km/h * 5/18 = 15 im/s

VB = 81 km/h * 5/18 = 22.5 jm/s

The relative velocity is:

\(v_{AB} = v_{A} - v_{B} =\) (15 i -22.5 j ) m/s

The acceleration of the cars are:

\(a_{A} = v^{2}_{A} = 15^{2} /150 =1.5 jm /s^{2}\)

The relative acceleration is:

\(a_{AB} = a_{A} - a_{B} =\) 1.5- (-3), negative - because - the - car - is- slowly - down

\(a_{AB\) = 4.5 jm/ \(s^{2}\)

The relative velocity is defined as the velocity of an object with respect to another observer. It is the time pace of change of relative position of one object with respect to another object.

The relative acceleration (also or ) is the acceleration of an object or observer B in the rest casing of another object or observer A. Acceleration of B relative to An is =ab−aa.

to know more about relative velocity click here:

https://brainly.com/question/17228388

#SPJ4

The third rule of a series circuit states that the sum of the individual voltages equals the total voltage.

These demonstrate the truth of the previous claim, which reads, "When voltage sources are connected in series, the overall voltage is equal to the algebraic sum of the individual voltages?"

The supply voltage is equal to the total of the component voltages in a series circuit. The voltages across each component in a series circuit are proportional to their resistances. This means that when two similar components are linked in series, the supply voltage is divided equally.

To learn more about voltage visit;

https://brainly.com/question/29445057

#SPJ4

Answer: Phytoplankton

Explanation:

Answer:

I'm not quite sure, but i believe you should be able to

Explanation:

l

Explanation:

Given that,

Mass, m = 1.8 kg

Compression, x = 2.6 cm

We know that,

Force on spring = weight

So,

\(mg=kx\)

Where

k is spring constant

\(k=\dfrac{mg}{x}\\\\k=\dfrac{1.8\times 9.8}{2.6\times 10^{-2}}\\\\k=678.46\ N/m\)

(2) If m = 5.2 kg

\(x=\dfrac{mg}{k}\\\\x=\dfrac{1.8\times 9.8}{678.46}\\\\x=2.6 \ cm\)

Hence, this is the required solution.

Answer: The rock

Explanation:

A 300 kw load has a trailing power factor of 0.65. The reactive power required in order to improve the power factor to unity is 346.12kVAR

Given Power of load (P) = 300KW

power factor = 0.65

We know that kVA = KW/cosФ

kVAR = kVA sinФ

sin = √1-cos^2Ф

here kVA is apparent power(S)

KW is active power(P)

kVAR is reactive power(Q)

cosФ = power factor

S = P/cosФ = 300/0.65 = 461.5kVA

sinФ  = √1-(0.65)^2 = 0.75

Now Q = (461.5)(0.75) = 346.12kVAR lagging

the additional load is capacitive

To learn more about capacitive click here https://brainly.com/question/12644355

#SPJ4

A bicycle with 26-inch diameter wheels is traveling at 20 mi/h. Find the angular speed of the wheels in rad/min.

______radians per minute

___revolutions per minuteω =  1320 rad/min, Revolution per minutes = 210.084

Diameter = 24 in

Radius r = 24/2 in = 12 in = 12 × 2.54 cm = 30.48‬ cm = 0.3048 m

Linear Speed v= 15 mi/hr = 15 * 1609.34 / 3600 m/s = 6.7056 m/s

Angular Speed ω = v / r = (6.7056 m/s) / (0.3048 m) = 21.999945

ω =  22 rad/s = 22× 60 rad/min = 1320 rad/min

∴ 1 revolution = 2π rad

⇒1 rad = 1 / 2π rev

so 1320 rad/min = 1320 / 2π  rev/min = 210.0845 rev/min

Revolution per minutes = 210.0845

To learn more about angular speed visit : https://brainly.com/question/14882921

#SPJ9

Answer:

Approximately \(3.1\; \rm m \cdot s^{-1}\).

Explanation:

The content of this pail is in a centripetal motion because its path forms part of a vertical circle. Let \(m\) denote the mass of the contents of this pail, let \(v\) denote the (linear) velocity of the content, and let \(r\) denote the radius of this circle. The net force on the contents of this pail will thus be:

\(\displaystyle F(\text{net}) = \frac{m\, v^2}{r}\) towards the center of the circle.

Assume that there is no friction between the content and walls of the pail. The only two possible forces on the contents pail towards the center would be:

Note that at the top of the circle, both the gravitational pull and the normal force from the bottom point towards the center of the circle. On the other hand, the normal force from the walls of the pail would be perpendicular to the line towards the center of the circle. At that point in the circle, there's no upward force to support the content of the pail. The uniform rotation will be sufficiently fast if it could allow the content to stay in the pail at the top of the circle.

Let \(g\) denote the gravitational field strength at the top of this circle. The size of the gravitational pull on the content would be \(m\cdot g\). Let \(F(\text{normal})\) denote the normal force from the bottom of the pail on the contents. The sum of these two forces should be equal to the vertical net force on the contents of this pail. That is:

\(F(\text{net}) = m\cdot g + F(\text{normal})\).

From the centripetal motion of the content:

\(\displaystyle \frac{m\, v^2}{r} = m\cdot g + F(\text{normal})\).

Rearrange to obtain an expression for the normal force:

\(\displaystyle F(\text{normal}) = \frac{m\, v^2}{r} - m\cdot g\).

Note, that the normal force the bottom of the pail exerts on the contents should be greater than or equal to zero. While the pail is at the top of the circle, the normal force from the bottom of the pail cannot pull the contents upwards. Hence:

\(\displaystyle \frac{m\, v^2}{r} - m\cdot g = F(\text{normal}) \ge 0\).

\(\displaystyle \frac{m\, v^2}{r} - m\cdot g \ge 0\).

Rearrange and simplify to obtain:

\(\displaystyle \frac{v^2}{r} - g \ge 0\).

\(v^2 \ge g\cdot r\).

\(v \ge \sqrt{g \cdot r}\).

In other words, if the gravitational field strength is \(g\) and the radius of the circle is \(r\), the minimum linear velocity required to keep the content in the pail at the top of the circle is \(\sqrt{g \cdot r}\).

If \(g = 9.81\; \rm N \cdot kg^{-1} = 9.81\; \rm m \cdot s^{-2}\) and \(r = 0.82 \; \rm m + 0.185\; \rm m \approx 1.005\; \rm m\), then the minimum value of \(v\) would be approximately:

\(\sqrt{9.81 \; \rm m \cdot s^{-1} \times 1.005\; \rm m} \approx 3.1\; \rm m \cdot s^{-1}\).

Answer:

The workdone is \(W_v = - 20 \ J\)

Explanation:

From the question we are told that

    The mass of the object is \(m = 2.5 \ kg\)

     The speed of fall is \(v = 2.5 \ m/s\)

     The depth of fall is  \(d = 80\ cm = 0.8 \ m\)

Generally according to the work energy theorem

      \(W = \frac{1}{2} mv_2^2 - \frac{1}{2} mv_1^2\)

Now here given the that the velocity is  constant  i.e  \(v_1 = v_2 = v\) then

We have that

    \(W = \frac{1}{2} mv^2 - \frac{1}{2} mv^2 = 0 \ J\)  

So in terms of workdone by the potential energy of the object and that of the viscous liquid we have

       \(W = W_v - W_p\)

Where  \(W_v\) is workdone by viscous liquid

             \(W_p\) is the workdone by the object which is mathematically represented as

            \(W_p = mgd\)

So  

       \(0 = W_v + mgd\)

=>    \(W_v = - m * g * d\)

substituting values

       \(W_v = - (2.5 * 9.8 * 0.8)\)

      \(W_v = - 20 \ J\)

Ans:

C

Explanation:

because it is moving down the stairs

A spring travelling down a flight of stairs has kinetic energy.

Since it is moving, it would be a spring moving down a set of steps. A stretched spring, a compressed spring, and a spring at the top of a set of steps are all in motion.

Kinetic energy is a type of power that a moving object or particle possesses.

An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force. A moving object or particle has kinetic energy, which depends on both its mass and its rate of motion.

Therefore, A spring travelling down a flight of stairs has kinetic energy.

To learn more about Kinetic energy, refer to the link:

https://brainly.com/question/26472013

#SPJ5

1 Therefore, the kinetic frictional force experienced by the block on the incline is 6.784 N.

2 The magnitude of the velocity of the block at point B is approximately 5.11 m/s.

1. The formula for the kinetic frictional force is given by f = μN, where μ is the coefficient of kinetic friction and N is the normal force. Since the block is on an incline, the normal force can be calculated as N = mg * cos(θ), where θ is the angle of inclination.

N = 4 kg * 9.8 m/s² * cos(30°) = 33.92 N

f = 0.2 * 33.92 N

= 6.784 N

2. Potential energy at point A = mgh, where h is the vertical height of the incline.

Potential energy at point A = 4 kg * 9.8 m/s² * 3 m * sin(30°)

= 58.8 J

The work done by friction is given by W = f * d, where d is the distance traveled along the incline (3 m).

Work done by friction = 6.784 N * 3 m = 20.352 J

Since the work done by friction is negative (opposite to the direction of motion), the total work done on the block is:

Total work = Potential energy at A - Work done by friction

Total work = 58.8 J - 20.352 J = 38.448 J

According to the work-energy theorem, this work done on the block is equal to the change in its kinetic energy. Therefore, we have:

38.448 J = 0.5 * 4 kg * B²

Solving for B, we find:

B = √(38.448 J / (0.5 * 4 kg)) ≈ 5.11 m/s

Therefore, the magnitude of the velocity of the block at point B is approximately 5.11 m/s.

Learn more about force on

https://brainly.com/question/12970081

#SPJ1

The net force acting on the future space vehicle be 81750 Newton.

The definition of force in physics is: The push or pull on a mass-containing item changes its velocity.

An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

Given parameters:

Mass of the  future space vehicle: m =  250,00 kilograms.

Net acceleration of the   future space vehicle: a = 3.27 m/s^2.

Hence, according to Newton's 2nd law of motion:

Net force acting on the future space vehicle: F = mass × acceleration

= 25000 × 3.27 Newton

= 81750 Newton.

Hence, net force acting on the future space vehicle be 81750 Newton.

Learn more about force here:

https://brainly.com/question/13191643

#SPJ1

Answer:

Explanation shown below.

Explanation:

1.The number of images formed by 2 parallel mirrors is an infinite number of images.

2. The characteristics of a plane mirror is such that the object distance equals the image distance.

Hence the object distance is 8.7cm from the right; the image formed would be 8.7cm behind the mirror.

Now a second image is going to be formed by the left mirror which is going to have an image distance of 30.1cm behind the mirror.

Now this image would be reflected on the right side to form a new image which is going to be seen as 38.8 +30.1 = 68.9cm behind the right Mirror .

Hence the shortest distances are 8.7cm and 68.9cm

3. The number of reflections is infinite for both cases.

Answer:

1.931 kilometres is the answer of 1.2 miles

Answer and Explanation:

1 mile = 1.609 km

Set up a fraction to cancel the miles to get the kilometers.

\(\frac{1.2mi}{?km} *\frac{1.609}{1mi} = 1.9308km\) <- This is the answer.

#teamtrees #PAW (Plant And Water)

The Title of the interviews Questionnaire is : Community Members' Right to Safe and Healthy Living Questionnaire. The Questionnaire is attached.

A tool for research known as a questionnaire comprises a series of questions formulated to extract data from individuals or a collective of individuals. A systematic approach in acquiring data, which permits researchers to obtain uniform feedback and perspectives from respondents, is termed as structured data collection.

Questionnaires have multiple applications such as conducting surveys, holding interviews, performing assessments, and carrying out evaluations.

Learn more about   questionnaire  from

https://brainly.com/question/25257437

#SPJ1

Answer:

Bath CD jshchdhdhfhfhhfhd jpg de f for frr for gi Jhong GO by be jr jpg be

34 bc of the velocity of the arrow will come down with less dense

Answer:

The ball would be traveling fastest at point (C).

Point (D) appears to be the same height as point (C) but the ball would probably have lost some speed on rebounding,

The radii of the first five Fresnel zones is 3.6 mm.

Distance from the light source to the wave surface, d₁ = 1 m

Distance from the wave surface to the observation point, d₂ = 1 m.

Wavelength of the light used, λ = 5 x 10⁻⁶m = 5 μm

The expression for the radius of the Fresnel zones is given by,

rₙ = √[nλd₁d₂/(d₁ + d₂)]

Therefore, the radii of the first five Fresnel zones is,

r₅ = √[5 x 5 x 10⁻⁶x 1 x 1/(1 + 1)]

r₅ = √(25 x 10⁻⁶/2)

r₅ = 3.6 x 10⁻³m

r₅ = 3.6 mm

To learn more about Fresnel zones, click:

https://brainly.com/question/19672328

#SPJ1

Answer:

This is because the X chromosome is large and contains many more genes than the smaller Y chromosome. In a sex-linked disease, it is usually males who are affected because they have a single copy of X chromosome that carries the mutation.

Explanation:

Centripetal acceleration=  6.25m/s/s for a ball that is attached to 1 m string and swung around in a horizontal circle.

A change in velocity is called acceleration. Centripetal acceleration is the change in velocity due to circular motion . Centripetal acceleration is calculated by taking the square of linear velocity divided by the radius of the circle the object is travelling.

Centripetal acceleration can be calculated when length and speed are given. Centripetal acceleration is measured in meters per second per second (m/s/s) and  the equation can be written as a = v^2 / r.

Centripetal acceleration is greater at high speeds and in sharp curves as can be noticed while driving a car.

To know more about centripetal acceleration, refer

https://brainly.com/question/898360

#SPJ10