A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the size of our Moon but has the mass of our Sun.
Estimate gravity on the surface on this star.

I know the solution of this question it is as the picture shows but I only need to add *10^3 to the lower part of the division lower part to get the correct answer. But I don't know why I should add it can anyone explain?

Answer:

cos 0 = 1.

Fs = 7×8 = 56 J

Explanation:

Answer:

Explanation:

Well they told you the exact formula to use. Work is the force multiplied by  the distance through which its applied.

W = (20N)(35m)

= 700 Joules

13.) Power is the amount of work done over the time through which the work is being done.

P = W/t

= 10J/2s

= 5J/s

Answer:

y-coordinate where the monkey catches the coconut is 13.664 m.

Explanation:

Given;

time taken for the monkey to catch the coconut, t = 1.4 s

initial upward speed of the coconut, Uy = 2.9 m/s

acceleration due to gravity, g = 9.8 m/s²

The y-coordinate where the monkey catches the coconut is calculated as;

[tex]h_y = U_yt +\frac{1}{2} gt^2\\\\h_y = (2.9\times 1.4) +\frac{1}{2} (9.8)(1.4^2)\\\\h_y = 4.06 + 9.604\\\\h_y = 13.664 \ m[/tex]

Therefore, y-coordinate where the monkey catches the coconut is 13.664 m.

Answer:

75 rad/s

Explanation:

The angular acceleration is the time rate of change of angular velocity. It is given by the formula:

α(t) = d/dt[ω(t)]

Hence: ω(t) = ∫a(t) dt

Also, angular velocity is the time rate of change of displacement. It is given by:

ω(t) = d/dt[θ(t)]

θ(t) = ∫w(t) dt

θ(t) = ∫∫α(t) dtdt

Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:

θ(t) = ∫∫α(t) dtdt

θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt

θ(t) = ∫[2t³]dt = t⁴/2 rad

θ(t) = t⁴/2 rad

At θ(t) = 10 rev = (10 *  2π) rad = 20π rad, we can find t:

20π = t⁴/2

40π = t⁴

t = ⁴√40π

t = 3.348 s

ω(t) = ∫α(t) dt = ∫6t² dt = 2t³

ω(t) = 2t³

ω(3.348) = 2(3.348)³ = 75 rad/s

The required value of angular velocity of wheel is 75 rad/s.

Given data:

The angular acceleration of the wheel is, [tex]\alpha (t) = 6.0 t^{2} \;\rm rad/s^{4}[/tex].

The turning rate of wheel is, n = 10 rev.

The angular acceleration is the time rate of change of angular velocity. It is given by the formula:

α(t) = d/dt[ω(t)

Clearly, the angular velocity is the single integral of angular acceleration.  Then,

ω(t) = ∫a(t) dt

Also, angular velocity is the time rate of change of displacement. It is given by:

ω(t) = d/dt[θ(t)]

θ(t) = ∫w(t) dt

θ(t) = ∫∫α(t) dt dt

Since,

α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴.

Then solve by substituting the values as,

θ(t) = ∫∫α(t) dt dt

θ(t) = ∫∫6t² dt dt

θ(t) =∫[∫6t² dt]dt  

θ(t) = ∫[2t³]dt = t⁴/2 rad

θ(t) = t⁴/2 rad

At θ(t) = 10 rev = (10 *  2π) rad = 20π rad, we can find t:

20π = t⁴/2

40π = t⁴

t = ⁴√40π

t = 3.348 s

Also,  

ω(t) = ∫α(t) dt = ∫6t² dt = 2t³

ω(t) = 2t³ ......................................................(1)

Substitute the value of time in equation (1) as,

ω( t = 3.348) = 2(3.348)³ = 75 rad/s

Thus, we can conclude that the required value of angular velocity of wheel is 75 rad/s.

Learn more about the angular velocity here:

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Answer:

7.5Watts

Explanation:

Given parameters:

Force of lift  = 25N

Height  = 3.6m

Time  = 12s

Unknown:

Power output  = ?

Solution:

Power is the rate at which work is done ;

     Power  = [tex]\frac{force x height }{time}[/tex]  

   Power  = [tex]\frac{25 x 3.6}{12}[/tex]    = 7.5Watts

Answer:

22.5 m

Explanation:

Using v² = u² - 2gy where u = initial velocity of BB = 21 m/s, v = final velocity of BB = 1 m/s (since this is the required speed of BB in which it will not harm the birds), g = acceleration due to gravity = 9.8 m/s² and y = minimum height of BB above the gun at which the birds may safely fly.

Substituting the values of the variables into the equation, we have

v² = u² - 2gy

(1 m/s)² = (21 m/s)² - 2(9.8 m/s²)y

collecting like terms, we have

(1 m/s)² - (21 m/s)² = - 2(9.8 m/s²)y

1 m²/s² - 441 m²/s² = -(19.6 m/s²)y

simplifying, we have

- 440 m²/s² = -(19.6 m/s²)y

dividing through by -19.6 m/s², we have

y = - 440 m²/s² ÷ -19.6 m/s²

y = 22.45 m

y ≅ 22.5 m

Answer:

A. Retina

Explanation:

Answer:

its always active

Explanation:

Answer:

104.59 m

Explanation:

From the question given above, the following data were obtained:

Horizontal velocity (u) = 13 m/s

Horizontal distance (s) = 60 m

Height of cliff =?

Next, we shall determine the time taken for the car to hit the ground. This can be obtained as follow:

Horizontal velocity (u) = 13 m/s

Horizontal distance (s) = 60 m

Time (t) =?

s = ut

60 = 13 × t

Divide both side by 13

t = 60 /13

t = 4.62 s

Finally, we shall determine the height of the cliff. This can be obtained as follow:

Time (t) = 4.62 s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

h = ½gt²

h = ½ × 9.8 × 4.62²

h = 4.9 × 21.3444

h = 104.59 m

The, the height of the cliff is 104.59 m

Answer:

B. The rod must be an insulator

Explanation:

We have that this rod must be an insulator. After this child rubbed the balloon, the balloon acquired static charge. So holding a rid against it is going to cause it to repel, this is to say it is repelling because the rod also is carrying some static charges. If this rid was a conductor, there would be no charge in its surface. The charge would have passed through her hand as it comes in contact.

Answer:

550V

Step  - Down transformer

Explanation:

Given parameters:

Number of turns in primary coil  = 300 turns

Secondary turns  = 120 turns

Output voltage  = 220V

Unknown:

Input voltage  = ?

Type of transformer =  ?

Solution:

To solve this problem, we use the expression below:

    [tex]\frac{V_{out} }{V_{in} }[/tex]    = [tex]\frac{Ns}{Np}[/tex]  

So insert the parameters and find Vin;

      [tex]\frac{220}{Vin}[/tex]   = [tex]\frac{120}{300}[/tex]  

    120Vin  = 220 x 300

           Vin  = [tex]\frac{220 x 300}{120}[/tex]   = 550V

Since the input voltage is greater than the output voltage, this is step - down transformer.

Answer:

mining

Explanation:

because they use the most heavy duty machines

Answer:

a)

Mass flow rate of core =  [tex]m_{e}[/tex] = 60.94 Kg/s

Mass flow rate of fan =  [tex]m_{s}[/tex]  = 73.12 kg/s

TSFC = 3.301 x [tex]10^{-5}[/tex]

b)

Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]

Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]

Explanation:

Data Given:

Height = 25000 ft

Vehicle velocity = [tex]u_{a}[/tex] = 815 ft/s = 248.41 m/s

[tex]m_{s} = 1.2m_{e}[/tex]

[tex]m_{f}[/tex] = 0.0255[tex]m_{e}[/tex]  

Where,

[tex]m_{s}[/tex] = Mass flow rate of fan

[tex]m_{e}[/tex] = Mass flow rate of core

F = Thrust

Density of core = [tex]D_{e}[/tex] = 0.000578 slugs/[tex]ft^{3}[/tex] = 0.2979 kg/[tex]m^{3}[/tex]

Density of fan = [tex]D_{s}[/tex] = 0.00154 slugs/[tex]ft^{2}[/tex] = 0.7937 kg/[tex]m^{3}[/tex]

Ambient Pressure of Fan = [tex]P_{s}[/tex] = 10.07 Psi = 69430.21 Pa

Ambient Pressure of core = [tex]P_{e}[/tex] = 10.26 Psi = 70740.2 Pa

Thrust = F = 10580 lbf = 47062.2 N

Velocity of fan = [tex]u_{s}[/tex] = 1147 ft/s = 349.6 m/s

Velocity of core = [tex]u_{e}[/tex] = 1852 ft/s = 564.5 m/s

At the height of 25000 ft, P = 37600 [tex]P_{a}[/tex]

Now,

we have:

[tex]m_{e}[/tex] = [tex]u_{e}[/tex] x  [tex]D_{e}[/tex]  x [tex]A_{e}[/tex]

Plugging in the values, we get:

[tex]m_{e}[/tex]  = 168.16 [tex]A_{e}[/tex]   Equation 1

And,

[tex]m_{s}[/tex]  = [tex]D_{s}[/tex]  x [tex]A_{s}[/tex] x  [tex]u_{s}[/tex]

[tex]m_{s}[/tex]  = 277.5 [tex]A_{s}[/tex]  Equation 2

As, we know,

[tex]m_{s} = 1.2m_{e}[/tex]  

[tex]m_{s}[/tex]  = 277.5 [tex]A_{s}[/tex]

And now for Thrust, we have:

F = [tex]A_{e}[/tex] x ([tex]P_{e}[/tex]  - [tex]P_{a}[/tex] ) + [tex]A_{s}[/tex] x ([tex]P_{s}[/tex] - [tex]P_{a}[/tex] ) + [tex]m_{e}[/tex]x ([tex]u_{e}[/tex]  - [tex]u_{a}[/tex] ) + [tex]m_{s}[/tex] x ([tex]u_{s}[/tex]  -  [tex]u_{a}[/tex] ) Equation 3

Now, substitute equation 1 and 2 in equation 3, we get:

Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]

Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]

Mass flow rate of core =  [tex]m_{e}[/tex] = 60.94 Kg/s

Mass flow rate of fan =  [tex]m_{s}[/tex]  = 73.12 kg/s

TSFC = Thrust Specific Fuel Consumption = fuel mass flow rate  / Thrust

TSFC = [tex]m_{f}[/tex]/F

And,

[tex]m_{f}[/tex] = 0.0255[tex]m_{e}[/tex]  

[tex]m_{e}[/tex]   =  60.94

[tex]m_{f}[/tex]  = 0.0255 x 60.94

[tex]m_{f}[/tex]  = 1.55397

TSFC = [tex]m_{f}[/tex]/F

TSFC = 1.55397/47062.2

TSFC = 3.301 x [tex]10^{-5}[/tex]

Low TSFC = High efficiency

High TSFC = Low efficiency

a)

Mass flow rate of core =  [tex]m_{e}[/tex] = 60.94 Kg/s

Mass flow rate of fan =  [tex]m_{s}[/tex]  = 73.12 kg/s

TSFC = 3.301 x [tex]10^{-5}[/tex]

b)

Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]

Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]

Answer:

15km/h

Explanation:

→ Speed = Distance ÷ Time

30 ÷ 2 = 15km/h

Answer:

Mass and Acceleration

Explanation:

The typical Force equation is:

F = ma

where m = mass, and a=acceleration.

Answer:

Gravity is the main force responsible for mass movements. Gravity is a force that acts everywhere on the Earth's surface, pulling everything in a direction toward the center of the Earth

Explanation:

Answer:

a) Time period is 3.3 seconds

b) The frequency is 0.3030 Hz

c) amplitude is 0.25 m

d) maximum speed is 0.476 m/s

Explanation:

Given the data in the question;

a) Period

Time Period T = Time taken for one oscillation

T = 33s / 10 = 3.3 seconds

Therefore, Time period is 3.3 seconds

b) Frequency

we know that frequency is the inverse of time period

so;

Frequency f = 1/T = 1 / 3.3 s

Frequency f = 0.3030 Hz

Therefore, The frequency is 0.3030 Hz

c) amplitude

amplitude A = [tex]\frac{1}{2}[/tex]( 60 cm - 10 cm )

A = [tex]\frac{1}{2}[/tex] × 50 cm

A = 25 cm

A = 0.25 m

Therefore, amplitude is 0.25 m

d) maximum speed of the glider

maximum speed [tex]V_{max}[/tex] = ωA

and ω = 2π/T

so maximum speed [tex]V_{max}[/tex] = [tex]\frac{2\pi }{T}[/tex]A

so we substitute

so maximum speed [tex]V_{max}[/tex] = [tex]\frac{2\pi }{3.3}[/tex] × 0.25 m

so maximum speed [tex]V_{max}[/tex] = 0.476 m/s

Therefore, maximum speed is 0.476 m/s

The time period is 3.3 s, frequency is 0.303 Hz, amplitude is 0.25 m, and maximum speed is 0.47575 m/s.

Based on the given information,

• The air-track glider connected with a spring oscillates between the 10 cm mark and the 60 cm mark.  

• In 33 seconds, the glider completes 10 oscillations.  

There is a need to find, the period, frequency, amplitude, and maximum speed of the glider.

a) Time period (T) for one oscillation is,

[tex]\frac{33s}{10} = 3.3 s[/tex]

b) The frequency (f) is the reciprocal of the time period,

[tex]f = \frac{1}{T} =\frac{1}{3.3S} = 0.303 Hz[/tex]

c) The amplitude (A) is,

[tex]A = \frac{1}{2} (60 cm-10cm) = 25cm[/tex]

[tex]A = 0.25 m[/tex]

d) Maximum speed of the glider is,

[tex]Vmax = \frac{2\pi }{T} (0.25 m)[/tex]

[tex]Vmax = 0.47575 m/s[/tex]

Thus,  time period is 3.3 s, frequency is 0.303 Hz, amplitude is 0.25 m, and maximum speed is 0.47575 m/s.

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Answer:

B

Explanation:

I'm pretty sure I the third option C.

Explanation:

sorry if I'm wrong

Answer:

it will option option A hope it helps

Answer:

a) ω₁ = ω₂ = 3.7 rad/sec

b) Δθ₁ = Δθ₂ = 18.5 rad

c) d₁ = 14.5 m  d₂ = 57.5 m

d) Fc1 = 273.9 N Fc2 = 1069.8 N

e) The boy near the outer edge.

Explanation:

a)

       [tex]v = \omega*r (1)[/tex]

       [tex]\omega = \frac{v_{out} }{r_{out} } = \frac{11.5m/s}{3.14m} = 3.7 rad/sec (2)[/tex]

b)

        ⇒  Δθ₁ = Δθ₂ = 18.5 rad.

c)

       [tex]v_{inn} = \omega * r_{inn} = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)[/tex]

      vout is a given of the problem ⇒ vout = 11. 5 m/s

       [tex]d_{inn} = v_{inn} * t = 2.9m/s* 5.0 s = 14.5 m (5)[/tex]

      [tex]d_{out} = v_{out} * t = 11.5 m/s* 5.0 s = 57.5 m (6)[/tex]

d)

       [tex]F_{c} = m*\frac{v^{2}}{r} (7)[/tex]

      [tex]F_{cin} = m*\frac{v_{in}^{2}}{r_{in}} = 25.4 kg* \frac{(2.9m/s)^{2} }{0.78m} = 273.9 N (8)[/tex]

      [tex]F_{cout} = m*\frac{v_{out}^{2}}{r_{out}} = 25.4 kg* \frac{(11.5m/s)^{2} }{3.14m} = 1069.8 N (9)[/tex]

e)

Answer:

the rate at which the speaker produces energy is 510.51 J/s

Explanation:

Given;

sound intensity of the speaker, I = 6.5 W/m²

distance, r = 2.5 m

The rate at which the speaker produces energy on a spherical surface is given as;

P = IA

P = I(4πr²)

P = 6.5 (W/m²) x 4π x (2.5 m)²

P = 510.51 J/s

Therefore, the rate at which the speaker produces energy is 510.51 J/s

Answer:

It means when you look into the lens your vision magnifies by x1

Explanation:

Answer: 350 ms

Explanation:

Just took the quiz:)

Answer:

B. 350

Explanation:

Answer:

0.0003

Explanation:

In the rules of Sig Figs, all zeros before with decimals are not sigificant. I.E. 0.00000000000000009. Despite how many 0's there are, only the 9 is significant. Zeros before a number is not significant. In 100, only the one is signficant in 100. with a dot at the end, the one and the two zeros are significant. hope this helps.

Answers:

the second option

Explanation:

Answer:

The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.

Explanation:

Let suppose that RC car-slingshot-steelpellet is a conservative system, that is, that non-conservative forces (i.e. friction, air viscosity) can be neglected. The velocity of the steel pellet can be found by means of the Principle of Energy Conservation and under the consideration that change in gravitational potential energy is negligible and that the RC car travels at constant velocity:

[tex]\frac{1}{2}\cdot (m_{C}+m_{P})\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{C}\cdot v_{o}^{2} + \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]

[tex]\frac{1}{2}\cdot m_{P}\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]

[tex]m_{P}\cdot v_{o}^{2} + k\cdot x^{2} = m_{P}\cdot v^{2}[/tex]

[tex]v^{2} = v_{o}^{2} + \frac{k}{m_{P}}\cdot x^{2}[/tex]

[tex]v = \sqrt{v_{o}^{2}+\frac{k}{m_{P}}\cdot x^{2} }[/tex] (1)

Where:

[tex]v_{o}[/tex] - Initial velocity of the steel pellet, measured in meters per second.

[tex]v[/tex] - Final velocity of the steel pellet, measured in meters per second.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]m_{P}[/tex] - Mass of the steel pellet, measured in kilograms.

[tex]m_{C}[/tex] - Mass of the RC car, measured in kilograms.

[tex]x[/tex] - Initial deformation of the spring, measured in meters.

If we know that [tex]v_{o} = 5.6\,\frac{m}{s}[/tex], [tex]k = 85\,\frac{N}{m}[/tex], [tex]m_{P} = 0.025\,kg[/tex] and [tex]x = 0.035\,m[/tex], then the velocity of the steel pellet relative to the ground when it leaves the sling shot is:

[tex]v = \sqrt{\left(5.6\,\frac{m}{s} \right)^{2}+\frac{\left(85\,\frac{N}{m} \right)\cdot (0.035\,m)^{2}}{0.025\,kg} }[/tex]

[tex]v \approx 5.960\,\frac{m}{s}[/tex]

The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.

Answer:

The influence of Media and Information over elections, politics and, governance.    

Explanation:

Once we used to have politicians travel to interact with citizens and meet them face to face. That is slowly fading away in place of townhalls that take place online.

Video conferencing, video calls, social media apps, and technology now make it easy to connect with a lot of people at once, across various timezones, without changing location.

Once upon a time, the Television media, Radio Media, Cable Television (all categorized as traditional media) used to be the custodian of information, hence, the chief influencers of politics and governance.

Today, the internet has made it possible for individuals or a group of people without much tech-spend to set up and garner a following that has enough power to influence governance.

Online petitions and Wikileaks, the proliferation of terrorist organizations that have strong information and technology as well as social media skills have modified forever, the way public office holders and other stakeholders involved used to look at governance.

Answer:

v = 8.1 m/s

θ = -36.4º (36.4º South of East).

Explanation:

        [tex]p_{ox} = p_{fx} (1)[/tex]

         ⇒ [tex]m_{tr} * v_{tr}= (m_{olds} + m_{tr}) * v_{fx} (2)[/tex]

       [tex]v_{fx} = \frac{m_{tr}*v_{tr} }{(m_{tr} + m_{olds)} } = \frac{4146kg*9.7m/s}{2028kg+4146 kg} = 6.5 m/s (3)[/tex]

        [tex]p_{oy} = p_{fy} (4)[/tex]

        ⇒[tex]m_{olds} * v_{olds}= (m_{olds} + m_{tr}) * v_{fy} (5)[/tex]

       [tex]v_{fy} = \frac{m_{olds}*v_{olds} }{(m_{tr} + m_{olds)} } = \frac{2028kg*(-14.5)m/s}{2028kg+4146 kg} = -4.8 m/s (6)[/tex]

       [tex]v_{f} = \sqrt{v_{fx} ^{2} +v_{fy} ^{2} }} = \sqrt{(6.5m/s)^{2} +(-4.8m/s)^{2}} = 8.1 m/s (7)[/tex]

       [tex]\frac{v_{fy} }{v_{fx} } = tg \theta (8)[/tex]

      ⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)

      ⇒ θ = tg⁻¹ (-0.738) = -36.4º

Answer:

Personal attitudes play a role in conflict resolution and include prejudices, biases, prior experience, and the level of importance of a topic to the parties involved. It is important to note that you can control your own attitude, but you cannot control the attitudes of others; it is necessary to focus on your attitudes and assumptions before you address the other party's. Next, you need to know what the nature of the conflict is, or what the conflict is about. It is necessary to know how severe the conflict is, how easily it can be resolved, how much sacrifice it will take to overcome, and how heated it is likely to get. Lastly, one needs to understand the capabilities of the people involved. Is everyone on the same level and do they have what they need to work on the conflict, i.e. money, time, and support?

Explanation:

Answer:

The guy above is correct I'm just trying to make the post more reliable if two people say it then it's good plus I checked also Thanks other guy.

Explanation:

The element that has the same number of valence electrons as krypton (Kr) is known as Neon (Ne). Thus, the correct option for this question is C.

Valence electrons may be defined as the number of electrons present in the outermost shell of any atom during the time of electronic configuration. For example, Sodium has an atomic number of 11, its electronic configuration is 2, 8, and 1. It means that the valence electron on sodium is 1.

Neon and Krypton being the members of the same family which is known as noble gases or inert gases occupy the same valence electrons of eight in their outermost shell. It is group 8A of the periodic table. Apart from Ne, and Kr, other elements like argon, Xenon, and Radon also belong to the same family and occupy the same valence electron except for helium.

Therefore, the element that has the same number of valence electrons as krypton (Kr) is known as Neon (Ne). Thus, the correct option for this question is C.

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