How long does it take a plane, traveling at a constant speed of 200 m/s, to fly once around a circle whose radius is 2750 m

Answer:

2hr 21 min

Explanation:

0 days 2 hours 20 minutes 0 seconds

+ 0 days 0 hours 0 minutes 60 seconds

= 0 days 2 hours 21 minutes 0 seconds

Answer:

physical science

Explanation:

i guess??

P= 10m = 10 x 115 = 1150 N

work done while lifting a 115 kg sofa 1.5 m off the ground :

A= Ph = 1150 x 1.5 = 1725 J

P/s: I think the weight first is 115kg

ok done. Thank to me :>

Answer:

it is correct

Explanation:

Though no rounded numbers can be defined as accurate, if we were going by people's discovery, and research, we can define that the number, g = 9.8m/s^2, is accurate

Answer:

[tex]4 ms^{-2}[/tex]

Explanation:

Normal force - should be the reaction from the surface it's on - nullifies the weight of the object. At this point the only force is of [tex]22-14 = 8N[/tex] from whoever is pushing the square and the friction, towards the right. At this point we divide the force by the mass of the object to obtain an acceleration of [tex]8/2 = 4 ms^{-2}[/tex]

Answer:

the electric energy travels through the plug or wire or whatever is connecting the energy source to the lightbulb, which conducts electricity, and to the light bulb. the electric energy is transfered into light energy and thermal energy

Explanation:

i hope i understood this question and answered it correctly XD

Answer:

46 degrees

Explanation:

Answer: Physical changes only change the appearance of a substance, not its chemical composition. Chemical changes cause a substance to change into an entirely substance with a new chemical formula. Chemical changes are also known as chemical reactions.

Answer:

just give the other person brainlyest

Explanation:

Answe16000/3500(-1.5-2-)225000= 1.1

Explanation:

The mass of the mixture will be 2.19 ×10⁻³ kg. The unit of mass is kg and it is a scalar quantity.

Mass is a numerical measure of inertia, which is a basic feature of all matter. It is, in effect, a body of matter's resistance to a change in speed or position caused by the application of a force.

In the International System of Units (SI), the kilogram is the unit of mass.

The given data in the problem is;

The initial temperature of the mixture was, [tex]\rm T_i = 20 ^0 \ C[/tex]

The mixture froze at the temperature of, [tex]\rm T_F = -1.5 ^0 \ C[/tex]

The internal energy is,[tex]\rm E = \rm 165 \ kJ[/tex]

The specific heat capacity of the mixture is,[tex]\rm C = 3500 J/kg ^0C[/tex]

The specific latent heat of fusion of the mixture is,[tex]\rm l = 255 000 J/kg[/tex]

m is the mass of the mixture,

The latent heat is equal to the heat transfer. Because the net work done is zero.

The formula for the heat transfer is found as;

[tex]\rm Q = mCdt \\\\ m = \frac{Q}{Cdt} \\\\ m = \frac{165 }{3500(20-(-1.5)} \\\\ m = 2.19 \times 10^{-3} \ kg[/tex]

Hence the mass of the mixture will be 2.19 ×10⁻³ kg.

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Light - emitting diode lightbulbs known as LED lightbulbs are energy saving lightbulbs. They consume very small amount of energy compare to other lightbulbs. Therefore, LED lightbulbs are the most energy efficient way to light your home, as they use about ten percent (10%) of the energy of a comparable compact fluorescent bulb.

There are varieties of different light bulbs. some of them are  incandescent, compact fluorescent (CFL), halogen, and light-emitting diode (LED) to mention a few.

Before the advent of light - emitting diode (LED) bulbs, some of the bulbs consume a lot of energy. For instance, we have incandescent bulbs of 20 watts, 50 watts, 100 watts and even 200 watts.

But with the advent of light - emitting diode (LED) bulbs, consumers can save energy because light - emitting diode (LED) bulbs save a lot of energy and they are very bright.

We can therefore conclude that light - emitting diode lightbulbs known as LED lightbulbs are the most energy efficient way to light your home, as they use about ten percent (10%) of the energy of a comparable compact fluorescent bulb.

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Formulas change from F to degree C : C = (F – 32) / 1.8

so we have (176-32) /1.8= 80 oC

Formulas change from F to degree K : K = 5 / 9 x (F – 32) + 273.15

so we have 5/9 x (176-32) + 273.15 = 353 oK

ok done. Thank to me :>

Answer:

5.7km/h north and 5.8km/h west are instantaneous speeds, while 8.1km/h Northwest is the average speed.

Answer:

it might be b maybe

Explanation:

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

[tex] E_{t} = KE_{i} + PE_{i} [/tex]

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

[tex] E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J [/tex]

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

[tex] PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J [/tex]

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

[tex] KE_{A} + PE_{A} = KE_{B} + PE_{B} [/tex]

[tex] KE_{B} = KE_{A} + PE_{A} - PE_{B} [/tex]

Since

[tex] KE_{A} + PE_{A} = KE_{i} + PE_{i} [/tex]

we have:

[tex] KE_{B} = KE_{i} + PE_{i} - PE_{B} =  19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J [/tex]

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

[tex] PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J [/tex]

5) The kinetic energy of the roller coaster at point C is:

[tex] KE_{i} + PE_{i} = KE_{C} + PE_{C} [/tex]            

[tex] KE_{C} = KE_{i} + PE_{i} = 19820 J [/tex]      

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

[tex] KE_{C} = \frac{1}{2}mv_{C}^{2} [/tex]

[tex] v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s [/tex]

Hence, the velocity of the roller coaster at point C is 19.91 m/s.

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The conservation of the momentum allows to find the velocity of the second body after the elastic collision is:

           [tex]v_f = \frac{2u_o}{1- \frac{m_2}{m_1} }[/tex]  

the momentum is defined by the product of the mass and the velocity of the body.

        p = mv

The bold letters indicate vectors, p is the moment, m the mass and v the velocity of the body.

If the system is isolated, the forces during the collision are internal and the it  is conserved. Let's find the momentum is two instants.

Initial instant. Before crash.

      p₀ = m₁ u₀ + 0

Final moment. After crash.

      [tex]p_f = m_1 u_f + m_2 v_f[/tex]  

The momentum is preserved.

      p₀ = [tex]p_f[/tex]  

      [tex]m_1 u_o = m_1 u_f + m_2 v_f[/tex]  

Since the collision is elastic, the kinetic energy is conserved.

      K₀ = [tex]K_f[/tex]

      ½ m₁ u₀² = ½ m₁ [tex]u_f^2[/tex]  + ½ m₂  [tex]v_f^2[/tex]  

Let's write our system of equations.

       [tex]m_1 u_o = m_1 u_f + m_2 v_f \\m_1 u_o^2 = m_1 u_f^2 + m_2 v_f^2[/tex]

Let's solve

       [tex]u_f = u_o - \frac{m_2}{m_2} \ v_f \\u_f^2 = u_o^2 - \frac{m_2}{m_1} \ v_f^2[/tex]

       [tex]( u_o - \frac{m_2}{m_1} v_f)^2 = u_o - \frac{m_2}{m_1} \ v_f^2 \\u_o^2 - 2 \frac{m_2}{m_1} \ u_o v_f + (\frac{m_2}{m_1} )^2 v_f^2 = u_o^2 - \frac{m_2 }{m_1} \ v_f^2[/tex]  

        [tex]2 \frac{m_2}{m_1} \ u_o = \frac{m_2}{m_1} v_f \ ( 1 - \frac{m_2}{m_1}) \\v_f = \frac{2u_o}{1-\frac{m_2}{m_1} }[/tex]

In conclusion, using the conservation of momentum, we can find the velocity of the second body after the elastic collision is:

           [tex]v_f = \frac{2u_o}{1-\frac{m_2}{m_1} }[/tex]  

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[tex]\huge \bf༆ Answer ༄[/tex]

First, find the force exerted by the box.

[tex] \sf \: F = ma [/tex]

The required terms are :

Calculate the force : -

_____________________________________

Pressure can be calculated using this formula : -

The stated terms are :

Calculate the Pressure now : -

[tex]꧁  \:  \large \frak{Eternal \:  Being } \: ꧂[/tex]

Answer:

please check the image

Explanation:

I hope this image helps you please follow the steps. Thank you

Answer:

1kg/```````2

Explanation:

solution

The amount of work done to stop the car is equal to 322,099.85 Joules.

Given the following data:

Conversion:

Velocity = 95 km/h = [tex]\frac{95 \times 1000}{60 \times 60} = 26.39 \;m/s^2[/tex]

To determine the amount of work done to stop the car, we would apply the work-energy theorem:

Note: The amount of work done must balance the kinetic energy possessed by the car due to its motion.

From the work-energy theorem, we have:

[tex]Work\done = \Delta K.E\\\\Work\done = \frac{1}{2} MV^2[/tex]

Substituting the given parameters into the formula, we have;

[tex]Work\;done = \frac{1}{2} \times 925 \times (26.39)^2\\\\Work\;done =462.5 \times 696.4321[/tex]

Work done = 322,099.85 Joules

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Answer:

at the center of the galaxy

Explanation:

as there r more new born stars found

Answer:

No work is done on the weight. :3

To solve this we must be knowing each and every concept related to work and its calculation. Therefore, 250J is the work done in joule by the rope on the weight.

Work in physics is the energy delivered to or out of an item by applying force across a displacement. It is frequently expressed in its most basic form as the combination of displacement and force.

When a force is applied, it is said to produce positive work if it has a proportion in the orientation of the movement of a site of application. Work done is positive when the direction of force acting on the object and displacement of the object both are in the same direction.  

Mathematically,    

Work= force ×displacement

force =20 N

displacement=2.00 m

displacement=2×3.14×2=12.5m

substituting all the given values in the above equation, we get

Work= 20 ×12.5

Work =250J

Therefore, 250J is the work done in joule by the rope on the weight.

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Answer:

C

Explanation:

Fusion and fission are similar in that they both release large amounts of energy. Nuclear fusion is a process in which two nuclei join to form a larger nucleus. Nuclear fission is a process in which a nucleus splits into two smaller nuclei.

Answer:

The formula for potential energy depends on the force acting on the two objects. For the gravitational force the formula is P.E. = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m / s2 at the surface of the earth) and h is the height in meters.

Answer:

Velocity/time =acceleration

Acceleration x time = velocity

Velocity/acceleration=time

we know 1kg=9.8N​ so 30 kg= 30 x 9.8 = 294 N

work is done when a 30 kg mass is to be lifted through a height 6m :

A = 294 x 6 = 1764 J

ok done. Thank to me :>

Answer:

it is 1 / square root of the change in distance

Explanation:

since the equation of the gravitational force is

[tex]F=G{\frac{m_1m_2}{r^2}}[/tex]

then F and r are inversely related, by the equation

[tex]F \alpha \frac{1}{r^2}[/tex]

this means when F is multiplied by 2 then r^2 is multiplied by half

so when r is multiples by half we first square the number (0.5) ^ 2 = 0.25

so

[tex]F \alpha \frac{1}{(0.25)r^2}[/tex]

which means F is multiples by 1/0.25 or 4 when the radius is halfed

Answer:

18.66m

Explanation:

This was actually fun! Let set up the problem first: As you drop the sphere, it will accelerate till it hits the water with a given speed [tex]\dot z[/tex]. Once sinking the ball is subjected to two forces: its own weight [tex]m_sg = \rho_sVg[/tex] directed towards the bottom of the lake, and buoyancy (Archimedes law), ie a force upward equal to the volume of displaced water, [tex]m_wg=\rho_wVg[/tex]. At this point it's the classical "how high can I toss a ball before it falls down" with a water twist. Please note that I'm using z as the height of the sphere, and [tex]\dot z; \ddot z[/tex] represent the velocity (with one dot) and the acceleration (with two dots) in the z direction .Let's assume the sign of all quantities being positive if directed upwards, and negative if towards the bottom of the lake.

Let's first determine how fast the ball hit the water. For me the easiest way is saying "at 8m it has a given potential energy, and 0 kinetic energy. When it hits the water it loses all potential energy and has only kinetic energy". In numbers:

[tex]m_sgz=\frac12m_s\dot z_0^2[/tex] Let's divide by the masses, g is a known value ([tex]9.81 ms^{-2}[/tex]), z is 8 meters, and we get that [tex]\dot z_0 = 4\sqrt g =12.52 m/s[/tex]

At this point, let's determine the force acting on the sphere, courtesy of Newton second law and the debate we had earlier:

[tex]m_s \ddot z = -m_sg + m_wg\\\rho_sV \ddot z = -\rho_sVg + \rho_wV g[/tex]

At this point we can divide by the volume of the sphere, and make use of the fact that [tex]\rho_s = 0.7 \rho_w[/tex]

[tex]0.7\rho_w \ddot z = -.7\rho_w g+ \rho_wg \rightarrow 0.7 \ddot z = 0.3 g\\\ddot z = \frac {0.3}{0.7} g= \frac 37g = 4.2 ms^-2[/tex]

We're almost there. At this point we can write the law of motion for the sphere.

[tex]\dot z = \dot z_0 + \ddot z t[/tex] that we will use to find the time for the ball to stop sinking - ie reaches 0 velocity.

[tex]0 = -12.52 +4.2t \rightarrow t= 12.52/4.2 \approx 3s[/tex]

At this point we can use the other law of motion to find out the distance traveled

[tex]z= z_0 + \dot z_ot + \frac12 \ddot z t^2\\z= 0 -12.52 (3) + \frac12 (4.2)(3)^2 = -37.56+18.90 = -18.66 m[/tex]

as usual, please double and triple check all the calculations, it's almost 1.30 am here and I am not the most confident number cruncher even when fully awake.

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The velocity for the entire trip is 0.4 m/s as It takes her 500 seconds to make the round trip and 60 kg • m/s2 = (45 kg • v) + (20 kg • 3 m/s).

Path 1 = 400 m В B. A Path 2 = 200 m Path 3 = 300 m. Thus, option C is correct.

A particle's settling velocity known as the rate at which is travels through a still fluid. The specific gravity of the particles, their size, and their shape all have an impact on settling velocity.

A particle in still air will gravitationally settle and reach its terminal velocity fairly quickly. A particle's terminal velocity in a still fluid is referred to as the settling velocity (also known as the "sedimentation velocity").

Understanding variations in the hydraulic regime and interactions between sediment and fluid in the surf zone depends heavily on the particle settling velocity at the foreshore region. In contrast to sedimentation, which is the end product of the settling process, settling is the movement of suspended particles through the liquid.

Therefore, Thus, option C is correct.

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