It is common to see birds of prey rising upward on thermals. The paths they take may be spiral-like. You can model the spiral motion as uniform circular motion combined with a constant upward velocity. Assume a bird completes a circle of radius 6.00 m every 5.00 s and rises vertically at a rate of 3.00 m/s. Part APart complete Find the speed of the bird relative to the ground. Express your answer using three significant figures. v

Answer:

They are used in periscopes, for signalling, in kaleidoscopes, to see round dangerous bends, in meters, as mirror tiles, in a sextant, in an overhead projector, an SLR camera, car wing mirrors, in microscopes and as reflecting number plates to mention only some!

Explanation:

Hope it is helpful....

Answer:

two uses are:

Answer:

the height of the bridge above the water is 49.6 m.

Explanation:

Given;

initial velocity of the stone, u = 15 m/s

time of motion of the stone, t = 2 s

The height of the bridge above the water is calculated from the following kinematic equation as follows;

h = ut + ¹/₂gt²

h = (15 x 2) + ¹/₂(9.8)(2²)

h = 30 + 19.6

h = 49.6 m

Therefore, the height of the bridge above the water is 49.6 m.

Answer:

Explanation:

Let the tension in ropes be T₁ and T₂ . Ropes are making angle of 52 and 40 degree with the horizontal . The vertical component of tension will add up together to balance the weight of the decoration

T₁ sin52 + T₂ sin40 = 3 x 9.8

.788 T₁ + .643 T₂ = 29.4

The horizontal component of tension will add up to zero because the decoration piece is at rest .

T₁ cos52 + T₂ cos40 = 0

.615 T₁ -  .766 T₂ = 0

T₁ =  1.24 T₂

Substituting this value of T₁ in earlier equation , we have

.788 x 1.24 T₂ + .643 T₂ = 29.4

1.62 T₂ = 29.4

T₂ = 18.15 N

T₁ = 1.24 x 18.15 = 22.51 N .

Explanation:

You do the radius times the circumference of the earth

Answer:

How can you describe the orbital period (planetary year) of each planet?

A year is defined as the time it takes a planet to complete one revolution of the Sun, for Earth this is just over 365 days. This is also known as the orbital period. Unsurprisingly the the length of each planet's year correlates with its distance from the Sun.

Explanation:

Answer: The orbital period is how long it takes a plane to fulfill its  "revolution"(365 days)

Explanation:

Answer:

a)    F = 21.16 N,  b)     a = 3.17 10²⁸ m / s

Explanation:

a) The outside between the alpha particles is the electric force, given by Coulomb's law

          F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]

in that case the two charges are of equal magnitude

          q₁ = q₂ = 2q

let's calculate

         F = [tex]9 \ 10^9 \ \frac{ (2 \ 1.6 \ 10^{-19} )^2 }{ (6.60 \ 10^{-15} )^2 }[/tex]

         F = 21.16 N

this force is repulsive because the charges are of the same sign

b) what is the initial acceleration

         F = ma

         a = F / m

         a = [tex]\frac{21.16}{4.0026 \ 1.67 \ 10^{-27} }[/tex]21.16 / 4.0025 1.67 10-27

         a = 3.17 10²⁸ m / s

this acceleration is in the direction of moving away the alpha particles

Answer:

Explanation:

For resistance of a wire , the formula is as follows

R = ρ L / S

where ρ is specific resistance , L is length and S is cross sectional area

Given L = 14 000 m ,

S = 4.8 x 10⁻⁴ m²

specific resistance of aluminum = 2.8 x 10⁻⁸ ohm-meter

Putting the values in the formula

R = 2.8 x 10⁻⁸ x 14 x 10³ /  (4.8 x 10⁻⁴ )

R = 0.8167 ohm .

= .82 ohm .

Answer:

Explanation:

Let the skateboarder's movement in x -direction be taken into consideration .

a )

initial velocity in x direction u = 0

acceleration a = 2.6 m /s²

time t = .6 s

displacement in x direction

s = ut + 1/2 a t²

= 0 + .5 x 2.6 x .6²

= .468 m

= 46.80 cm

c )

velocity after .6 s

v = u + at

= 0 + 2.6 x .6

= 1.56 m /s

Let the skateboarder's movement in y -direction be taken into consideration .

b ) initial velocity in y direction u = - 4.2 m /s ( velocity is towards south or - y direction )

acceleration a = 0

time t = .6 s

displacement in y direction

s = ut + 1/2 a t²

= - 4.2 x .6  + 0

= - 2.52  m

2.52 m towards south .

d )

velocity after .6 s

v = u + at

= - 4.2  + 0

= - 4.2  m /s

or 4.2 m /s towards south .

Answer:

If the thrust is increased, the aircraft accelerates and the velocity increases. This is the second part sited in Newton's first law; a net external force changes the velocity of the object. The drag of the aircraft depends on the square of the velocity. So the drag increases with increased velocity.

Answer:

vegetables and legumes or beans

fruit

lean meats and poultry, fish, eggs, tofu, nuts and seeds, legumes or beans

grain (cereal) foods, mostly wholegrain or high cereal fibre varieties

milk, yoghurt, cheese or alternatives, mostly reduced fat.

Explanation:

Foods are grouped together because they provide similar amounts of key nutrients. For example, key nutrients of the milk, yoghurt, cheese and alternatives group include calcium and protein, while the fruit group is a good source of vitamins, especially vitamin C.

Answer:

The answer is (A)

Explanation:

We know this because The suns energy is entirely re-radiated back into space.

We know also that the answer is not (D) It makes Earth too hot for plants and animals to survive.

Atmospheric gases are gases located in the Earth's atmosphere

The green house effect is a natural process that warms the Earth's surface

I also know the answer because I took the test... FLVS exam 3.09 right?

I took it and got this question right... So i know you will! GOOD LUCK!!

Mark me brainlyest please:)

Answer:

200

Explanation:

20m/s*10sec=200

Answer:

i hope this helps i did'nt quiet understand the secound one. I hope youcan see my picture well .

Explanation:

sincerily, MEMC3891

Answer:

26.9444m/s

pls brainliest

Answer:

a. 3

Explanation:

Answer:

A. 3

Explanation:

AP3X

Answer:

1.875 x 10⁶ m /s .

Explanation:

Force on electron = E e where E is electric field and e is charge on electron

acceleration generated = Ee / m where m is mass of the electron .

Putting the values

acceleration generated = 5 x 1.6 x 10⁻¹⁹ / 9.1 x 10⁻³¹

= .879 x 10¹² m /s²

v² = u² + 2 as , initial velocity u = 0 , displacement s = 2 m

v² = 0 + 2 x .879 x 10¹² x 2

v = 1.875 x 10⁶ m /s .

Answer:

y = 2.56  10⁻² m

Explanation:

The resolution of this telescope is given by the Rayleigh criterion, for the phenomenal diffraction the first minimum for a linear slit is in

              a sin θ = λ

in general the angles are very small, so we approximate

              sin θ = θ

we substitute

              θ = λ / a

in the case of circular slits we must use polar coordinates, which introduces a numerical factor, leaving the equation

             θ = 1.22 [tex]\frac{\lambda }{D}[/tex]

where D is the diameter of the circular opening

In this case they indicate the lens diameter D = 2.4 m, the observation distance r = 90.4 km = 90.4 10³ m

how angles are measured in radians

            θ = y / r

we substitute

              y / r = 1.22\frac{\lambda }{D}

              y = 1.22 \frac{\lambda r }{D}

let's calculate

             y = [tex]1.22 \frac{ 557 \ 10^{-9} \ \ 90.4 \ 10^{3} }{2.4}[/tex]

             y = 2.56  10⁻² m

this is the minimum distance that can differentiate two objects on Earth

Answer:

[tex]v = 0.7071c[/tex]

Explanation:

Given

Distance to the planet = 35 light years. So, the entire distance is: 2 * 35 = 70.

[tex]\triangle{x'} = 70[/tex]

[tex]T_0 = 70\ years[/tex] i.e time of travel of the ship.

For the observer on earth, the time is:

[tex]T' = \gamma T_0[/tex]

The required speed so that it does not take more than 70 years is then calculated using:

[tex]\triangle x' = vT'[/tex]

Substitute [tex]T' = \gamma T_0[/tex]

[tex]\triangle x' = v\gamma T_0[/tex]

[tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]

So, we have:

[tex]\triangle x' = \frac{vT_0}{\sqrt{1 - v^2/c^2}}[/tex]

Make v the subject of formula.

Square both sides

[tex]\triangle x'^2 = \frac{v^2T^2_0}{1 - v^2/c^2}[/tex]

Cross Multiply

[tex](1 - \frac{v^2}{c^2}) *\triangle x'^2 = v^2T^2_0[/tex]

Divide both sides by [tex]\triangle x'^2[/tex]

[tex](1 - \frac{v^2}{c^2}) = \frac{v^2T^2_0}{\triangle x'^2}[/tex]

Divide through by [tex]v^2[/tex]

[tex](\frac{1}{v^2} - \frac{v^2}{v^2*c^2}) = \frac{v^2T^2_0}{v^2\triangle x'^2}[/tex]

[tex]\frac{1}{v^2} - \frac{1}{c^2} = \frac{T^2_0}{\triangle x'^2}[/tex]

Make [tex]\frac{1}{v^2}[/tex] the subject

[tex]\frac{1}{v^2} = \frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}[/tex]

Inverse both sides

[tex]v^2 = \frac{1}{\frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}}[/tex]

Take square root of both sides

[tex]v = \sqrt{\frac{1}{\frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}}}[/tex]

Substitute values for [tex]T_0[/tex] and [tex]\triangle x[/tex]

[tex]v = \sqrt{\frac{1}{\frac{70^2}{(70c)^2} + \frac{1}{c^2}}}[/tex]

[tex]v = \sqrt{\frac{1}{\frac{70^2}{70^2*c^2} + \frac{1}{c^2}}}[/tex]

[tex]v = \sqrt{\frac{1}{\frac{1}{c^2} + \frac{1}{c^2}}}[/tex]

[tex]v = \sqrt{\frac{1}{\frac{2}{c^2}}}[/tex]

[tex]v = \sqrt{\frac{c^2}{2}}[/tex]

[tex]v = c\sqrt{\frac{1}{2}}[/tex]

[tex]v = c * 0.7071[/tex]

[tex]v = 0.7071c[/tex]

Answer:

3.333

Explanation:

Acceleration is force divided by mass. So divide the force, 2, by the mass, 0.60, and you will get 3.333. I hope this helps :)

The acceleration of an object is the force divided by time. The acceleration of the ball of 0.60 kg when a force of 2N is applied, is 3.33 m/s².

Acceleration of an object is the measure of rate of change in its velocity. Like velocity acceleration is a vector quantity having both magnitude and direction.

Acceleration is defined as the ratio of change in velocity to the change in time. However, according to Newton's second law of motion fore applied on an object is the product of its mass and acceleration.

Hence, F = m a .

Given that, force applied on the ball = 2 N

mass of the ball = 0.60 Kg.

Then acceleration a = force/mass

a = 2 N/ 0.60 Kg = 3.33 m/s²

Therefore, the acceleration of the ball is 3.33 m/s².

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Answer:

31.68 m/s

Explanation:

The law of conservation of energy states that energy is not lost or gained it is just converted, in this example, since it is not given any resistance from the wind, you'd have two variables Speed on the Y-axis and on the X-axis, since both of them would result in the same decrease and increase with against gravity, it doesn't matter the value of both.

As the stone continues to go upwards it will continue to lose speed due to de-acceleration from the gravity acting on it, similarly, it will continue to gain Potential energy, instead of kinetic energy, when it reaches its highest point the speed on Y will be "0" and the free fall will start, since the up and down movement will be equal in time the and acceleration would be equal -9.81 m/s and 9.81 m/s because the only acceleration you have is gravity, you only need to calculate how much speed will gain a rock accelerating at 9.81 m/s falling 20 m:

[tex]H=\frac{1}{2}g*t^{2} \\\frac{20}{1/2*9.81} =t^{2} \\\\t^{2} =4.08\\t=\sqrt{4.08} \\t=2.21[/tex]

Now we just add the time accelerating:

[tex]Vf=Vo+at\\Vf=10 m/s+ 2.21*9.81\\Vf=10 m/s+21.68\\Vf=31.68 m/s[/tex]

Answer:

D

Explanation:

Took it on edg

The speed of the charge at the given magnetic force and field is determined as 1.1 x 10⁴ m/s.

The speed of the charge is calculated as follows;

F = qvBsinθ

v = F/qBsinθ

where;

v = (3.5 x 10⁻²)/(8.4 x 10⁻⁴ x 6.7x 10⁻³ x sin35)

v = 10,842.33 m/s

v ≅ 1.1 x 10⁴ m/s

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Answer:

Explanation:

[tex]KE=\frac{1}{2}mv^2[/tex]

[tex]784=\frac{1}{2}(40)v^2[/tex]

[tex]784=20v^2[/tex]

[tex]39.2=v^2[/tex]

[tex]v=6.26m/s[/tex]

Answer:

Option A. 180000 Kgm/s.

Explanation:

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

For Train Car A:

Mass of train car A = 45000 Kg

Velocity of train car A = 4 m/s

Momentum of train car A =?

For Train Car B:

Mass of train car B = 45000 Kg

Velocity of train car B = 0 m/s

Momentum is simply defined as the product of mass and velocity. Mathematically, it can be expressed as:

Momentum = mass × velocity

With the above formula, the momentum of train car A before collision can be obtained as follow:

Mass of train car A = 45000 Kg

Velocity of train car A = 4 m/s

Momentum of train car A =?

Momentum = mass × velocity

Momentum = 45000 × 4

Momentum of train car A = 180000 Kgm/s

Answer:

c

Explanation: