Mechanical advantage allows you to apply a force over a what distance to what the distance an object moves

In an experiment, a shearwater (a seabird) was taken from its nest, flown a distance 5220km away, and released. It found its way

denis-greek [22]

Answer:

-4.72005 m/s

0 m/s

Explanation:

Displacement = 5220 km

Time = 12.8 days

The average velocity is given by

$v_a=\dfrac{0-5220\times 10^3}{12.8\times 24\times 60\times 60}\\\Rightarrow v_a=-4.72005\ m/s$

The average for the return flight is -4.72005 m/s

When the total displacement is divided by the total time of a journey then we get the average velocity.

Displacement is the minimum distance between the initial and final points of the journey.

Here, the displacement of the whole episode is 0 as the initial and final point is zero.

Hence, the average velocity for the whole episode is 0 m/s

3 0

2 years ago

When light strikes an opaque material, which of the following accurately describes what happens to the light rays? Some of the l

ExtremeBDS [4]

None of the light passes through it; some of the light is absorbed as heat but most is reflected off the surface. This is how you see objects. reflected light from them hits your eye. (Opaque means not transparent)

8 0

3 years ago

Read 2 more answers

In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 3.40 mm . what is the differenc

Strike441 [17]

The solution would be like this for this specific problem:

Given:

diffraction grating slits = 900 slits per centimeter

interference pattern that is observed on a screen from the grating = 2.38m

maxima for two different wavelengths = 3.40mm

slit separation .. d = 1/900cm = 1.11^-3cm = 1.111^-5 m 

Whenas n = 1, maxima (grating equation) sinθ = λ/d
Grant distance of each maxima from centre = y ..
As sinθ ≈ y/D y/D = λ/d λ = yd / D 

∆λ = (λ2 - λ1) = y2.d/D - y1.d/D
∆λ = (d/D) [y2 -y1]

∆λ = 1.111^-5m x [3.40^-3m] / 2.38m .. .. ►∆λ = 1.587^-8 m

6 0

3 years ago

Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are

Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

$d_{12} = \sqrt{3^{2}+3^{2} } = \sqrt{18}$ m : distance from q₁ to q₂

(d₁₂)² = 18 m²

$d_{13} =\sqrt{1^{2}+3^{2} } = \sqrt{10}$ m : distance from q₁ to q₃

(d₁₃)² = 10 m²

$d_{14} =\sqrt{3^{2}+2^{2} } = \sqrt{13}$ m : distance from q₁ to q₄

(d₁₄)² = 13 m²

K= 8.98755 × 10⁹ N *m²/C²

q₁= 7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45° , β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

$F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }$

$F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }$

Fn₁= 23.95*10⁻⁹ N

8 0

3 years ago

Write an essay on basketball history and how the game has changed over time

jarptica [38.1K]

Basket ball has gone through some changes and hence got into shape as what we play or watch nowadays. Overall rules have the same fundamental principles as set in 1891 by the founder of this game.

December 21st, in 1891, the introductory basketball game was played in Springfield, Massachusetts. It was first brought into shape by a Canadian-by birth, Dr. James Naismith. The basic idea of the new game was to keep the sports loving students in shape during the winters or in between the outdoor game seasons.

'The basket ball' which had a set of thirteen basic rules set up a strong foundation to the game and still played almost in the same style with few modifications.In the very year of 1891, after mixing and changing theme of several other played games of those days, basket ball was born.

Initially, basket ball with the 13 rules was played with 2 peach baskets setup as goals, which now a days are the baskets on a high poles although modified but with same basic idea. In the very first game played in Springfield, the set of players were able to score a single point only, in the whole game.

By comparing Naismith’s basketball and basketball of today we see a characteristic of the original game which had no dribbling, instead a player had to pass the ball to a another player of his team ,from his the very spot where he caught it. The second thing which has now changed is the fouls, in the original game if any team made 3 consecutive foul play ,the oponent received a goal in reward. Although this scoring technique doesn't exist in the modern basketball. Rather nowadays, if any team makes five fouls in one quarter, the offending team is in the penalty ,and shoot free throws are awarded against them.

All other rules are quiet intact in the modern play as well. As the game today,has extended to over 2 hundered countries and enjoyed by many on television screens, with modern umpiring and technologies being used in the game, but the original idea of the founder is still endured.

5 0

3 years ago