I believe it is D I’m sorry if it’s wrong
The initial velocity of the bird before the gust of wind : <u>4 m/s</u>
<h3>Further explanation</h3>
An equation of uniformly accelerated motion
v = vo + at
Vt² = vo² + 2a (x-xo)
x = distance on t
vo/vi = initial speed
vt/vf = speed on t /final speed
a = acceleration
Acceleration is a change in speed within a certain time interval
a = Δv /Δ t
Let complete the task :
A bird is flying to the right when a gust of wind causes the bird to accelerate leftward at 0.5 m/s² for 3 s. After the wind stops, the bird is flying to the right with a velocity of 2.5 m/s.
Assuming the acceleration from the wind is constant, what was the initial velocity of the bird before the gust of wind?
we can use formula :
vf = vi + a.t
vf = final velocity = 2.5 m/s
vi = asked
a = - 0.5 m/s²(leftward=negative)
t = 3 s
then :
<h3>Learn more</h3>
The car reach the end of the road
brainly.com/question/13750982
Keywords: uniformly accelerated motion, distance, velocity, acceleration
#LearnwithBrainly
If the polarity of a magnet is flipped while its strength remains constant. There is no change in the electromagnetic force since there is no change in the current in the wire.
<h3>What is the faraday law of electromagnetic induction?</h3>
According to Faraday's law of electromagnetic induction, the rate of change of magnetic flux linked with the coil is responsible for generating emf in the coil resulting in the flow of the amount of current.
If a magnet's polarity were switched while maintaining the same magnetic field intensity. There is no change in the electromagnetic force, and there is no change in the current in the wire.
Hence the current in the wire will be the same as in the first part.
To learn more about the electromagnetic induction refer to:
brainly.com/question/26334813
#SPJ1
Answer: the work done by the force is 0
Explanation:
F (x², xy)
121 = 11²
so R = x² + y² = 11²
p = x². Q = xy
Δp/Δy = 0, ΔQ/Δx
using Green's theorem
woek = c_∫F.Δr = R_∫∫ ΔQ/Δx - Δp/Δy) ΔA
= (x² + y² = 121)_∫∫ yΔA
now let x = rcosФ, y = rsinФ
ΔA = rΔrΔФ
so r from 0 to 11
and Ф from 0 to 2π
= 0_∫^2π 0_∫^11 rsinФ × rΔrΔФ
= 0_∫^2π SinФΔФ 0_∫^11 r²Δr
= [ -cosФ]^2π_0 [r³/3]₀¹¹ = ( -cos2π + cos0) (11³/3) = 0
therefore the work done by the force is 0
You don't. Although tons of stories have been written about it,
and loads of scientific speculation about how it maybe possibly
might be done, it's never been done.