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3 years ago

Why a flying areoplane has more kinectic energy than a flying insect

1 answer:

6 0

Answer: Because the angle of attack is so high, a lot of momentum is transferred downward into the flow. These two features create a large amount of lift force as well as some additional drag.

Brainlest would be much appreciated.

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4. A ball is thrown vertically upward from the ground with a velocity of 30m/s. (a) how long will it take to rise to the highest

yarga [219]

All the answers are:

a) The time that will it take to rise to the highest point is 3.06 seconds.

b) The ball will rise to a height of 45.87 meters.

c) The time at which the ball will have a velocity of 10 m/s upward is 2.04 seconds.

The time when the ball has 10 m/s downward is 1.02 seconds.

d) The displacement of the ball will be zero at 6.12 seconds.

e) The time when the magnitude of the ball's velocity is equal to half its velocity of projection is 1.53 seconds.

f) The ball's displacement is equal to half the maximum height to which it rises after 0.90 seconds.

g) In each moment (upward and downward) the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Let's calculate the values for each case.

a) At the highest point, the final velocity is 0, so we can use the following equation.

$v_{f}=v_{i}-gt$ (1)

Where:

v(i) is the initial velocity
v(f) is the final velocity
g is the acceleration due to gravity (9.81 m/s²)

We know that v(i) = 30 m/s.

$0=30-9.81t$

Solve it for t:

$t=3.06\: s$

Hence, the time is 3.06 s.

b) At the highest point, the final velocity is 0, so we can use the following equation.

$v_{f}^{2}=v_{i}^{2}-2gh$ (2)

$0=v_{i}^{2}-2gh$

We know that the initial velocity is 30 m/s.

$0=30^{2}-2gh$

Solving it for h we have:

$h=\frac{30^{2}}{2*9.81}$

$h=45.87 \: m$

Then, the height is 45.87 m.

c) Using equation (1) we can find the time (t).

$10=30-(9.81t)$

So, the time elapsed to get 10 m/s is:

$t_{upward}=2.04\: s$

We know the upward time is equal to the downward time. So the time from v=10 m/s to v=0 m/s will be.

$t_{upward}=2.04+t$

$t=1.02\: s$

This is the time when the ball has 10 m/s downward.

Therefore, the time upward is 2.04 s, and the time downward is 1.02 s.

d) It will be when the ball returns to the ground.

$t=2t_{upward}$

$t=2*3.06$

$t=6.12\: s$

The displacement will be zero after 6.12 s.

e) Here we need to find the time when v(f) is 15 m/s

$15=30-gt$

$t=\frac{15}{9.81}$

$t=1.53\: s$

The time when the v(f) is 15 m/s is 1.53 s.

f) Here, we need to find t when h = 45.87/2 m = 22.94 m

We can use the next equation:

[tex]h=v_{i}t-0.5gt^{2}/tex]

[tex]22.94=30t-0.5*9.81*t^{2}/tex]

Solving this quadratic equation, t will be:

[tex]t=0.90\: s/tex]

Hence, the ball's displacement is equal to half the maximum h, at 0.90 s.

g) In each moment the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Learn more about vertical motion here:

brainly.com/question/13966860

I hope it helps you!

3 0

3 years ago

Which band of the electromagnetic spectrum has the lowest frequency

Murrr4er [49]

Answer:

Radio waves

Explanation:

Electromagnetic waves are produced by the oscillations of electric and magnetic field. They are transverse waves, which means that the oscillations occur in a direction perpendicular to the direction of propagation of the wave, and they are the only type of waves that can travel through a vacuum.

Electromagnetic waves are classified into 7 different types, depending on their frequencies. From lowest to highest frequencies, we have:

Radio waves

Microwaves

Infrared

Visible light

Ultraviolet

X-rays

Gamma rays

Radio waves are the electromagnetic waves with lowest frequency. They are used, for examples, for satellites, telecommunication, broadcasting.

5 0

4 years ago

A batter hits a baseball so that it leaves the bat with an initial speed of 60m/s at an initial angle of 42 with the horizontal

Elza [17]

Answer: 24.06°

Explanation:

So,we can say after t if it reaches height h then,

h = (37 sin 53)t - 1/2 * 9.8t^2 (as,vertical component of velocity is 37 sin 53)

Given t = 2s

So, h = 39.5m

And horizontal displacement will be

r = 37 cos 53 *2 = 44.52m

So,after 2s the baseball will be lying 39.57m

above its point of projection and 44.52m ahead of its point of projection.

Now.let the vertical component of velocity will become Vy after time 2s

So, Uy = 37 sin 53- 9.8* 2

or, U = 9.95m/s

And.horizontal component of velocity remains

constant i.e Vx = 37 cos 53 = 22.27m/s

So.magnitude of velocity after 2s is

Square root of (Vx^2 + Vy^2)= 24.4m/s

Making an angle of tan 22.27/9.95 = 24.06°

6 0

3 years ago

An electric dipole is formed from two charges, ±q, spaced 0.800 cm apart. The dipole is at the origin, oriented along the y-axis

Simora [160]

Answer: q = 2.781e-9C = 2.781nC

E=200C

Explanation:

E = Qd/(2πEor^3)

Where

E=Electric field intensity

Q=Charge

d=distance between the dipole=0.008m

Eo=permitivitty

400 N/C = Q(0.80e-2 m)/(2πε*(10e-2 m)^3)

Q= (400* 2* 3.142 * 8.85 x 10-12 * 0.1^3)/0.008

q = 2.781e-9C = 2.781nC

Though the dipole are two separate charges. And since the point is on the x-axis, the electric field strengths are equivalent. The magnitude of the vector sum is:

E = kq*2sin θ/r^2

= 2(8.99e9 N*m^2/C^2)(2.781e-9 C)*sin(arctan(.4/10))/(10e-2 m)^2

= 2(8.99e9) * (2.781e-9) * sin(2.290)/(10e-2 m)^2

=200 C

3 0

3 years ago

Billy pulls his friend Sue on a sled at a constant velocity with a horizontal force of 200 N. The weight of the sled and Sue is

Nezavi [6.7K]

Pulls would be the logical answer.

6 0

3 years ago