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Nataly [62]
3 years ago
15

An airplane is flying horizontally with a constant momentum during a time interval ?t. (a) Is there a net impulse acting on the

plane during this time? Use the impulse-momentum theorem to guide your thinking. (b) In the horizontal direction, both the thrust generated by the engines and air resistance act on the plane. Considering your answer to part (a), how is the impulse of the thrust related in magnitude and direction to the impulse of the force due to the air resistance?
Physics
1 answer:
goldenfox [79]3 years ago
7 0

Answer:

Explanation:

Given

Airplane is flying with horizontally with a constant momentum during time interval \Delta t

Impulse is given by change in momentum, so there is no net impulse on the Plane because momentum is constant

(b) As there is no change in momentum therefore impulse of thrust and air drag is balanced i.e. both are equal in magnitude but act in opposite direction                            

You might be interested in
A sinusoidal sound wave moves through a medium and is described by the displacement wave function
ZanzabumX [31]

By applying the wave equation we know that the maximum speed of the element's oscillatory motion is 1716 micrometer / s.

We need to know about wave equations to solve this problem. The displacement of the wave on the y-axis can be explained by the wave equation

y = A cos (kx - ωt)

where y is y-axis displacement, A is amplitude, k is wave number, x is x-axis displacement, ω is angular speed and t is time.

the wavenumber and angular speed of the wave equation can be determined respectively by

k = 2π / λ

ω = 2πf

where k is the wavenumber, λ is wavelength and f is frequency.

From the question above, we know that:

y = 2.00cos (15.7x - 858t)

v = dy / dt

v = d(2.00cos (15.7x - 858t)) / dt

v = -858 x (-2.00sin(15.7x - 858t))

v = 1716 sin(15.7x - 858t) micrometer/s

maximum velocity can be reached when (sinθ = 1), hence

v = 1716 sin(15.7x - 858t)

v = 1716 x 1

v = 1716 micrometer / s

For more on wave equation on: brainly.com/question/25699025

#SPJ

4 0
1 year ago
According to Newton's 3rd Law, an object's momentum depends on it's velocity and mass. 4 dog-sled teams competed to see who coul
podryga [215]

Answer:

C.) Sled Team C 28 kg moving at 12m/s

I'm pretty sure.

7 0
3 years ago
Which force attracts all matter to each other
likoan [24]

Gravity is the force that attracts all matter to each other.

Explanation:

Sir Isaac Newton discovered Gravity when he saw a falling apple while thinking about the forces of nature.

Gravity is a fundamental force that causes objects to have weight. Gravity acts on all matter and is a function of both mass and distance. Each object attracts every other object with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force of attraction is, however, negligible between most objects because of their small size.

Gravitational force is given as:

F = \frac{Gm1m2}{r^{2} }

Where G is gravitational constant and is equal to 6.674×10−11 m³⋅kg⁻¹⋅s⁻²

m₁ and m₂ are the masses of the two objects.

r is the distance between the two objects.

The gravity is what makes an apple fall on the ground and gravity is the force that keeps us on the ground.

Keywords: gravity, Newton, Force, weight

Learn more about gravitational force from brainly.com/question/14321566

#learnwithBrainly

8 0
3 years ago
(b) Which has more mass: 50 cm of gold or 50 cm of aluminum? Explain.
monitta
Gold’s molar mass is about 196 while aluminum is about 27, thus 50cm of gold has more mass
8 0
3 years ago
THIS MARCIN
nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

f

1

=

v

1

−

u

1

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

h

h

′

=

u

v

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

h

h

′

=

u

v

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

f

1

=

v

1

−

(−25)

1

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

v

1

=

f

1

+

(−25)

1

=

10

1

−

25

1

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

v

1

=

250

25−10

=

250

15

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

15

250

=

3

50

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

5

h

′

=−

25

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

25

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

5 0
2 years ago
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