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lutik1710 [3]
2 years ago
5

Find the acute angles between the curves at their points of intersection. (The angle between two curves is the angle between the

ir tangent lines at the point of intersection. Give your answers in degrees, rounding to one decimal place. Enter your answers as a comma-separated list.) y = 8x2, y = 8x3
Physics
1 answer:
ladessa [460]2 years ago
3 0

Answer:

1.190390345 degrees.

Explanation:

As stated in the problem that the angle between the two curves is also the angle between slopes of the two cures at the point of intersection, which is 1 (when we set two equation equal and solve for x).

We know, if you don't  you could verify it for yourself and it will be a nice mathematical exercise for you, that when there are two lines with slopes m_1 and m_2 then the angle between them and their two slopes has following relationship.

tan(\alpha ) = |\frac{m_2-m_1}{1+m_1m_2}|   where \alpha is angle between the two lines.

As it is clear that we can easily get the angle between the two curves if we know slopes at that point, which you will see in second is very straight forward to calculate.

Slopes are simply derivatives evaluated at the point of intersection and the two derivatives are D_1 16x and D_2 =24x^2, substituting x =1 we get, m_1 = 16 and m_2 = 24.

Now put m_1 and m_2 in this relationship, rearrange and solve for angle, it will come out to be

\alpha = 1.190390345 degrees.

that is our angle that we want to know between the two cures at the point of their intersection.

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May you help me answer this​
Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is 2.2 m/s^2

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

1)

There are three Kepler's law of planetary motion:

  1. 1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
  2. 2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
  3. 3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, T^2 \propto r^3, where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

M g sin \theta - T = Ma (1)

where

Mg sin \theta is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

g=9.8 m/s^2 (acceleration of gravity)

\theta=35^{\circ}

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

T-mg sin \theta = ma (2)

where

m = 3.5 kg (mass of the block)

\theta=35^{\circ}

From (2) we get

T=mg sin \theta + ma

Substituting into (1),

M g sin \theta - mg sin \theta - ma = Ma

Solving for a,

a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2

2b)

The tension in the string can be calculated using the equation

T=mg sin \theta + ma

where

m = 3.5 kg (mass of lighter block)

g=9.8 m/s^2

\theta=35^{\circ}

a=2.2 m/s^2 (acceleration found in part 2)

Substituting,

T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

U=mgh

where

m is the mass of the object

g is the strength of the gravitational field

h is the heigth of the object relative to the ground

3b)

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the object in this problem,

m = 500 g = 0.5 kg

v = 3 m/s

Substituting, we find its kinetic energy:

K=\frac{1}{2}(0.5)(3)^2=2.25 J

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

And about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

7 0
3 years ago
Which of the following objects is accelerating?
Schach [20]

Answer: A flower pot falling

Explanation:

3 0
3 years ago
Read 2 more answers
in a one dimensional collision, a 4kg object and 6 kg onject have initial velocity. calculate the magnitude of impulse
Greeley [361]

in a one dimensional collision, a 4kg object with 5ms^1 and 6 kg object with 2ms^1 have initial velocity, the magnitude of impulse is 12 , 18

given,

mass 1 = 4kg

mass 2 = 6kg

velocity 1 = 5ms^1

velocity 2 = 2ms^1

impulse 1 = 4*(5-2)

= 12

Impulse 2 = 6*(5-2)

= 18

Learn more about impulse here

brainly.com/question/16980676

#SPJ4

8 0
1 year ago
The average velocity of blood flowing in a certain 4-mm-diameter artery in the human body is 0.28 m/s. The viscosity and density
OLga [1]

Answer:

V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s

Explanation:

The volume flow rate of the blood in the artery can be given by the following formula:

V = Av

where,

V = Volume flow rate = ?

A = cross-sectional area of artery = πd²/4 = π(0.004 m)²/4 = 1.26 x 10⁻⁵ m²

v = velcoity = 0.28 m/s

Therefore,

V = (1.26\ x\ 10^{-5}\ m^2)(0.28\ m/s)

<u>V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s</u>

4 0
3 years ago
An object of mass 3 kg, moving with and sticks to an object of mass 2.00 kg with an initial velocity of 3.00 m/s. Find the final
Aleonysh [2.5K]

Answer:

1.8 m/s

Explanation:

momentum = mass × velocity

initial momentum = m1v1+m2v2

                             = 3×3 +2×0 = 9+0= 9 kg m/s

let combined velocity be V

HENCE  

final momentum = total mass × velocity

                            = (3+2) × V = 5V

According to law of conservation of momentum

final momentum = initial momentum

5V = 9

V =9/5

V = 1.8 m/s

8 0
3 years ago
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