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2 years ago

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A disc with a mass of 1 kg moves horizontally to the right with a speed of 7 m/s on a table with negligible friction when it col

lides with a second disc with a mass of 9 kg. The second disc is moving horizontally to the right with a speed of 1 m/s at the moment of impact. The two discs stick together upon impact. What is the speed of the composite body immediately after the collision? (round your answer to the nearest tenth)

1 answer:

Elodia [21]2 years ago

7 0

Answer:

1.6 m/s

Explanation:

First you need to find the momentums of each disc by multiplying their velocities with mass.

disc 1: 7*1= 7 kg m/s

disc 2: 1*9= 9 kg m/s

Second, you need to find the total momentum of the system by adding the momentums of each sphere.

9+7= 16 kg m/s

Because momentum is conserved, this is equal to the momentum of the composite body.

Finally, to find the composite body's velocity, divide its total momentum by its mass. This is because mass*velocity=momentum

16/10=1.6

The velocity of the composite body is 1.6 m/s.

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Solve the following quadratic equations:2x² - 4x - 9 = 0 x₊ = 1 + √22/2Answersx_ = 1 - √22/2

Pavel [41]

The root quadratic equation of $2x^{2} - 4x - 9 = 0$ is $x=\frac{2+\sqrt{22}}{2},\:x=\frac{2-\sqrt{22}}{2}$

What is the formula used to calculate the quadratic equation?

To solve the quadratic equation use the formula of $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Find the quadratic equation by using the formula ?

On comparing with $2x^{2} - 4x - 9 = 0$

We get , $a=2 , b=-4 , c=-9$

$D=b^{2} +7\\\\\\=(-4)^{2} +7*2*(-9)\\\\=-16-126\\\\=110$

By using the formula,

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$X==\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\c* \:2\left(-9\right)}}{2\c* \:2}\\\\\sqrt{\left(-4\right)^2-4\cdot \:2\left(-9\right)}\\\\$

Apply rule,

$-(-a)=a\\\\=\sqrt{\left(-4\right)^2+4\cdot \:2\cdot \:9}\\\\$

Apply exponent rule,

$(-a)^{n} =a^{n}$ if n is even

$(-4)^{2} =4^{2}$

$\sqrt{4^2+4\cdot \:2\cdot \:9}$

Multiply the numbers :

$4*2*9=72\\\\\sqrt{4^2+4\cdot \:2\cdot \:9}\\\\4^{2} =16\\\\=\sqrt{16+72}\\\\$

Add the number $=\sqrt{88}$

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2 years ago

If love is the answer, then what is the question?

kipiarov [429]

Answer:

what is happiness

Explanation:

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3 years ago

Read 2 more answers

At some instant and location, the electric field associated with an electromagnetic wave in vacuum has the strength 71.9 V/m. Fi

kirza4 [7]

Answer:

a) Magnetic field strength, B = 2.397 * 10⁻⁷ T

b) Total energy density, U = 4.58 * 10⁻⁸ J/m³

c) Power flow per unit area, S = 13.71 W/m²

Explanation:

a) Electric field strength, E = 71.9 V/m

The relationship between the Electric field strength and the magnetic field strength in vacuum is:

E = Bc where c = 3.0 * 10⁸ m/s

71.9 = B * 3.0 * 10⁸

B = 71.9 / (3.0 * 10⁸)

B = 23.97 * 10⁻⁸

B = 2.397 * 10⁻⁷ T

b) Total Energy Density:

$U = \frac{1}{2} \epsilon_0E^2 + \frac{1}{2} \frac{B^2}{\mu_0} \\U = \frac{1}{2}* 8.85 * 10^{-12}*71.9^2 + \frac{1}{2} \frac{(2.397*10^{-7})^2}{4\pi*10^{-7}}\\U = 2.29 * 10^{-8} + 2.29 * 10^{-8}\\U = 4.58 * 10^{-8} J/m^3$

c)Power flow per unit area

$S = \frac{1}{\mu_0} EB\\S = \frac{1}{4\pi * 10^{-7} } * 71.9 * 2.397 * 10^{-7}\\S = 13.71 W/m^2$

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3 years ago

What is the top of a wave called?

ohaa [14]

Answer:

The highest part of the wave is called the crest.

Explanation:) hope this helps

The highest part of the wave is called the crest. The lowest part is called the trough. The wave height is the overall vertical change in height between the crest and the trough and distance between two successive crests (or troughs) is the length of the wave or wavelength.

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1 year ago

A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f

Delvig [45]

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, $F=70\ N$

Mass of the dog is, $m=25\ kg$

Angle of inclination is, $\theta =30$ °

Acceleration due to gravity is, $g=9.8\ m/s^2$

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

$F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N$

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

$F_g=mg=25\times 9.8=245\ N$

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

$N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N$

Therefore, the normal force acting on the dog is 210 N.

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3 years ago