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3 years ago

A person pushes two boxes with a horizontal force F of magnitude of 100 N.

Physics

1 answer:

Monica [59]3 years ago

7 0

The magnitude of the action–reaction pair between the two boxes (A and B) will be "18.2 N".

According to the question,

Mass of box A,

$m_A = 9\ kg$

Mass of box B,

$m_B = 2 \ kg$

Horizontal force,

$F_{app} = 100 \ N$

From the Newton's law,

→ $F_{app} = (\frac{F_{app}}{m_A+m_B} )a$

or,

→ $a = \frac{F_{app}}{(m_A+m_B)}$

Bu substituting the values, we get

→ $= \frac{100}{9+2}$

→ $= \frac{100}{11}$

→ $9.10 \ m/s^2$

We can see that between the two boxes, the action-reaction pair exist.

then,

→ $F_{action-reaction} = m_b \ a$

→ $=2\times 9.10$

→ $= 18.2 \ N$ (magnitude)

Thus the above solution is appropriate.

Learn more about the magnitude here:

brainly.com/question/13545862

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<h2>Potential energy lost by 10 N rock will be greater</h2>

Explanation:

Two rocks of 5N and 10N falls from the same height . Thus they will loose the potential energy.

The potential energy lost = mass x acceleration due to gravity x height

The potential energy lost by first 5 N rock = 5 h

Because weight of rock m g = 5 N

Similarly , the potential energy lost by 10 N Rock = 10 h

here weight of rock m g = 10 N

Thus comparing these two , the potential energy lost by 10 N rock is greater than that of 5 N rock .

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4 years ago

A rocket in deep space has an exhaust-gas speed of 2000 m/s. When the rocket is fully loaded, the mass of the fuel is five times

notka56 [123]

Answer:

$v_{f}$ = 1,386 m / s

Explanation:

Rocket propulsion is a moment process that described by the expression

$v_{f}$ - v₀ = $v_{r}$ ln (M₀ / Mf)

Where v are the velocities, final, initial and relative and M the masses

The data they give are the relative velocity (see = 2000 m / s) and the initial mass the mass of the loaded rocket (M₀ = 5Mf)

We consider that the rocket starts from rest (v₀ = 0)

At the time of burning half of the fuel the mass ratio is that the current mass is

M = 2.5 Mf

$v_{f}$ - 0 = 2000 ln (5Mf / 2.5 Mf) = 2000 ln 2

$v_{f}$ = 1,386 m / s

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3 years ago

During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 31.7 rad/s. Find the angular displ

Crank

Answer:

$\Delta \theta = 3116.11\,rad$ and $\Delta \theta = 495.944\,rev$

Explanation:

The tub rotates at constant speed and the kinematic formula to describe the change in angular displacement ( $\Delta \theta$ ), measured in radians, is:

$\Delta \theta = \omega \cdot \Delta t$

Where:

$\omega$ - Steady angular speed, measured in radians per second.

$\Delta t$ - Time, measured in seconds.

If $\omega = 31.7\,\frac{rad}{s}$ and $\Delta t = 98.3\,s$ , then:

$\Delta \theta = \left(31.7\,\frac{rad}{s} \right)\cdot (98.3\,s)$

$\Delta \theta = 3116.11\,rad$

The change in angular displacement, measured in revolutions, is given by the following expression:

$\Delta \theta = (3116.11\,rad)\cdot \left(\frac{1}{2\pi} \frac{rev}{rad} \right)$

$\Delta \theta = 495.944\,rev$

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3 years ago

A 60 newton force is directed in a southwest direction. What are the magnitudes of the force components in the south and west di

JulijaS [17]

Hello
This question is to be tackled using vectors. When we look at the Southwest direction, we know it lies exactly between the South and West directions, producing an angle of 45 degrees. When we take sin(45) and multiply it by the original force, we obtain the component towards West; that is:
60 * sin (45) = 42.4 Newtons
And cos(45) gives us the force towards the South direction; that is:
60 * cos(45) = 42.4 Newtons
This can also be checked by using the formula of the magnitude of a vector and squaring 42.4, adding it to the square of 42.4 and then taking the square root of the answer.
sqrt(42.4^2 + 42.4^2) = sqrt(3600) = 60

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3 years ago

A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/s2 . It runs out of fuel at the end of 4.00s

Murrr4er [49]

Answer:

a) The velocity and position of the rocket at the end of 4 seconds are 117.6 meters per second and 235.2 meters, respectively, b) The maximum height reached by the rocket is 940.296 meters, c) The rocket crashes on the ground at a velocity of -96.030 meters per second.

Explanation:

The complete statement is:

A rocket moves straight upward , starting from rest with an acceleration of 29.4 m/s2. it runs out of fuel at the end of 4.00 s and countinues to coast upward , reaching a maximum height before falling back to earth . a) find the rocket's velocity and position at the end of 4.00 s . b) Find the maximum height the rocket reaches. c) find the velocity on the instant before the rocket crashes on the ground.

a) The rocket accelerates uniformly, whose equations of motion are:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot a \cdot t^{2}$

$v = v_{o} + a\cdot t$

Where:

$y$ - Final position, measured in meters.

$y_{o}$ - Initial position, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

$t$ - Time, measured in second.

$a$ - Acceleration, measured in meters per square second.

$v$ - Final velocity, measured in meters per second.

If we know that $y_{o} = 0\,m$ , $v_{o} = 0\,\frac{m}{s}$ , $t = 4\,s$ and $a = 29.4\,\frac{m}{s^{2}}$ , the velocity and position of the rocket are, respectively:

$y = 0\,m+\left(0\,\frac{m}{s} \right)\cdot (4\,s)+\frac{1}{2}\cdot \left(29.4\,\frac{m}{s^{2}} \right) \cdot (4\,s)^{2}$

$y = 235.2\,m$

$v = 0\,\frac{m}{s} + \left(29.4\,\frac{m}{s^{2}}\right)\cdot (4\,s)$

$v = 117.6\,\frac{m}{s}$

The velocity and position of the rocket at the end of 4 seconds are 117.6 meters per second and 235.2 meters, respectively.

b) Now, the rocket experiments a free-fall motion. The maximum height of the rocket is obtained by equalizing the equation of velocity to zero and evaluating the equation of position later. That is:

$y = 235.2\,m+\left(117.6\,\frac{m}{s} \right)\cdot t+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}$