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3 years ago

9

Two forces are applied to a 17 kg box, as shown. The box is on a smooth surface. Which statement best describes the acceleration

of the box? A. The box accelerates at 1.0 m/s2 to the right because the net force is 17 N to the right. B. The box accelerates at 1.9 m/s2 to the right because the greater force is to the right. C. The box accelerates at 3.0 m/s2 because the combined forces cause the box to accelerate. D. The box does not accelerate, because neither force is large enough to move the box.

1 answer:

RoseWind [281]3 years ago

7 0

To the picture the answer is A. I can’t answer the typed question because I need the picture for the box

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Can you access an instance variable from a static method? explain why or why not.

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3 years ago

Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur

Alenkasestr [34]

Answer:

the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$

the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be $\phi$ and $\beta$ respectively.

The velocity of the block to be = v

The equivalent mass of the system = $m_e$

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

$K.E = \frac{1}{2} m_ev^2$ --------------- equation (1)

The angular velocity of the cylinder = $\omega$ : &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

$v = \omega R$

$\omega =\frac{v}{R}$

The kinetic energy of the system in terms of individual mass can be written as:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2$

By replacing $\omega$ with $\frac{v}{R}$ ; we have:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2$

$K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2$ ------------------ equation (2)

Equating both equation (1) and (2); we have:

$m_e = m_1+m_2+\frac{I}{R^2}$

Therefore, the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$ which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

$f_{block }=m_1gsin \beta$

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

$f_{cylinder} = m_2gsin \phi$

Equating the above both equations; we have the equation of motion of the equivalent system to be

$m_e \bar x = f_{cylinder}-f_{block}$

which can be written as follows from the previous derivations

$(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Finally; the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

8 0

3 years ago

ASAP WILL GIVE BRAINLIEST Take a piece of paper, some nails and a magnet bar, place a magnet at the centre of the paper, and spr

Zanzabum

Answer: find the answer in the explanation

Explanation:

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Other phenomena you can observe are:

1.) The nails will align themselves and show some lines of forces which is equivalent to the magnetic field lines

2.) The direction of the line of forces

3.) The strength of the magnetic field pole.

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3 years ago

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nasty-shy [4]

<u>Answer:</u> The ball is travelling with a speed of 5.5 m/s after hitting the <u>bottle.</u>

<u>Explanation:</u>

To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

$m_1,u_1\text{ and }v_1$ are the mass, initial velocity and final velocity of ball.

$m_2,u_2\text{ and }v_2$ are the mass, initial velocity and final velocity of bottle.

We are given:

$m_1=0.4kg\\u_1=18m/s\\v_1=?m/s\\m_2=0.2kg\\u_2=0m/s\\v_2=25m/s$

Putting values in above equation, we get:

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Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.

5 0

3 years ago