A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's initial speed is 1.37 m/s and it stops after 2.8 s. Determine the magnitude of the friction force exerted on the box.A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's initial speed is 1.37 m/s and it stops after 2.8 s. Determine the magnitude of the friction force exerted on the box.

Answers

Answer 1

Answer:

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_(o)+a\cdot t$ ), we expand the previous expression:

$-f = \left((W)/(g) \right)\cdot \left((v-v_(o))/(t)\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left((W)/(g) \right)\cdot \left((v-v_(o))/(t)\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_(o)$ - Initial speed, measured in meters per second.

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

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,(m)/(s^(2))$ , $v_(o) = 1.37\,(m)/(s)$ , $v = 0\,(m)/(s)$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left((52.4\,N)/(9.807\,(m)/(s^(2)) ) \right)\cdot \left((0\,(m)/(s)-1.37\,(m)/(s) )/(2.8\,s) \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

Answer 2

Answer:

Final answer:

The magnitude of the friction force acting on the box is determined by calculating the box's acceleration, establishing its mass based on its weight information, and applying these values in Newton's second law. The calculated value is 2.62 N.

Explanation:

To determine the magnitude of the friction force, we first have to compute the acceleration of the box. Acceleration (a) can be found using the formula 'final velocity - initial velocity / time'. Since the final velocity is 0 (the box stops), and the initial velocity is 1.37 m/s, and the time is 2.8 s, we get: a = (0 - 1.37) / 2.8 = -0.49 m/s^2. The negative sign indicates deceleration.

Next, we use Newton's second law, which states that the net force acting on an object is equal to its mass times its acceleration. The net force in this case is the frictional force because there is no other force acting on the box in the horizontal direction. However, we do not know the mass of the box, but we do know its weight, and weight = mass x gravitational acceleration (g). So mass = weight/g = 52.4N / 9.8m/s^2 = 5.35 kg.

Lastly, we substitute the mass and deceleration into Newton's second law to find the frictional force (f): f = mass x deceleration = 5.35kg x -0.49m/s^2 = -2.62 N. Again, the negative sign indicates that the force acts opposite to the direction of motion. Thus, the frictional force magnitude is 2.62 N.

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Nicki rides her bike at a constant speed for 6 km. That part of her ride takes her 1 h. She then rides her bike at a constant speed for another 9km. That part of her trip takes her 2 h. What is her average speed?

Answers

Answer:

24km/h

Explanation:

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Suppose that a barometer was made using oil with rho=900 kg/m3. What is the height of the barometer at atmospheric pressure?

Answers

Hey there!

The pressure under a liquid column can be , calculated using the following formula :

P = p x g x h

P atm = 1.013 x 10⁵ Pa

g = 9.8 m/s²

h = ?

h = P / ( p x g ) =

h= ( 1.013 x 10⁵ Pa ) / ( 900 x 9.8 ) =

h = ( 1.013 x 10⁵ ) / ( 8820 ) =

h = 11.48 m ≈ 11.50 m

Hope this helps!

A turntable with a rotational inertia of 0.0120 kg∙m2 rotates freely at 2.00 rad/s. A circular disk of mass 200 g and radius 30.0 cm, and initially not rotating, slips down a spindle and lands on the turntable. (a) Find the new angular velocity. (b) What is the change in kinetic energy?

Answers

To solve this problem it is necessary to apply the related concepts to the moment of inertia in a disk, the conservation of angular momentum and the kinematic energy equations for rotational movement.

PART A) By definition we know that the moment of inertia of a disk is given by the equation

$I = (1)/(2) MR^2$

Where

M = Mass of the disk

R = Radius

Replacing with our values we have

$I = (1)/(2) (0.2)(0.3)^2$

$I = 9*10^(-3)kg\cdot m^2$

The initial angular momentum then will be given as

$I = I_1 \omega_1$

$I = 0.012*2$

$I = 0.024kg\cdot m^2/s$

Therefore the total moment of inertia of the table and the disc will be

$I_2 = 9*10^(-3)+0.012$

$I_2 = 0.021kg\cdot m^2$

The angular velocity at the end point will be given through the conservation of the angular momentum for which it is understood that the proportion of inertia and angular velocity must be preserved. So

$I_1 \omega_1 = I_2\omega_2$

$(0.012)(2)=(1.08*10^(-4))\omega_2$

$\omega_2 = (0.012*2)/(0.021)$

$\omega_2 = 1.15rad/s$

Therefore the new angular velocity is 1.15rad/s

PART B) Through the conservation of rotational kinetic energy we can identify that its total change is subject to

$\Delta KE = (1)/(2)I_1\omega_1^2-(1)/(2)I_2\omega^2$

$\Delta KE = (1)/(2)(I_1\omega_1^2-I_2\omega^2)$

$\Delta KE = (1)/(2)(0.024*2^2-0.021*1.15^2)$

$\Delta KE = 0.034J$

Therefore the change in kinetic energy is 0.034J

The acrylic plastic rod is 20 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. Eₚ = 2.70 GPa, vₚ = 0.4.

Answers

Given Information:

diameter = d = 15 mm

Length = L = 20 mm

Axial load = P = 300 N

Eₚ = 2.70x10⁹ Pa

vₚ = 0.4

Required Information:

Change in length = ?

Change in diameter = ?

Answer:

Change in length = 0.01257 mm

Change in diameter = -0.003772 mm

Explanation:

Stress is given by

σ = P/A

Where P is axial load and A is the area of the cross-section

A = 0.25πd²

A = 0.25π(0.015)²

A = 0.000176 m²

σ = 300/0.000176

σ = 1697792.8 Pa

The longitudinal stress is given by

εlong = σ/Eₚ

εlong = 1697792.8/2.70x10⁹

εlong = 0.0006288 mm/mm

The change in length can be found by using

δ = εlong*L

δ = 0.0006288*20

δ = 0.01257 mm

The lateral stress is given by

εlat = -vₚ*εlong

εlat = -0.4*0.0006288

εlat = -0.0002515 mm/mm

The change in diameter can be found by using

Δd = εlat*d

Δd = -0.0002515*15

Δd = -0.003772 mm

Therefore, the change in length is 0.01257 mm and the change in diameter is -0.003772 mm

Sometimes, in an intense battle, gunfire is so intense that bullets from opposite sides collide in midair. Suppose that one (with mass M = 5.12 g moving to the right at a speed V = [08]____________________ m/s directed 21.3° above the horizontal) collides and fuses with another with mass m = 3.05 g moving to the left at a speed v = 282 m/s directed 15.4° above the horizontal. a. What is the magnitude of their common velocity (m/s) immediately after the collision? b. What is the direction of their common velocity immediately after the collision? (Measure this angle in degrees from the horizontal.) c. What fraction of the original kinetic energy was lost in the collision?

Answers

The magnitude of the speed is 83.0325 m\s, the direction is 62.7 degrees, and the fraction of kinetic energy lost is 0.895.

What is collision?

The collision is the phenomenon when two objects come in direct contact with each other. Then both the bodies exert forces on each other.

The mass, angle, and velocity of the first object are 5.12 g, 21.3°, and 239 m/s.

And the mass, angle, and velocity of the second object be 3.05 g, 15.4°, and 282 m/s.

The momentum (P₁) before a collision will be

$\rm P_1 = (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) \hat{x} + (m_1 u_1 sin \theta _1+ m_2 u_2 sin \theta _2) \hat{y}$

The momentum (P₂) after a collision will be

$\rm P_2 = (m_1 + m_2) u \ cos\ \theta \ \hat{x} \ + (m_1 + m_2) u \ sin \ \theta \ \hat{y}$

Applying momentum conservation, we have

$\rm (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) = (m_1 + m_2) u \ cos\ \theta \ \n\n$ ...1

$\rm (m_1 u_1 sin \theta _1+m_2 u_2 sin \theta _2) \ =(m_1 + m_2) u \ sin \ \theta$ ...2

From equations 1 and 2, we have

$\rm \theta = tan \ ^(-1) ( (m_1 u_1 cos \theta _1 +m_2 u_2cos \theta _2))/( (m_1 u_1 sin \theta _1 - m_2 u_2 sin \theta _2))\n\n\n\theta = tan \ ^(-1) (5.12*239*cos21.3+3.05*282*cos15.4)/(5.12*239*sin21.3-3.05*282*sin15.4)\n\n\n\theta = 62.7^o$

From equation 1, we have

$\rm u = ((m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) )/( (m_1 + m_2) \ cos\ \theta ) \n\n\nu = (5.12*239*cos21.3 - 3.05*282*cos15.4)/((5.12+3.05)cos62.2)\n\n\nu = 83.0325 m/s$

Then the change in kinetic energy, we have

$\rm \Delta KE = (1)/(2)m_1u_1^2+(1)/(2)m_2u_2^2-(1)/(2)(m_1+m_2)u^2\n\n\n\Delta KE = (1)/(2) * 5.12 * 239^2 + (1)/(2)*3.05*282^2 - (1)/(2)(5.12+3.05)*83.032^2\n\n\n\Delta KE = 239.34 \ J$

The fraction of kinetic energy lost will be

$\rm Energy \ lost = (239.34)/(267.5) = 0.895$

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1. Explain the change of state from solid dry ice to carbon dioxide gas.2. The motion of the particles in dry ice and carbon dioxide gas.

3. Explain how the original mass of dry ice compares with the mass of carbon dioxide gas.

Answers

1. Since the solid carbon dioxide never become liquid on melting under normal pressure. Thus through the process of sublimation, the solid carbon dioxide changes to gas

2. The molecules in dry ice, are in caged like structure just as the normal water ice but as it melts, the CO2 molecules having high affinity for gaseous state converts into a gas.

3. Dry ice is heavier than its gaseous form. Density of dry ice = 97.6 lb/cu.ft.

Density of carbon dioxide gas = 0.1144 lb/cu.ft.

With higher density and a fixed volume, mass of dry ice is higher than the CO2 gas

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