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MAXImum [283]
3 years ago
8

A classmate is attempting to draw a free body diagram for a box being pushed across the floor at a constant speed. The image bel

ow is what
they have drawn so far.
Which (if any) of their forces have been drawn incorrectly
O A. The applied force should be the same size as the friction force
B. There is nothing wrong with the free body diagram
OC. The normal force should be larger than the force and gravity
OD. The friction force should be in the same direction as the applied force

Physics
1 answer:
aivan3 [116]3 years ago
7 0

Answer:

A. The applied force should be the same size as the friction force

Explanation:

Whenever we apply a force to an object it moves if the force applied to that object is unbalanced and there is no force or a lesser force to counter it. According to Newton's Second Law of motion, when an unbalanced force is applied to an object it produces an acceleration in the object in its own direction. So, the two forces acting on this box are the frictional force and the applied force in horizontal direction. In order to move the box at constant speed, the applied force must first, overcome the frictional force, so the object can start its motion. Since, the motion has constant velocity, it means no acceleration. So, the force must be balanced in order to avoid acceleration as a consequence of Newton's Second Law of motion. Therefore, the correction in this case will be:

<u>A. The applied force should be the same size as the friction force</u>

You might be interested in
Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 5
LekaFEV [45]

Answer:

  • |\vec{F}_3| = 102.92 \ N
  • \theta = 57 \° 24 ' 48''

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

\vec{a} = \vec{0}.

As the net force equals acceleration multiplied by mass , this must mean:

\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}.

So, the sum of the three forces must be zero:

\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0},

this implies:

\vec{F}_3  = - \vec{F}_1 - \vec{F}_2.

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector \hat{i} pointing east,  with the y axis unit vector \hat{j} pointing south, so the positive angle is south of east. For this, we got for the first force:

\vec{F}_1 = 83.7 \ N \ (-\hat{j}),

as is pointing north, and for the second force:

\vec{F}_2 = 59.9 \ N \ (-\hat{i}),

as is pointing west.

Now, our third force will be:

\vec{F}_3  = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})

\vec{F}_3  =  83.7 \ N \ \hat{j}  + 59.9 \ N \ \hat{i}

\vec{F}_3  =  (59.9 \ N , 83.7 \ N )

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

|\vec{R}| = \sqrt{R_x^2 + R_y^2}

|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}

|\vec{F}_3| = 102.92 \ N

this is the magnitude.

To find the direction, we can use:

\theta = arctan(\frac{F_{3_y}}{F_{3_x}})

\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })

\theta = 57 \° 24 ' 48''

and this is the angle south of east.

7 0
3 years ago
The fatty layer contains many nerves and blood vessels. True or False
mylen [45]

Answer:

False , The fatty layer do not contains many nerves and blood vessels. The fatty layer is related to skin layers.

Explanation:

The fatty layer is the most under a layer of skin. It made up of a system of collagen and fat cells. It helps maintain the body's temperature and shields the body from harm by serving as a shock absorber.

The dermis is the central layer of the skin. The dermis is maintained collectively by a protein termed collagen. This layer gives skin elasticity and power. The dermis also holds shock and feel receptors.

5 0
3 years ago
A constellation’s changing position in the sky, at the same time of the evening, over a period of several weeks is evidence that
Goryan [66]

Answer:

d- Earth revolves around the sun.

Explanation:

Earth rotation can be defined as the amount of time taken by planet earth to complete its spinning movement on its axis.

This ultimately implies that, the rotation of earth refers to the time taken by earth to rotate once on its axis. One spinning movement of the earth on its axis takes approximately 24 hours to complete with respect to the sun.

On the other hand, earth revolution can be defined as a complete trip along a path around the sun. This path is known as an orbit and it typically takes the Earth 365¼ days to complete it's journey around the Sun.

When a constellation (stars) changes its position in the sky, at the same time of the evening and over a period of several weeks; it ultimately implies or is an evidence that Earth revolves around the sun.

8 0
3 years ago
Consider the classic problem with holiday lights, one little bulb goes out and the whole string goes out. First consider a strin
Flura [38]

Answer:

<em>a. 0.33 Amp</em>

<em>b. 2.4 Volt</em>

<em>c. 0</em>

<em>d. 2.45 Amp</em>

<em>e. Infinite</em>

<em>f. Series is safer</em>

Explanation:

<u>Series Connection of Resistors</u>

When two or more resistors are connected in series, the current through each one of them is the same, and the voltage divides depending on the particular value of each resistance. If all the resistances are equal, then the voltage is equally divided.

a. The string of 50 bulbs is connected to a 120 VAC outlet and consumes 40 W. The power of a circuit is given by

P=V.I

Solving for I

\displaystyle I=\frac{P}{V}=\frac{40}{120}=0.33\ Amp

Since all the bulbs are connected in series the current is the same for all of them.

b. The voltage is equally divided, so each bulb has  120/50= 2.4 V

c. If one of the bulbs burns out and its resistance becomes infinite, then the series circuit is open and no current flows through it, neither through the rest of the bulbs. The typical case of the whole string going out.

d. If one of the bulbs short circuits, the resistance of that bulb is zero and the voltage is distributed by the 49 remaining bulbs. Thus the new current is

\displaystyle I=\frac{V}{R}=\frac{120}{49}=2.45\ A

e. If the bulbs were connected in parallel, all of them would have the same voltage, and the total current will be equally divided among them. In that case, a short circuit in one of the bulbs will cause a parallel short, theoretically producing an infinite current and making the short circuit protection blow up.

f. The condition described above makes the strings be made of series-connected bulbs which is safer than the parallel circuit. If a single bulb shorts, the entire string goes out in a series connection, but the breaker would trigger disconnection of the house circuit if it's a parallel connection. That is why we must deal with unusable strings instead of burning cables.

6 0
3 years ago
NEED HELP ASAP
Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:  

p_{o}=p_{f} (1)  

Where:  

p_{o}=m_{1} V_{o} + m_{2} U_{o} (2)  

p_{f}=(m_{1} + m_{2}) V_{f} (3)

m_{1}=110 kg is the mass of the first football player

V{o}=-7 m/s is the velocity of the first football player (to the south)

m_{2}=75 kg  is the mass of the second football player

U_{o}=4 m/s is the velocity of the second football player (to the north)

V_{f} is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f} (4)  

Isolating V_{f}:

V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}} (5)

V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg} (6)

V_{f}=-2.54 m/s (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy \Delta K is defined as:

\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2} (8)

Simplifying:

\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2}) (9)

\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2}) (10)

Finally:

\Delta K=-2351.25 J (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0
3 years ago
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