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earnstyle [38]
1 year ago
8

a college student rests a backpack upon his shoulder. the pack is suspended motionless by one strap from one shoulder.

Physics
1 answer:
Minchanka [31]1 year ago
4 0

The student's shoulder supports the weight of the bag.

<h3>What is the free body diagram?</h3>

Free-body diagrams are utilized to display the relative direction and strength of all forces that are being applied to an item in a certain scenario. A unique illustration of the geometric diagrams that were covered in a previous lesson is the free-body diagram. We will make use of these graphics throughout the entire study of physics.

A university student is carrying a backpack. One strap is hanging the rucksack immobile from one shoulder.

The weight of the backpack is balanced by the shoulder of the student.

The free-body diagram is attached below.

More about the free body diagram link is given below.

brainly.com/question/24087893

#SPJ4

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A car whose total mass is 800kg travelling with a uniform velocity of 20m/s suddenly observes a stationary dog 50m ahead on its
maxonik [38]

Answer:

The driver hits the stationery dog because the applied force is less than required force

Explanation:

Kinetic energy will be given by

KE=0.5mv^{2} where m is the mass of the vehicle and v is the speed/velocity of the vehicle.

Substituting 800 Kg for m and 20 m/s for v we obtain

KE=0.5*800*(20 m/s)^{2}=160,000

Frictional force by vehicle pads is given by

Fr=\frac {KE}{d} where d is the distance moved

Substituting 160000 for KE and 50 m for d we obtain

Fr=\frac {160000}{50}=3200 N

Therefore, the vehicle hits the dog since the required force is 3200N but the driver applied only 2000 N

7 0
3 years ago
In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be
LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

   Therefore we can say that

               m_1 * 15cm  = m_2 * xcm

Making x the subject of the formula  

                x = \frac{m_1 * 15}{m_2}

                    = \frac{0.42 * 15}{0.47}

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

                m_3 = 0.59 kg

3 0
3 years ago
What type of charge does an electron have?
GarryVolchara [31]

An electron has a negative charge. Hope this helps.

4 0
3 years ago
Read 2 more answers
A man fires a silver bullet of mass 2g with a velocity of200m/sec into wall. What is the temperature changeof the bullet? Note:
Sphinxa [80]

F=nmv

where;

n=no. of bullets = 1

m=mass of bullets=2g *10^-3

V=velocity of bullets200m/sec

F=1

loss in Kinetic energy=gain in heat energy

1/2MV^2=MS∆t

let M council M

=1/2V^2=S∆t

M=2g

K.E=MV^2/2

=(2*10^-3)(200)^2/2

2 councils 2

2*10^-3*4*10/2

K.E=40Js

H=mv∆t

(40/4.2)

40Js=40/4.2=mc∆t

40/4.2=2*0.03*∆t

=158.73°C

7 0
3 years ago
Which interaction of nature depends on the distance through which it acts and is involved in beta decay? a. strong c. electromag
USPshnik [31]

The answer is weak.  The interaction of nature that will depend on the distance through the way it acts and involved in beta decay is the weak interaction or the weak force. This interaction is the responsible for radioactive decay which also plays a significant role in nuclear fission.

5 0
3 years ago
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