1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ikadub [295]
3 years ago
15

On a part time job you are asked to bring a cylindrical iron rod of density 7800 kg/m^3, length 81.2 cm and diameter 2.75 cm fro

m a storage room to a machinist. Calculate the weight of the rod, w. The acceleration due to gravity, g= 9.81 m/s^2.
w= 36.9 N Part A

Part B) Now that you know the weight of the rod, do you think that you will be able to carry the rod without a cart? Yes or No?
Physics
1 answer:
Hitman42 [59]3 years ago
7 0

Answer:

3.76188 kg

36.9040428 N

Yes

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

L = Length of rod = 81.2 cm

d = Diameter of rod = 2.75 cm

r = Radius = \frac{d}{2}=\frac{2.75}{2}=1.375\ cm

A = Area of the cylinder =\pi r^2

v = Volume = L\times A

m = Mass

\rho = Density of rod = 7800 kg/m³

Density

\rho=\dfrac{m}{v}\\\Rightarrow m=\rho\times v\\\Rightarrow m=\rho\times L\times A\\\Rightarrow m=7800\times 0.812\times \pi\times 0.01375^2\\\Rightarrow m=3.76188\ kg

The mass of the rod is 3.76188 kg.

As the the mass of the rod is 3.76188 kg I will be able to carry the rod.

Weight

W=mg\\\Rightarrow W=3.76188\times 9.81\\\Rightarrow W=36.9040428\ N

The weight of the rod is 36.9040428 N

You might be interested in
A certain unbalanced force gives a 5kg object an acceleration of 15 m/s2. What would be the acceleration if the same force was a
stira [4]

Answer:

a2 = 2.5 m/s2

Explanation:

F1 = m1 a1 We use the same force so F1 = F2

= 5kg × 15m/s2 F2 = m2 a2

= 75N a2 is required

a2 = F2 / m2

= 75N / 30 kg

= 2.5 m/s2

7 0
3 years ago
An archer puts a 0.30-kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m. Assuming th
vovikov84 [41]

Answer:

41.74 m/s

Explanation:

The energy used to draw the bowstring = the kinetic energy of the arrow.

Fd = 1/2mv²................................ Equation 1

Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.

make v the subject of the equation

v = √(2Fd/m)...................... Equation 2

Given: F = 201 N, m = 0.3 kg, d = 1.3 m.

Substitute into equation 2

v = √(2×201×1.3/0.3)

v = √(1742)

v = 41.74 m/s.

Hence the arrow leave the bow with a speed of 41.74 m/s

3 0
3 years ago
Read 2 more answers
Which planet has a storm called the Great Dark Spot swirling in its atmosphere?
Viktor [21]
Im thinking D. Neptune..
3 0
3 years ago
Read 2 more answers
What is the power of a student that has done a work of 10 joules in 10 seconds​
Alla [95]

Answer:

1 Watt

Explanation:

P=W/t

P=10/10

P=1 Watt

8 0
3 years ago
You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown
Kipish [7]

Answer:

a

The mass is  m_2 =21.75*10^{-27} \ kg

b

The velocity is  v_2 = 3.0*10^{6} m/s

Explanation:

From the question we are told that

     The speed of the protons is  u_1 =  2.10*10^{7} m/s

     The mass of the protons is  m

     The speed of the rebounding protons are v_1 =  -1.80 * 10^{7} \ m/s

The negative sign shows that it is moving in the opposite direction

     

Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as

        m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1

Where m_1 is the mass of a single proton

          So substituting values

       m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1

        m_2 =13 m_1

The mass of on proton is  m_1 = 1.673 * 10^{-27} \ kg

So     m_2 =13 ( 1.673 * 10^{-27} )

        m_2 =21.75*10^{-27} \ kg

Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as

      m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2

Now  u_2 because before collision the the nucleus was at rest

So

        m_1 u_1 =  m_1 v_1 + m_2v_2

=>    v_2 =  \frac{m_1(u_1 -v_1)}{m_2}

Recall that m_2 =13 m_1

So

       v_2 =  \frac{m_1(u_1 -v_1)}{13m_1}

=>         v_2 =  \frac{(u_1 -v_1)}{13}

substituting values

              v_2 =  \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}

              v_2 = 3.0*10^{6} m/s

   

7 0
3 years ago
Other questions:
  • Which of the following rocks would most likely be found on the ocean floor
    11·2 answers
  • 10. The energy of moving objects is called
    5·2 answers
  • What is the kinetic energy of an object moving 40m/s with a mass of 20kg?
    7·1 answer
  • Choose the scientific notation that best represents the standard notation given. 6,840,000,000 m 68.4 x 108 m 6.84 x 108 m 6.84
    12·2 answers
  • What is the strength of an electric field that will balance the weight of a proton?
    12·1 answer
  • Si tenemos tres cubos del mismo tamaño (hierro, madera e icopor). ¿Qué diferencias puede encontrar entre ellos?
    6·1 answer
  • Heeeelllllllpppp I need this right now
    13·2 answers
  • Which of the following are correct statements about the way an atom is put
    5·1 answer
  • 4. How long does it take a car traveling at 45 km/h to travel 100.0 m?<br> 4500m
    7·1 answer
  • 3. A 40-gram ball of clay is dropped from a height, h, above a cup which is attached to a spring of spring force constant, k, of
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!