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alukav5142 [94]
3 years ago
6

Y’all I have best describes the purpose of a bar graph

Physics
1 answer:
Sholpan [36]3 years ago
3 0

so you can see all the different categories at once. both as a whole and on an individual scale.

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Now let’s apply the work–energy theorem to a more complex, multistep problem. In a pile driver, a steel hammerhead with mass 200
andrew11 [14]

Answer:

a) v = 7.67

b) n = 81562 N

Explanation:

Given:-

- The mass of hammer-head, m = 200 kg

- The height at from which hammer head drops, s12 = 3.00 m

- The amount of distance the I-beam is hammered, s23 = 7.40 cm

- The resistive force by contact of hammer-head and I-beam, F = 60.0 N

Find:-

(a) the speed of the hammerhead just as it hits the I-beam and

(b) the average force the hammerhead exerts on the I-beam.

Solution:-

- We will consider the hammer head as our system and apply the conservation of energy principle because during the journey of hammer-head up till just before it hits the I-beam there are no external forces acting on the system:

                                   ΔK.E = ΔP.E

                                  K_2 - K_1 = P_1- P_2

Where,  K_2: Kinetic energy of hammer head as it hits the I-beam

             K_1: Initial kinetic energy of hammer head ( = 0 ) ... rest

             P_2: Gravitational potential energy of hammer head as it hits the I-beam. (Datum = 0)

             P_1: Initial gravitational potential energy of hammer head      

- The expression simplifies to:

                                K_2 = P_1

Where,                     0.5*m*v2^2 = m*g*s12

                                v2 = √(2*g*s12) = √(2*9.81*3)

                                v2 = 7.67 m/s

- For the complete journey we see that there are fictitious force due to contact between hammer-head and I-beam the system is no longer conserved. All the kinetic energy is used to drive the I-beam down by distance s23. We will apply work energy principle on the system:

                               Wnet = ( P_3 - P_1 ) + W_friction

                               Wnet = m*g*s13 + F*s23

                               n*s23 = m*g*s13 + F*s23

Where,    n: average force the hammerhead exerts on the I-beam.

               s13 = s12 + s23

Hence,

                             n = m*g*( s12/s23 + 1) + F

                             n = 200*9.81*(3/0.074 + 1) + 60

                             n = 81562 N

                               

                                                   

6 0
3 years ago
A cylindrical container closed of both end has a radius of 7cm and height of 6cm A.)find the total surface area of the container
Natalka [10]

Given:

A cylindrical container closed of both end has a radius of 7cm and height of 6cm.

Explanation:

A.) Find the total surface area of the container.

  • A = 2πrh + 2πr²
  • A = 2(3.14)(7)(6) + 2(3.14)(7 × 7)
  • A = 263.76 + 307.72
  • A = 571.48

B.) Find the volume of the container.

  • V = πr²h
  • V = (3.14)(7×7)(6)
  • V = 923.16

Not sure huhuness.

#CarryOnLearning

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stich3 [128]
I'm pretty sure its transverse waves 
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6. Draw conclusions: Newton’s first law states that an object in motion will travel at a constant velocity unless acted upon by
gavmur [86]
You can test if it’s true by holding a pencil in mid air over a table and the table is supposed to be the unbalanced forced that stopped the pencil from moving at the constant velocity it was going by.
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Why would a flat sheet of paper and a wad of paper with the same mass not fall through the air at the same rate?
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The flat sheet of paper has more surface area than the crumpled ball
7 0
3 years ago
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