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bekas [8.4K]
3 years ago
6

A heavy ball with a weight of 110 N is hung from the ceiling of a lecture hall on a 4.9-m-long rope. The ball is pulled to one s

ide and released to swing as a pendulum, reaching a speed of 5.0 m/s as it passes through the lowest point.
Required:
What is the tension in the rope at that point?
Engineering
1 answer:
katovenus [111]3 years ago
5 0

Answer:T = 167.3 N

Explanation:

Given that the

Weight mg = 110 N

The mass m of the ball will be

m = 110/9.8 = 11.22 kg

As the direction of the ball’s velocity is changing, the force responsible for this is centripetal force F. And

F = mV^2/r

Where

V = 5.0 m/s

r = L = 4.9 m

m = 11.22

Substitute all these parameters into the formula

F = (11.22 × 5^2)/4.9

F = 280.6/4.9

F = 57.27 N

Tension T = F + mg

Substitute F and mg into the formula

T = 57.27 + 110

T = 167.3 N

Therefore, the tension in the rope at that point is 167.3 N

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You are a designer of a new processor. You have to choose between two possible implementations (called M1 and M2) of the same ar
Kaylis [27]

Answer:

A ) CPI : M1 = 2.4 , M2 = 2.65

B ) MIPS : M1 = 1083, M2 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is  by 27 million number of instructions per sec

Explanation:

A) The CPI for each machine

CPI = ( Total number of execution cycles ) / ( instruction counter executed )

For Machine 1 ( M1 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2.5 , number of times C was executed = 2.5, Number of times D was executed = 1. and this was based on the frequency given above

hence CPI for M1 =[ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] / 10

       CPI  for M1 = 2.4

For Machine 2 ( M2 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2.  number of times C was executed = 1.5, Number of times D was executed = 2.5 times. and this was based on the frequency given above

Hence CPI  for M1 = [ ( 2 * 4 ) + ( 2 * 2 ) + ( 3 * 1.5 ) + ( 4 * 2.5 ) ] / 10

            CPI for M2 = 2.65

B ) Calculate the native MIPS  ratings for M1 and M2

MIPS = ( instruction counts ) / ( Execution time * 10^6 )

For M1

Assumptions : number of instructions executed = 10

                        each clock cycle = 0.3846 * 10^-9.      frequency = 2.6 Ghz

first we calculate the total execution time which is equal to :

= [ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] * 0.3846 * 10 ^-9

= 9.2304 * 10 ^-9 secs

therefore the MIPS for M1

= 10 / ( 9.2304 * 10^-9 ) * 10^6  = 1083

                                         

For M2

Assumptions : number of instructions executed = 10

                        each clock cycle = 0.3846 * 10^-9.      frequency = 2.8 Ghz

first we calculate the total execution time which is equal to :

= [ (2*4) + (2*2) + (3 * 1.5 ) + ( 4 * 2.5 ) ] * 0.3846 * 10^-9 = 9.4631 * 10^-9 secs

therefor the MIPS for M2

= 10 / ( 9.4631*10^-9) * 10^6 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is by 27 million number of instructions per sec

8 0
3 years ago
If the bolt head and the supporting bracket are made of the same material having a failure shear stress of 'Tra;i = 120 MPa, det
Nina [5.8K]

Answer:

P=361.91 KN

Explanation:

given data:

brackets and head of the screw are made of material with T_fail=120 Mpa

safety factor is F.S=2.5

maximum value of force P=??

<em>solution:</em>

to find the shear stress

                            T_allow=T_fail/F.S

                                         =120 Mpa/2.5

                                         =48 Mpa

we know that,

                               V=P

<u>Area for shear head:</u>

                              A(head)=π×d×t

                                           =π×0.04×0.075

                                           =0.003×πm^2

<u>Area for plate:</u>

                               A(plate)=π×d×t  

                                            =π×0.08×0.03

                                            =0.0024×πm^2

now we have to find shear stress for both head and plate

<u>For head:</u>

                                   T_allow=V/A(head)

                                    48 Mpa=P/0.003×π                 ..(V=P)

                                             P =48 Mpa×0.003×π

                                                =452.16 KN

<u>For plate:</u>

                                   T_allow=V/A(plate)

                                    48 Mpa=P/0.0024×π                 ..(V=P)

                                             P =48 Mpa×0.0024×π

                                                =361.91 KN

the boundary load is obtained as the minimum value of force P for all three cases. so the solution is

                                                P=361.91 KN

note:

find the attached pic

7 0
3 years ago
A plate of an alloy steel has a plane-strain fracture toughness of 50 MPa√m. If it is known that the largest surface crack is 0.
Ivahew [28]

Answer:

option B is correct. Fracture will definitely not occur

Explanation:

The formula for fracture toughness is given by;

K_ic = σY√πa

Where,

σ is the applied stress

Y is the dimensionless parameter

a is the crack length.

Let's make σ the subject

So,

σ = [K_ic/Y√πa]

Plugging in the relevant values;

σ = [50/(1.1√π*(0.5 x 10^(-3))]

σ = 1147 MPa

Thus, the material can withstand a stress of 1147 MPa

So, if tensile stress of 1000 MPa is applied, fracture will not occur because the material can withstand a higher stress of 1147 MPa before it fractures. So option B is correct.

8 0
3 years ago
Identify the right components for gsm architecture that consists of the hardware or physical equipment such as digital signal pr
sergiy2304 [10]

The right components for gsm architecture that consists of the hardware or physical equipment such as digital signal processors, radio transceiver, display, battery, case and sim card is the Mobile station.

<h3>What are the 4 main components?</h3>

In GSM, a cell station includes 4 fundamental additives: Mobile termination (MT) - gives not unusualplace features consisting of: radio transmission and handover, speech encoding and decoding, blunders detection and correction, signaling and get right of entry to to the SIM. The IMEI code is connected to the MT.

Under the GSM framework, a cell tele cell smartphone is called a Mobile Station and is partitioned into  wonderful additives: the Subscriber Identity Module (SIM) and the Mobile Equipment (ME).

Read more about the mobile station:

brainly.com/question/917245

#SPJ4

6 0
2 years ago
The intake and exhaust processes are not considered in the p-V diagram of Otto cycle. a) true b) false
vovangra [49]

Answer:

b) false

Explanation:

We know that Otto cycle is the ideal cycle for all petrol working engine.In Otto cycle all process are consider is ideal ,means there is no any ir-reversibility in the processes.

It consist four processes

1-2:Reversible adiabatic compression

2-3:Constant volume heat addition

3-4:Reversible adiabatic expansion

3-4:Constant volume heat rejection

Along with above 4 processes intake and exhaust processes are parallel to each other.From the P-v diagram we can see that all processes.

But actually in general we are not showing intake and exhaust line then it did not mean that in Otto cycle did not have intake and exhaust processes.

6 0
3 years ago
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