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Naddik [55]
3 years ago
12

A hole is punched at A in a plastic sheet by applying a 660-N force P to end D of lever CD, which is rigidly attached to the sol

id cylindrical shaft BC. Design specifications require that the displacement of D should not exceed 15 mm from the time the punch first touches the plastic sheet to the time it actually penetrates it. Determine the required diameter of shaft BC if the shaft is made of a steel with G.
Physics
1 answer:
olasank [31]3 years ago
4 0

Answer:

hello your question lacks some data and required diagram

G = 77 GPa,  т all = 80 MPa

answer : required diameter = 252.65 * 10-^3 m

Explanation:

Given data :

force ( P )  = 660 -N force

displacement = 15 mm

G = 77 GPa

т all = 80 MPa

i) Determine the required diameter of shaft BC

considering the vertical displacement ( looking at handle DC from free body diagram )

D' = 0.3 sin∅  ,   where D = 0.015

hence ∅ = 2.8659°

calculate the torque acting at angle ∅  of  CD on the shaft BC

Torque = 660 * 0.3 cos∅

             = 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N

hello attached is the remaining part of the solution

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What are the magnitude and direction of electric field at the center of square?
Kobotan [32]

Answer:

hello the diagram related to this question is missing attached below is the missing diagram

Answer :

The magnitude of the electric field = 4KQ / L^2

direction = 45° east to south

Explanation:

The magnitude of the electric field = 4KQ / L^2

direction = 45° east to south

6 0
3 years ago
Which force is represented by the arrow at A?
Nat2105 [25]

Answer:

B) Force of gravity

Explanation:

As it is the force which acts downward on the body towards the ground, it is clearly the force gravity represented by the arrow at A.

7 0
2 years ago
Read 2 more answers
Consider the hydrogen atom. How does the energy difference between adjacent orbit radii change as the principal quantum number i
Kisachek [45]

Answer:

the energy difference between adjacent levels decreases as the quantum number increases

Explanation:

The energy levels of the hydrogen atom are given by the following formula:

E=-E_0 \frac{1}{n^2}

where

E_0 = 13.6 eV is a constant

n is the level number

We can write therefore the energy difference between adjacent levels as

\Delta E=-13.6 eV (\frac{1}{n^2}-\frac{1}{(n+1)^2})

We see that this difference decreases as the level number (n) increases. For example, the difference between the levels n=1 and n=2 is

\Delta E=-13.6 eV(\frac{1}{1^2}-\frac{1}{2^2})=-13.6 eV(1-\frac{1}{4})=-13.6 eV(\frac{3}{4})=-10.2 eV

While the difference between the levels n=2 and n=3 is

\Delta E=-13.6 eV(\frac{1}{2^2}-\frac{1}{3^2})=-13.6 eV(\frac{1}{4}-\frac{1}{9})=-13.6 eV(\frac{5}{36})=-1.9 eV

And so on.

So, the energy difference between adjacent levels decreases as the quantum number increases.

5 0
3 years ago
A motorcycle accelerated from 10 metre per second to 30 metre per second in 6 seconds. How far did it travel in this time ?
Ber [7]
U=10 m/s
v=30 m/s
t=6 sec

therefore, a=(v-u)/t
                   =(30-10)/6
                   =(10/3) ms^-2

now, displacement=ut+0.5*a*t^2
                              =60+ 0.5*(10/3)*36
                              =120 m
And you can solve it in another way:

v^2=u^2+2as
or, s=(v^2-u^2)/2a
       =(900-100)/6.6666666.......
       =120 m

7 0
4 years ago
Two wheels with fixed hubs and radii 0.51 m and 1.9 m, each having a mass of 3 kg, start from rest. Forces 5 N and F2 are applie
Katarina [22]

Answer:

18.63 N

Explanation:

Assuming that the sum of torques are equal

Στ = Iα

First wheel

Στ = 5 * 0.51 = 3 * (0.51)² * α

On making α subject of formula, we have

α = 2.55 / 0.7803

α = 3.27

If we make the α of each one equal to each other so that

5 / (3 * 0.51) = F2 / (3 * 1.9)

solve for F2 by making F2 the subject of the formula, we have

F2 = (3 * 1.9 * 5) / (3 * 0.51)

F2 = 28.5 / 1.53

F2 = 18.63 N

Therefore, the force F2 has to 18.63 N in order to impart the same angular acceleration to each wheel.

3 0
3 years ago
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