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4 years ago

1. What is the linear distance traveled in one revolution of a 9 in diameter wheel?

Engineering

1 answer:

zmey [24]4 years ago

5 0

Explanation:

1. Circumference is pi times diameter.

C = πd

C = 9π inches

C ≈ 28.3 inches

2. Mechanical advantage is ratio of output force to input force.

MA = Fout / Fin

MA = 155 lb / 15 lb

MA ≈ 10.3

3. Mechanical advantage is ratio of input distance over output distance.

MA = din / dout

MA = 10.5 in / 0.85 in

MA ≈ 12.4

4. Mechanical advantage of pulley system is the number of strands.

MA = 8

5. Mechanical advantage is ratio of output force to input force.

MA = Fout / Fin

8 = Fout / 200 lb

Fout = 1600 lb

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A Wii remote flung from a hand through a TV, with a kinetic energy of 1.44J and a mass of 4.5kg. Whats the velocity?

Eva8 [605]

Answer:

0.8

Explanation:

7 0

3 years ago

At the instant shown car A is travelling with a velocity of 24 m/s and which is decreasing at 4 m/s2 along the highway. At the s

SVEN [57.7K]

(a) V(A/B) = (14 i - 17.32 J) m/s

(b) acc(A/B) = ( 5.11 i + 5.13 j ) m/s²

Explanation:

We will solve with respect to Cartesian vector form.

So,

V(A)= (24i) m/s

acc(A) = (4i) m/s²

There are two components of Car B, cos 60⁰ and sin 60⁰

V(B) = 20 cos 60° i + 20 sin 60° j

V(B) = (10 i + 17.32 j ) m/s

The car B moves along a curve, so it will have a tangential acceleration and a normal acceleration.

The tangential acceleration, a(t) = 5 m/s²

Normal acceleration, a(n) = $\frac{v^2}{p} \\\\$

So,

$a(n) = \frac{(20)^2}{250}\\ \\a(n) = 1.6 m/s^2$

For the tangential acceleration, the acceleration is slowing down. So,

a(t) = (-5 cos 60° i - 5 sin 60° j ) m/s²

a(t) = ( -2.5 i - 4.33 j) m/s²

For normal acceleration, it towards center. So,

a(n) = (1.6 sin 60° i - 1.6 cos 60° j) m/s²

a(n) = (1.39 i - 0.8 j ) m/s²

Total acceleration of Car B:

acc(B) = a(t) + a(n)

acc(B) = ( -2.5 i - 4.33 j) m/s² + (1.39 i - 0.8 j ) m/s²

acc(B) = (-1.11i - 5.13 j ) m/s²

(a) V(A/B) = ?

V(A) = V(B) + V(A/B)

(24i) m/s = (10 i + 17.32 j ) m/s + V(A/B)

V(A/B) = (14 i - 17.32 J) m/s

(b) acc(A/B) = ?

acc(A) = acc(B) + acc(A/B)

(4i) m/s² = (-1.11i - 5.13 j ) m/s² + acc(A/B)

acc(A/B) = ( 5.11 i + 5.13 j ) m/s²

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3 years ago

A car is about to start but it blows up. what is the problem with the car ?

ratelena [41]

Answer:

because there is a bomb

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3 years ago

Read 2 more answers

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=??

solution:

to find the shear stress

T_allow=T_fail/F.S

=120 Mpa/2.5

=48 Mpa

we know that,

V=P

Area for shear head:

A(head)=π×d×t

=π×0.04×0.075

=0.003×πm^2

Area for plate:

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

For head:

T_allow=V/A(head)

48 Mpa=P/0.003×π ..(V=P)

P =48 Mpa×0.003×π

=452.16 KN

For plate:

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

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3 years ago

On a piece of paper, sketch the x-y stress state and the properly oriented principal stress state. Use the resulting sketch to a

stealth61 [152]

Answer:

See explaination

Explanation:

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3 years ago