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Oksana_A [137]
3 years ago
15

What is a dynamic load? *

Engineering
1 answer:
alexira [117]3 years ago
8 0
Answer: Bricks in Building
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A. A compound gear train is composed of four gears: E,F,G,H. Gear E has 15 teeth and is meshed with gear F. Gear F has 25 teeth
cricket20 [7]

Answer:

What subject is this in?? That would help a lot knowing

Explanation:

5 0
3 years ago
Refectories are one of the types of ceramics that have low melting temperature. a)-True b)-False
MaRussiya [10]

Answer:

b). False

Explanation:

A refractory material is a type of material that can withstand high temperatures without loosing its strength. They are used in reactors, furnaces, kilns, etc.

    Refractory materials are certain super alloys and ceramics materials.

Properties of refractory materials :

1. Refractory materials have high melting point.

2.They acts barriers between high heat zone and low heat zone.

3. The specific heat of refractory material is very low.

4. Refractories that have high bulk densities are better in quality.

Hence, Refractory materials have a very high melting temperature.

3 0
3 years ago
A 2-cm-diameter vertical water jet is injected upward by a nozzle at a speed of 15 m/s. Determine the maximum weight of a flat p
Ede4ka [16]

Answer:58.28 N

Explanation:

Given data

dia. of nozzle \left ( d\right )=2 cm

initial velocity\left ( u\right )=15 m/s

height\left ( h\right )=2m

Now velocity of jet at height of 2m

v^2-u^2=2gh

v^2=15^2-2\left ( 9.81\right )\left ( 2\right )

v=\sqrt{185.76}=13.62 m/s

Now\ forces\ on\ plate\ are\ weight\left ( Downward\right ) and jet\ force\left ( upward\right )

equating them

W=\left ( \rho Av\right )v

W=10^{3}\times \frac{\pi}{4}\left ( 0.02\right )^2\times 13.62^2

W=58.28 N

7 0
3 years ago
The wheel and the attached reel have a combined weight of 50lb and a radius of gyration about their center of 6 A k in = . If pu
marishachu [46]

The complete question is;

The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb.ft, determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

The image of this system is attached.

Answer:

Velocity = 11.8 ft/s

Explanation:

Since the wheel at A rotates about a fixed axis, then;

v_c = ω•r_c

r_c is 4.5 in. Let's convert it to ft.

So, r_c = 4.5/12 ft = 0.375 ft

Thus;

v_c = 0.375ω

Now the mass moment of inertia about of wheel A about it's mass centre is given as;

I_a = m•(k_a)²

The mass in in lb, so let's convert to slug. So, m = 50/32.2 slug = 1.5528 slug

Also, let's convert ka from inches to ft.

So, ka = 6/12 = 0.5

So,I_a = 1.5528 × 0.5²

I_a = 0.388 slug.ft²

The kinetic energy of the system would be;

T = Ta + Tc

Where; Ta = ½•I_a•ω²

And Tc = ½•m_c•(v_c)²

So, T = ½•I_a•ω² + ½•m_c•(v_c)²

Now, m_c is given as 200 lb.

Converting to slug, we have;

m_c = (200/32.2) slugs

Plugging in the relevant values, we have;

T = (½•0.388•ω²) + (½•(200/32.2)•(0.375ω)²)

This now gives;

T = 0.6307 ω²

The system is initially at rest at T1 = 0.

Resolving forces at A, we have; Ax, Ay and Wa. These 3 forces do no work.

Whereas at B, M does positive work and at C, W_c does negative work.

When pulley B rotates, it has an angle of; θ_b = 5 revs × 2π rad/revs = 10π

While the wheel rotates through an angle of;θ_a = (rb/ra) • θ_b

Where, rb = 3 in = 3/12 ft = 0.25 ft

ra = 7.5 in = 7.5/12 ft = 0.625 ft

So, θ_a = (0.25/0.625) × 10π

θ_a = 4π

Thus, we can say that the crate will have am upward displacement through a distance;

s_c = r_c × θ_a = 0.375 × 4π

s_c = 1.5π ft

So, the work done by M is;

U_m = M × θ_b

U_m = 50lb × 10π

U_m = 500π

Also,the work done by W_c is;

U_Wc = -W_c × s_c = -200lb × 1.5π

U_Wc = -300π

From principle of work and energy;

T1 + (U_m + U_Wc) = T

Since T1 is zero as stated earlier,

Thus ;

0 + 500π - 300π = 0.6307 ω²

0.6307ω² = 200π

ω² = 200π/0.6307

ω² = 996.224

ω = √996.224

ω = 31.56 rad/s

We earlier derived that;v_c = 0.375ω

Thus; v_c = 0.375 × 31.56

v_c = 11.8 ft/s

3 0
3 years ago
A torque T 5 3 kN ? m is applied to the solid bronze cylinder shown. Determine (a) the maximum shearing stress, (b) the shethe 1
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