Ask question

Ask question

All categories

3 years ago

13

Determine the largest intensity w of the uniform loading that can be applied to the frame without causing either the average nor

mal stress or the average shear stress at section b–b to exceed σ=19 mpa and τ=24 mpa, respectively.

1 answer:

jeka943 years ago

8 0

.........................|||||||..............................

You might be interested in

Which explanation best describes a wave?

Gekata [30.6K]

Answer:

C. a disturbance that travels through a medium with a transfer of energy and without a transfer of matter

Explanation:

A wave is any disturbance that transfers energy from one location to the other via a substance called medium. It is important to note that a wave only conveys energy and not matter. For example, sound wave is a type of wave that carries sound energy from one place to another via mediums such as water, air etc.

Hence, according to this question, a wave can be described as a disturbance that travels through a medium with a transfer of energy and WITHOUT A TRANSFER OF MATTER.

5 0

3 years ago

John doe gets on the highway in his 1967 Shelby 427 Cobra starting from the dead stop at the bottom of the on ramp of it can be

masha68 [24]

Answer:

It will take john 4.477 seconds

7 0

3 years ago

How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s

poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

$W = \Delta KE$

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

Here,

m = mass

$v_{f,i}$ = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2$

$W_1 = 150(m) J$

Second case) When the particle goes from 20m/s to 30m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2$

$W_1 = 250(m) J$

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0

3 years ago

The upper body primal movement patterns are

kramer

Answer:

Bend to extend. Bending with a hips back movement, back straight, feet flat and forward. ...

Squat. This is a hips down motion. ...

Lunge. This is a long, linear stride, lowering your back knee to just above the ground, with a completely upright torso. ...

Rotate. ...

Push. ...

Pull. ...

Gait.

8 0

3 years ago

a mass of 4 kilograms is paced onto a spring snd it stretches 0.2 meters. Why is the spring constant?

Ray Of Light [21]

Answer:

fgch

Explanation:

5 0

3 years ago