A wood pipe having an inner diameter of 3 ft. is bound together using steel hoops having a cross sectional area of 0.2 in.2 The

Incomplete query. The complete inquiry is as follows

Calculate the torque exerted on the shaft of a vehicle transmitting 225 hp at a rotation speed of 3000 rpm.

Response:

Torque=0.51 Btu

Analysis:

Given Information

Power=225 hp

Revolutions =3000 rpm

To determine

T( torque )=?

Process

As an object is moved by force over a distance, work is performed on that object. Similarly, when torque rotates an object through an angle, work is also accomplished.

Answer:

A) and B) are valid.

Explanation:

When an object remains at rest, it is indicative that no net force acts upon it.

The downward gravitational force from Earth must be counterbalanced by an upward force of equal magnitude in order to maintain rest.

This upward force is provided by the normal force, which adjusts to satisfy Newton’s 2nd Law and is always perpendicular to the surface supporting the object (in this instance, the ground).

At the molecular level, this normal force comes from the ground's bonded molecules acting like tiny springs, compressed by the object’s molecules, providing an upward restorative force.

Thus, statements A) and B) are true.

Response: The width decreases by 2.18 × 10^(-6) m

Clarification:

Given data;

Shear Modulus; E = 207 GPa = 207 × 10^(9) N/m²

Force; F = 60000 N.

Poisson’s ratio; υ = 0.30

The initial width is 20 mm, and the thickness is 40 mm.

Area = 20 × 10^(-3) × 40 × 10^(-3)

Area = 8 × 10^(-4) m²

The formula for shear modulus is;

E = σ/ε_z

where σ represents stress calculated as Force(F)/Area(A)

while ε_z stands for longitudinal strain.

Thus;

E = (F/A)/ε_z

ε_z = (F/A)/E

ε_z = (60,000/(8 × 10^(-4)))/(207 × 10^(9))

ε_z = 3.62 × 10^(-4)

Next, the lateral strain is given by;

ε_x = - υ × ε_z

ε_x = -0.3 × 3.62 × 10^(-4)

ε_x = -1.09 × 10^(-4)

The change in width can be determined as;

Δw = w_o × ε_x

Where w_o denotes the original width = 20 × 10^(-3) m

So; Δw = 20 × 10^(-3) × -1.09 × 10^(-4)

Δw = -2.18 × 10^(-6) m

A negative sign indicates a reduction in width.

Therefore, the width decreases by 2.18 × 10^(-6) m

Answer:

A flea can attain a maximum elevation of 51 mm.

Explanation:

Hello!

The following equations describe the height and velocity of the flea:

During the jump:

h = h0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

In free fall:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = flea's height at time t.

h0 = initial height.

v0 = starting velocity.

t = time interval.

a = flea's acceleration while jumping.

v = flea's velocity at that specific time.

g = gravitational acceleration.

Initially, we need to determine the time taken for the flea to attain a height of 0.0005 m. This will help us calculate the flea's velocity during the jump:

h = h0 + v0 · t + 1/2 · a · t²

If we assume the ground as the origin, thus h0 = 0. Since the flea starts stationary, v0 = 0. Therefore:

h = 1/2 · a · t²

We need to find the value of t when h = 0.0005 m:

0.0005 m = 1/2 · 1000 m/s² · t²

0.0005 m / 500 m/s² = t²

t = 0.001 s

Next, we calculate the velocity achieved during that time:

v = v0 + a · t (v0 = 0)

v = a · t

v = 1000 m/s² · 0.001 s

v = 1.00 m/s

At a height of 0.50 mm, the flea's velocity stands at 1.00 m/s. This initial speed will reduce due to gravity's downward pull. When the speed reaches zero, the flea will have reached its peak height. Using the velocity equation, let's determine the time taken to reach maximum height (v = 0):

v = v0 + g · t

At peak height, v = 0:

0 m/s = 1.00 m/s - 9.81 m/s² · t

-1.00 m/s / -9.81 m/s² = t

t = 0.102 s

Now, we can compute the height attained by the flea during this time:

h = h0 + v0 · t + 1/2 · g · t²

h = 0.0005 m + 1.00 m/s · 0.102 s - 1/2 · 9.81 m/s² · (0.102 s)²

h = 0.051 m

A flea reaches a maximum height of 51 mm.

The mass of the block submerged in water measures 1.94 kg. We know the water's density to be 1 gm/cm³, with the initial scale reading at 19 N and a change in this reading amounting to 10 N. To determine the mass of the block when it is immersed in water, we apply the gravitational acceleration g, valued at 9.8.