All categories

3 years ago

A combination of two or more simple machines is a (n)

Physics

1 answer:

Lilit [14]3 years ago

8 0

A compound Machine is 2 machines that work together in order to make a task easier.

You might be interested in

An athlete competing in long jump leaves the ground with a speed of 9.14 m/s at an angle of 35.00 above the horizontal. What is

Lemur [1.5K]

Answer:

R = 8.01 m

Explanation:

We can solve this problem using the projectile launch equations. The jump length is the throw range

R = v₀² sin 2θ / g

in the exercise they give us the initial speed of 9.14 m / s and in the launch angle 35º

let's calculate

R = 9.14² sin (2 35) / 9.8

R = 8.01 m

this is the jump length

5 0

3 years ago

The maximum stress in a section of a circular tube subject to a torque is τmax = 27 MPa . If the inner diameter is Di = 3.75 cm

DIA [1.3K]

Answer:

$T_{max} = 4.735\,kN\cdot m$

Explanation:

The shear stress due to torque can be calculed by using the following model:

$\tau_{max} = \frac{T_{max}\cdot r_{ext}}{J_{tube}}$

The maximum torque on the section is:

$T_{max} = \frac{\tau_{max}\cdot J_{tube}}{r_{ext}}$

The Torsion Constant for the circular tube is:

$J_{tube} = \frac{\pi}{32}\cdot (D_{ext}^{4}-D_{int}^{4})$

$J_{tube} = \frac{\pi}{4}\cdot [(0.053\,m)^{4}-(0.038\,m)^{4}]$

$J_{tube} = 4.560\times 10^{-6}\,m^{4}$

Now, the require output is computed:

$T_{max} = \frac{(27\times 10^{3}\,kPa)\cdot (4.560\times 10^{-6}\,m^{4})}{0.026\,m}$

$T_{max} = 4.735\,kN\cdot m$

7 0

3 years ago

Read 2 more answers

A relaxed spring of length 0. 13 m stands vertically on the floor; its stiffness is 1180 N/m. You release a block of mass 0. 5 k

Daniel [21]

The length of spring when the block comes momentarily to rest on the compressed spring will be 0.054 m.The length of the spring by the letter x.

<h3>What is the potential energy of the spring?</h3>

The energy is stored in the spring when it is stretched or compressed by some length. It is the product of mass, gravity and distance compressed or stretched. Mathrmatically it is given by;

PE=mgh

The given data in the problem is ;

m is the mass of the block is 0.5 kg height,

h is the height is released is 0.7 m

x initial length of the spring = 0.13 m.

K is the force constant of the spring = 1180 N/m.

By the law of conservation of energy,

The potential energy of the spring gets converted

$\rm mgh=\frac{1}{2}KL^2 \\\\ \rm L= \sqrt{\frac{2mgh}{K} } \\\\ \rm L= \sqrt{\frac{2\times 0.5\times9.81 \times 0.7}{1180} \\\\ \\\\$