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3 years ago

Define wheel and axle in science terms

Physics

1 answer:

topjm [15]3 years ago

5 0

Answer:

A simple machine consisting of an axle to which a wheel is fastened so that torque applied to the wheel winds a rope or chain onto the axle, yielding a mechanical advantage equal to the ratio of the diameter of the wheel to that of the axle.

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In a race, you run 3000 meters east in 21 minutes. What is your velocity in km/min?

STALIN [3.7K]

velocity = distance / time v= d/t firstly you change 3000 m into km distance D = 3000/1000 = 3 km secondly time t = 21 min using velocity formula v = d/t = 3 km/21min =0.142km/min

6 0

3 years ago

Instead of rolling down the ramp over the step, imagine the ball falls off the back of the ramp directly to the ground. What is

lesya692 [45]

According to the law of conservation of energy, the potential energy is converted to kinetic energy. Remember, potential energy is calculated using height and weight. If the ball is on the ground, height is 0.

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3 years ago

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In a home's central heating system, when the room temperature measured by the thermostat reaches a set value, the thermostat sen

shutvik [7]

These actions are an example of feedback.
Given that the room has reached the desired temperature, there is no more need for it to be heated, at least until the temperature drops a bit. This is why the thermostat sends feedback about this situation to the heater, which immediately switches off until it is needed again.

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3 years ago

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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:

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

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

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3 years ago

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What muscles do you use to throw a ball

tamaranim1 [39]

You use your shoulder muscles aka the deltiods of the muscle which plays a huge part in the rotation of your arm
you also use your triseps
also, latissimus dorsi
abdominals
and your quadriceps

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3 years ago