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3 years ago

Describe the parts of the wheel and axle

Physics

2 answers:

suter [353]3 years ago

8 0

Answer:

A simple machine that may be used the most often is called the wheel and axle. The wheel and axle has two basic parts: wheel and axle. It has two circular objects which includes a larger disc and a small cylinder both joined at the center.

Explanation:

Natalka [10]3 years ago

3 0

Explanation:

A simple machine that may be used the most often is called the wheel and axle. The wheel and axle has two basic parts: wheel and axle. They can be found everywhere. It has two circular objects which includes a larger disc and a small cylinder both joined at the center.The large disc is the wheel, and the small cylinder or rod is the axle. There may be two wheels attached to the axle.

There are two basic ways a wheel and axle can work together to help move things.

1. The Force is applied to the Wheel

For example, a screwdriver is an example of a wheel and axle. The handle is the wheel where the force is applied. It turns or spins and increases the force of the shaft or axle, which helps turn the screw.

Another example of force being applied to the wheel is when a doorknob is turned. The wheel (doorknob) is turned and the locking mechanism connected to the shaft turns and the door can then be opened.

2. The Force is applied to the Axle

A Ferris wheel is an example of force being applied to the axle. When the axle turns it results in the giant wheel turning. The wheel is much larger than the axle and covers more distance and area. A ceiling fan works the same way.

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A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2. Compu

boyakko [2]

Answer:

0.42 m/s²

Explanation:

r = radius of the flywheel = 0.300 m

w₀ = initial angular speed = 0 rad/s

w = final angular speed = ?

θ = angular displacement = 60 deg = 1.05 rad

α = angular acceleration = 0.6 rad/s²

Using the equation

w² = w₀² + 2 α θ

w² = 0² + 2 (0.6) (1.05)

w = 1.12 rad/s

Tangential acceleration is given as

$a_{t}$ = r α = (0.300) (0.6) = 0.18 m/s²

Radial acceleration is given as

$a_{r}$ = r w² = (0.300) (1.12)² = 0.38 m/s²

Magnitude of resultant acceleration is given as

$a = \sqrt{a_{t}^{2} + a_{r}^{2}}$

$a = \sqrt{0.18^{2} + 0.38^{2}}$

$a$ = 0.42 m/s²

8 0

3 years ago

What happens to a circuit's resistance as more resistors are added in series?

zysi [14]

It increases

Explanation:

A circuit's resistance increases as more resistors are added to it in series.

Series connection of circuit provides an additive increase in the resistance offered by the overall circuit.

A resistor is a device in a circuit that impedes the flow of electric current. It simply uses electric current.

Examples are bulbs, radio, television.

The more of these devices connected one to another in series, the more the resistance increases.

learn more;

Voltage brainly.com/question/6949231

#learnwithBrainly

8 0

3 years ago

Plz answer this asap no explanation needed answer both questions first answer gets brainliest

Soloha48 [4]

Answer:

8) d

9) c

<em>Hope this helps! :)</em>

5 0

3 years ago

An air-gap, parallel plate capacitor with area A and gap width d is connected to a battery that maintains the plates at potentia

sergejj [24]

Answer:

The new potential energy decreases by the factor of 2 to the old potential energy.

Explanation:

Capacitance of a parallel plate capacitor is given by the relation :

C = (ε₀A)/d

Here ε₀ is vacuum permittivity, A is area of the capacitor plate and d is the distance between them.

Potential energy of the capacitor, U = $\frac{1}{2}CV^{2}$

Here V is the potential difference between the plates.

According to the problem, the distance between the plates get double but area remains same. So,

d₁ = 2d

Here d₁ is new distance between the plates.

Hence, new capacitance is :

C₁ = (ε₀A)/d₁ = (ε₀A)/2d = C/2

The capacitor have same potential difference that is V. Hence, the new potential energy is :

U₁ = $\frac{1}{2}C_{1} V^{2}$ = $\frac{1}{2}\frac{C}{2} V^{2}$

U₁ = U/2

$\frac{U_{1} }{U} = \frac{1}{2}$

7 0

3 years ago

If the work required to speed up a car from 11 km/h to 21 km/h is 6.0×103 J , what would be the work required to increase the ca

goblinko [34]

Explanation:

We need convert the velocities first to m/s and we get the following:

v2 = 21 km/hr = 5.8 m/s

v1 = 11 km/hr = 3.1 m/s

We need to find the mass of the car also for later use do using the work-energy theorem:

$delta \: w = \frac{1}{2} m(v \frac{2}{2} - v\frac{2}{1} )$

6.0x10^3 J = (0.5) m [(5.8)^2 - (3.1)^2]

m = 499.4 kg

Now we determine work needed delta W to change its velocity from 21 km/hr to 33 km/hr

v2 = 33 km/hr = 9.2 m/s

v1 = 21 km/hr = 5.8 m/s

delta W = (0.5)(499.4)[(9.2)^2 - (5.8)^2]

= 1.3 x 10^4 J

6 0

3 years ago