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3 years ago

What simple machine can be described as a shaft that is attached to the center of a wheel?

Physics

1 answer:

charle [14.2K]3 years ago

4 0

"Wheel & Axle" <span>can be described as a shaft that is attached to the center of a wheel

Hope this helps!</span>

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What type of tide occurs when the sun, moon, and earth form a right triangle in space?

stiks02 [169]

Neap tide is your answer

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3 years ago

An observer on the earth sees a spaceship approaching at 0.54c. The ship then launches an exploration vehicle that, according to

AURORKA [14]

Answer:

Explanation:

Expression for relative velocity

= $\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2} }$

= (.54 + .82 )c/ $1+ \frac{.54 \times.82}{1}$

= 1.36 c / 1.4428

= .94 c

β = .94

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3 years ago

Students with disabilities who require assistance in class are generally mainstreamed into regular classes rather than putting t

Brilliant_brown [7]

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3 years ago

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A diesel engine does not use spark plugs to ignite the fuel and air in the cylinders. Instead, the temperature required to ignit

Klio2033 [76]

Answer:

Compression ratio = 24.42

Explanation:

From Thermodynamic,

Adiabatic Equation ⇒ TV^{γ-1} = Constant.

⇒T₁V₁^{γ-1} = T₂V₂^{γ-1} ..................(1)

Where T₁= initial Temperature, V₁ = Initial Volume, T₂ = final Temperature, V₂ = Final Volume.

Given: T₁= 18°C we convert to Kelvin(K) by adding 273.

∴ T₁= 18°C + 273 = 291K

T₂ = 733° C also,we convert to Kelvin(K) by adding 273

∴ T₂ = 733° C +273 =1046K

γ = 7/5

∴ Rearranging equation(1), we have

(T₁/T₂) = (V₂/V₁)^{γ-1}................(2)

also rearranging equation(2) we have

(V₁/V₂)^{γ-1} = (T₂/T₁).

Where (V₁/V₂) = Compression ratio.

∴ (V₁/V₂)^{(7/5)-1} =( 1046/291)

simplifying the index in the equation

I.e (7/5)-1 = (7-5 )/5 = 2/5.

(V₁/V₂)^2/5 =(1046/291)^2/5

Multiplying the power on both side of the equation by 5/2.

∴(V₁/V₂)^(2/5)×(5/2) = (1046/291)^(5/2)

⇒ V₁/V₂= (1046/291)^2.5=( 3.59)^2.5

V₁/V₂ = 24.42.

∴ Compression ratio = 24.42

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3 years ago

Uphill escape ramps are sometimes provided to the side of steep downhill highways for trucks with overheated brakes. For a simpl

o-na [289]

Answer:

404.4 m

Explanation:

Converting the initial speed from km/h to m/s then

$140\times \frac {1000m}{3600s}=38.88888889 m/s \approx 38.89 m/s$

The acceleration is resolved as shown in the figure hence

deceleration of the truck along the inclined plane will be

$a=-g sin \theta$ where g is acceleration due to gravity

Substituting g with $9.81 m/s^{2}$ then

$a=-9.81 m/s^{2} sin 11^{\circ}=-1.871836245\approx -1.87 m/s^{2}$

Using kinematic equation

$v^{2}=u^{2}+2as$ and making s the subject then

$s=\frac {v^{2}-u^{2}}{2a}$ where v and u are final and initial velocities respectively

Substituting 0 for v, 38.89 m/s for u and $-1.87 m/s^{2}$ then

$s=\frac {0^{2}-38.89^{2}}{2\times -1.87}=404.3936096 m\approx 404.4 m$

3 0

3 years ago