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1 year ago

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An oil has a kinematic viscosity of 1.25 × 10^-4 and a specific gravity of 0.8. What is the viscosity of this oil in SI unit of

kg/m-s? In the unit of centipoise?

1 answer:

emmainna [20.7K]1 year ago

5 0

Step 1 of 3
Calculate the density of oil as follows:

Here, SG is the specific gravity and is the density of standard fluid (water).
Substitute 0.8 for SG and for .

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Calculate the watts of power expended when a force 400 N travels 12 meters over 10 seconds.

Pachacha [2.7K]

Answer:

Explanation:

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2 years ago

Which of the following statements is NOT correct?

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The answer is
D:if the

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2 years ago

What is the potential energy for 65kg climber on top of Mount Everest (8,800 m high)

Fed [463]

Answer:

PE= m * g *h

work:

PE= 65kg * 9.8 kg *8,800 m

PE=5605600 m/kg

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2 years ago

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

4 0

2 years ago

A halfback on an apparent breakaway for a touchdown is tackled from behind. If the halfback has a mass of 98 kg and was moving a

uranmaximum [27]

Answer:

The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

Explanation:

Given that,

Mass of halfback = 98 kg

Speed of halfback= 4.2 m/s

Mass of corner back = 85 kg

Speed of corner back = 5.5 m/s

We need to calculate their mutual speed immediately after the touchdown-saving tackle

Using conservation of momentum

$m_{h}v_{h}+m_{c}v_{c}=m_{h+c}v_{h+c}$

Where, $m_{h}$ = mass of halfback

$m_{c}$ =mass of corner back

$v_{h}$ = velocity of halfback

$v_{c}$ = velocity of corner back

Put the value into the formula

$98\times4.2+85\times5.5=(98+85)\times v$

$v=\dfrac{98\times4.2+85\times5.5}{98+85}$

$v=4.80\ m/s$

Hence, The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

3 0

2 years ago