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Ainat [17]
2 years ago
13

How can you determine the class of lever of simple machine?​

Physics
1 answer:
V125BC [204]2 years ago
4 0

Answer:

A first-class lever has the fulcrum between the load and the effort. A second-class lever has the load between the effort and the fulcrum. A third-class lever has the effort between the load and the fulcrum. A see-saw is an example of a first-class lever.

Explanation:

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How much heat do you need to raise the temperature of 150 g of oxygen from -30c to -15c?
Naddika [18.5K]
The amount of heat needed to raise the temperature of a substance by \Delta T is given by
Q=m C_s \Delta T
where
m is the mass of the substance
Cs is its specific heat capacity
\Delta T is the increase in temperature

For oxygen, the specific heat capacity is approximately 
C_s = 0.92 J/(g K)
The variation of temperature for the sample in our problem is 
\Delta T= -15^{\circ}C-(-30^{\circ} C)=+15^{\circ}C=15 K
while the mass is m=150 g, so the amount of heat needed is
Q=m C_s \Delta T=(150 g)(0.92 J/g K)(15 K)=2070 J
4 0
3 years ago
Read 2 more answers
r was thirsty and decided to mix up a pitcher of lemonade. She put lemon juice, water, and sugar into a pitcher and stirred it t
slamgirl [31]
She could tell by how many components she put in. The compounds, are like the ingredients. The Mixture is all the ingredients stirred together.
7 0
2 years ago
Read 2 more answers
An object is moving east with a constant speed of 30 m/s for 5 seconds. What is the object's acceleration
pentagon [3]

Answer:

<h3>The answer is 6 m/s²</h3>

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

a =  \frac{v}{t}  \\

where

v is the velocity

t is the time

We have

a =  \frac{30}{5}  \\

We have the final answer as

<h3>6 m/s²</h3>

Hope this helps you

5 0
3 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
3 years ago
Which part of an atom is mostly empty space?
Veronika [31]

Answer:

The electron cloud is mostly empty space

7 0
3 years ago
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