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3 years ago

If the car goes exits a freeway and goes from 65 mph to 35 mph is it accelerating?

Physics

1 answer:

Zolol [24]3 years ago

8 0

Answer:

No, the car is decelerating

Explanation:

No the car is decelerating if it exits a freeway and goes from 65

mph to 35 mph since the change in velocity is negative.

change in velocity = final - initial

change in velocity = 35 - 65

change in velocity = -30mph

Since the change in velocity is negative, hence the car is decelerating. Deceleration is a negative acceleration

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How many atoms are in 3MgCl2?

Bogdan [553]

Answer:

3MgCl2 has 9 atoms.

Explanation:

The Element Magnesium (Mg) has 3 atoms.

The Element Chloride (Cl) has 6 attoms.

Their fore 6 + 3 is 9 of course. 3MgCl2 has 9 atoms.

BTW: 3MgCl2 is a molecular compound as well as H2O and CO2.

8 0

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

3 years ago

A small, positively charged ball is moved close to a large, positively charged ball. Which describes how the small ball likely r

Angelina_Jolie [31]

Answer:c

Explanation:

8 0

3 years ago

Read 2 more answers

Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper

vodka [1.7K]

Explanation:

Expression to calculate thermal resistance for iron ( $R_{I}$ ) is as follows.

$R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}$

where, $L_{I}$ = length of the iron bar

$k_{I}$ = thermal conductivity of iron

$A_{I}$ = Area of cross-section for the iron bar

Thermal resistance for copper ( $R_{c}$ ) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where, $L_{c}$ = length of copper bar

$k_{c}$ = thermal conductivity of copper

$A_{c}$ = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

Q = $\frac{(100^{o} - 0^{o}}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

Putting the given values into the above formula as follows.

Q = $\frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

= $\frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}$

= 2.92 Joule

It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

P = $\frac{Q}{T}$

Here, T is 1 second so, power conducted is equal to heat transferred.

So, P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

7 0

3 years ago

NEED IT ASAP PLEASE 3 examples of how we use physics in our everyday life. Please explain throughly.

Flauer [41]

Answer:

Alarm Clock. The buzzing sound of an alarm clock helps you wake up in the morning as per your schedule. The sound is something that you can’t see, but hear or experience.
Cell Phones Cellphones have become like Oxygen gas in modern social life. Hardly, anyone would have been untouched by the effects of a cell phone. Whether conveying any urgent message or doing incessant gossips, cellphones are everywhere. But do you know how does a cell phone work? It works on the principle of electricity and the electromagnetic spectrum, undulating patterns of electricity and magnetism.
Walking.Now, when you get ready for your office/school, whatever medium of commutation is, you certainly have to walk up to a certain distance. You can easily walk is just because of Physics. While you have a walk in a park or on a tar road, you have a good grip without slipping because of a sort of roughness or resistance between the soles of your shoes and the surface of the road.

Explanation:

physical is related to things perceived through the senses as opposed to the mind; tangible or concrete.

3 0

3 years ago