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V125BC [204]
3 years ago
12

if a car travels 30 miles north for one half hour, 50 miles east for one hour, and 30 miles south for 30 minutes, what is the to

tal displacement of the car?
Physics
2 answers:
galina1969 [7]3 years ago
6 0
Since we're given the distances, we don't need the times.

(30 miles north) + (30 miles south)  =  zero vertical displacement

The total displacement after this grueling 2 hours of driving is  50 miles east .
egoroff_w [7]3 years ago
5 0
50 miles east. because the 30 miles north then south cancel each other out. 
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Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum
cestrela7 [59]

Answer:

T_{2}=278.80 K

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}.

Now, let's use the ideal gas equation to the initial and the final state:

\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}

Let's recall that the term nR is a constant. That is why we can match these equations.  

We can find a relation between the volumes of the initial and the final state.

\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}

Combining this equation with the first equation we have:

(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}

(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}

Now, we just need to solve this equation for T₂.

T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40

Finally, T2 will be:

T_{2}=278.80 K

6 0
3 years ago
A 5cm tall object is placed 4cm in front of a converging lens that has a focal length of 8cm. Where is the image located in ____
OverLord2011 [107]

Answer:

a. -8 cm

Explanation:

d_{o} = distance of the object = 4 cm

d_{i} = distance of the image = ?

f = focal length of the converging lens  = 8 cm

using the lens equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}

\frac{1}{4} + \frac{1}{d_{i}} = \frac{1}{8}

d_{i} = - 8 cm

4 0
3 years ago
What course the colour of silt soil?
Nata [24]

Answer:

Beige to black.

Explanation:

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3 0
3 years ago
A chart labeled table A : effect of height on temperature with initial temperature as 25 degrees Celsius, mass w is 1.0 kilogram
o-na [289]

Answer:

100m: 4.9 kJ

200m: 9.8 kJ

1000m: 49.1 kJ

Explanation:

edge2020

7 0
3 years ago
Read 2 more answers
A furlong is an old british unit of length equal to 0.125 mi, derived from the length of a furrow in an acre of ploughed land. a
castortr0y [4]

The speed of light is: c = 3x10^8 m/s <span>

or 

c = 186,000,000 miles/sec = 1.86x10^8 mi/s 

1 furlong = 0.125 mile 

1 fortnight = 2 weeks(7d/wk)(24h/d)(3600s/h) 
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Therefore, 

c =1.86x10^8 mi/s(1furl/0.125mi)(1.2096x10^6s/fort) 

<span>c = 18x10^14 furlong/fortnight = 18x10^8 Mfurlong/fortnight</span></span>

4 0
3 years ago
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