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Ghella [55]
2 years ago
14

A man makes a 27.0 km trip in 16 minutes. (a.) How far was the trip in miles? (b.) If the speed limit was 55 miles per hour, wa

s the driver speeding?
Physics
1 answer:
Alekssandra [29.7K]2 years ago
8 0

Answer:

(a) 16.777mi

(b)Yes, he was speeding

Explanation:

(a)

Let's do the proper operations in order to convert km to mi:

27km*\frac{1000m}{1km} *\frac{1mi}{1609.34m} =16.77706389mi

We can conclude that the trip length in miles was:

d=16.77706389mi

(b)

Let's calculate the speed of the man during the trip:

v=\frac{d}{t}

But first, let's do the proper operations in order to convert min to h:

16min*\frac{1h}{60min} =2.666666667h

Now, the speed is:

v=\frac{16.77706389mi}{2.666666667h} =62.91398959\frac{mi}{h}

As we can see:

62.91398959\frac{mi}{h}>55\frac{mi}{h}

So, we can conclude that the driver was speeding

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Imagine you derive the following expression by analyzing the physics of a particular system: M= (mv2r)(mGr2). Simplify the expre
alex41 [277]

Answer:

The simplified expression is M  =  \frac{v^2 r}{G}

Explanation:

From the question we are told that  

     M  = \frac{ \frac{m v^2}{r} }{\frac{ mG}{r^2 } }

So simplifying we have

    M  =    \frac{m v^2}{r} *  \frac{r^2 }{ mG }

    M  =  \frac{v^2 r}{G}

Thus the simplified formula is M  =  \frac{v^2 r}{G}

3 0
2 years ago
A person's height will increase from birth until about age 25, and it may decrease starting at about age 70. This is an example
dsp73
LOL idk sorry maybe it b! Oh wait it is !
5 0
2 years ago
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A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a
aleksley [76]

Answer:

 T₂ =602  °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602  °C

5 0
2 years ago
Calculate the force between charges of 5.0 x 10^-8 c and 1.0 x 10^-7 if they are 5.0 feet apart
weqwewe [10]

Answer:

F = 19.375 x 10^-6  N

Explanation:

This problem can be solved by applying Coulomb's Law, which lets us determine the force between two electrically charged particles.

It is defined as

F = (ke * q1 * q2)/ r^2

Where,

ke =  is Coulomb's constant ≈ 9×10^9 N⋅m^2⋅C^−2

q1 = 5.0 x 10^-8 C

q2 = 1.0 x 10^-7 C

r = 5 ft = 1,524 m

F  = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)

F  = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)

F = 19.375 x 10^-6  N

6 0
2 years ago
Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another
ioda

Answer:

6.1875\times 10^{-8}

Explanation:

Assuming uniform spread of sound with no significant reflections or absorption. We know that sound intensity varies I=\frac {k}{r^{2}} where r is the distance

Since intensity is given then when at 3 m

1.1\times 10^{-7}= \frac {k}{3^{2}}

k=3^{2}\times 1.1\times 10^{-7}= 9.9\times 10^{-7}

Since we have the constant then at 4m

Intensity, I= \frac {9.9\times 10^{-7}}{4^{2}}=6.1875\times 10^{-8}

8 0
2 years ago
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