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kramer
3 years ago
13

You drive a beat-up pickup truck along a straight road for 8.4 km at 70 km/h, at which point the truck runs out of gasoline and

stops. Over the next 30 min, you walk another 2.0 km farther along the road to a gasoline station. What is your average velocity from the beginning of your drive to your arrival at the gasoline station?
Physics
2 answers:
andreev551 [17]3 years ago
6 0

Answer:

v_avg = 37 km/h

Explanation:

To find the average velocity in the complete trajectory you use the following formula:

v_{avg}=\frac{v_1+v_2}{2}   ( 1 )

v1: velocity in the first part of the trajectory = 70 km/h

v2: velocity in the second part of the trajectory = ?

You calculate v2 by using the following equation for a motion with constant velocity:

v_2=\frac{2.0km}{30min}*\frac{60min}{1h}=4\frac{km}{h}

you replace the values of v1 and v2 in (1) and you obtain:

v_{avg}=\frac{70km/h+4km/h}{2}=37\frac{km}{h}

hence, the average velocity is 37 km/h

kkurt [141]3 years ago
3 0

Answer:

v = 16.8 km/h

Explanation:

The average velocity can be calculated usign the following equation:

v = \frac{\Delta x}{\Delta t}

<u>Where</u>:

Δx: is the change in the displacement

Δt: is the change in the time

The total displacement is:

x_{t} = 8.4 km + 2.0 km = 10.4 km

The initial time is:

t = \frac{x}{v} = \frac{8.4 km}{70 km/h} = 0.12 h  

The total time is:

t_{t} = 0.12 h + 30min*\frac{1 h}{60 min} = 0.62 h  

Finally, by taking:

Δt = 0.62 h

Δx= 10.4 km

The average velocity is:

v = \frac{10.4 km}{0.62 h} = 16.8 km/h        

Therefore, the average velocity is 16.8 km/h.

I hope it helps you!                  

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emmasim [6.3K]

In order to calculate the thermal energy, first let's calculate the power, using the formula:

P=\frac{V^2}{R}

For a voltage V = 9 Volts and a resistance R = 50 ohms, we have:

\begin{gathered} P=\frac{9^2}{50} \\ P=\frac{81}{50}=1.62\text{ W} \end{gathered}

Now, multiplying the power by the time (in seconds), we can find the energy:

\begin{gathered} E=P\cdot t \\ E=1.62\cdot7.5\cdot60 \\ E=729\text{ J} \end{gathered}

In scientific notation, we have an energy of 7.3 * 10^2 J, therefore the correct option is the fourth one.

4 0
1 year ago
A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance
seropon [69]

Answer: Option B. R = (1/2)gt^2

Explanation:

S = R (horizontal distance)

V^2 = 2gS

V^2 = 2gR

R = V^2 / 2g

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R = (gt)^2 / 2g

R = (g^2 x t^2) / 2g

R = gt^2 / 2

But t^2 = 2h/g

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4 0
3 years ago
North Dakota Electric Company estimates its demand trend line​ (in millions of kilowatt​ hours) to​ be: D​ = 75.0 ​+ 0.45​Q, whe
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Answer:

The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

Explanation:

Given that,

The demand trend line​ is

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

We need to calculate the demand forecast for winter

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times101)\times0.80

D=96.36\ millions\ KWH

We need to calculate the demand forecast for spring

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times102)\times1.20

D=145.08\ millions\ KWH

We need to calculate the demand forecast for summer

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times103)\times1.40

D=169.89\ millions KWH

We need to calculate the demand forecast for fall

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times104)\times0.60

D=73.08\ millions KWH

Hence, The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

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Where are the questions so that I can deliver a more accurate answer. 
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Activity 1 more or less, pls answer this ASAP​
AURORKA [14]

Answer:

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