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creativ13 [48]
3 years ago
11

An airplane flies 40km directly south in 10 minutes and 20 km directly east in five minutes, its average speed is:

Physics
2 answers:
blondinia [14]3 years ago
8 0
Average speed = total distance / total time
total distance = 40 + 20 = 60km
total time taken = 10 + 5 = 15 minutes
Average speed = 60/15 = 4km/min
Zina [86]3 years ago
3 0

Answer:

D. 4km/min

Explanation:

Since the question deals with "speed" and not "velocity", the direction does not matter.

the plane covers 40 + 20 km=60km in 10mins +5 mins=15mins

60 km - 15 mins

1min- 60÷15= 4 mins

therefore, your answer is D. 4km/min

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A positive charge and a negative charge held a certain distance apart are released. as they move, the force on each particle:___
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7 0
1 year ago
Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
Tatiana [17]

Answer:

The average velocity is

266\frac{m}{s},274\frac{m}{s} and 117\frac{m}{s} respectively.

Explanation:

Let's start writing the vertical position equation :

L(t)=2t^{3}+t^{2}-5t+1

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

V_{avg}=\frac{Displacement}{Time} = Δx / Δt = \frac{x2-x1}{t2-t1}

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

t2-t1=8s-5s=3s

For the position variation we use the vertical position equation :

x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m

x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

\frac{798m}{3s}=266\frac{m}{s}

For the second time interval :

t1 = 4 s → t2 = 9 s

x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m

x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

\frac{1370m}{5s}=274\frac{m}{s}

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

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x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

\frac{702m}{6s}=117\frac{m}{s}

5 0
3 years ago
Heather and Matthew walk with an average velocity of 0.98 m/s eastward. If it takes them 34 min to walk to the store, what is th
TEA [102]

Answer:

D= 1999.2 m

Explanation:

Given that

Average velocity ,v= 0.98 m/s

time ,t= 34 min

We know that

1 min  = 60 s

That is why

t= 34 x 60 =2040 s

We know that

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D= v t

Now by putting the values in the above equation

D= 0.98 x 2040 m

D= 1999.2 m (eastward)

The direction of the displacement will be towards eastward.

That is why the displacement will be 1999.2 m or we can say that 1.9992 km.

3 0
3 years ago
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