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yKpoI14uk [10]
3 years ago
9

if you take one hour drive at an average speed of 65 mph, is it possible for another car with an average speed of 55 mph to pass

you
Physics
2 answers:
SVETLANKA909090 [29]3 years ago
7 0
Nope, this is impossible because in order for a car to pass another, they the 55 mph car would have to be behind the 65 mph car (meaning that starting ahead of the 65 mph doesn't count as pass). Assuming that they both drive for one hour, it is impossible because the first will cover a distance of 65 mi and the second would cover a distance of 55 mi. One is obviously ahead of the other and is therefore impossible for the slow one to pass the first one unless the slow car keeps driving after an hour. In that case, it would take approximately 11 minutes for it to pass the other car. This was found by finding the distance needed to pass the first car : 65 - 55 = 10 mi and converted using 1 hr/ 55 mi = .18181818 hr x 60 min/ 1 hr = 11 seconds
I hope this helps :)
Aleks04 [339]3 years ago
3 0
Sure it is.  We only know the average speeds, but we don't know
how much of the time each car drove faster or slower. 

The averages only tell us who got to the end of their trip sooner. 
But there could have been a time ... or several times during the hour ...
when the 55mph car was blazing and passed the 65mph car.

In fact, the 55mph car could have been driving at 110 mph for the
first 29 minutes, then the last 1.83 miles at a speed of 3.5 mph.
His average would then be 55mph, even though he spent the first
29 minutes passing everybody and everything on the road in the
first 53.16 miles of his trip.
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The diagram shows four paths from point A to point B.
iVinArrow [24]
Path 2.

Displacement is the direction and magnitude of an object from its starting point, so path 2 is the direct route you would need to take to find direction and magnitude.
8 0
2 years ago
What is the average power output pf a weight lifter who can lift 250 kg 2.0 m in 2.0 s
liq [111]

Answer:

Power = 2.45Kw or 2450 Watts.

Explanation:

<u>Given the following data;</u>

Mass, m = 250kg

Height, h = 2m

Time, t = 2secs

We know that acceleration due to gravity, g is equal to 9.8m/s²

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

Power = \frac {Energy}{time}

But Energy = mgh

Substituting into the equation, we have

Power = \frac {mgh}{time}

Power = \frac {250*9.8*2}{2}

Power = \frac {4900}{2}

Power = 2450 Watts

To convert to kilowatt (Kw), we would divide by 1000

Power = 2450/1000

Power = 2.45Kw.

Therefore, the average power output of the weightlifter is 2.45 Kilowatts.

8 0
3 years ago
Which describes the position on a horizontal number line
Savatey [412]

Answer:

It is formed by a horizontal number line, called the x-axis, and a vertical number line, called the y-axis.

Explanation:

3 0
3 years ago
Assume (unrealistically) that a TV station acts as a point source broadcasting isotropically at 3.2 MW. What is the intensity of
Dahasolnce [82]

Answer:

I=1.5\times10^{-28}W/m^2

Explanation:

The intensity is related to the power and surface area by I=\frac{P}{A}=\frac{P}{4\pi r^2}. We need to calculate the surface area of a sphere of radius r=4.3ly.

Since 4.3ly is the distance light travels in 4.3 years at 299792458m/s, we can obtain it in meters by doing:

r=vt=(299792458m/s)(4.3\times365\times24\times60\times60s)=4.1\times10^{16}m

So we have:

I=\frac{P}{4\pi r^2}=\frac{3.2\times10^6W}{4\pi (4.1\times10^{16}m)^2}=1.5\times10^{-28}W/m^2

4 0
3 years ago
Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op
Juliette [100K]

Answer:

The fundamental frequency of can is 2.7 kHz.                          

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

7 0
3 years ago
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