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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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1. calculate the value of acceleration that objects gains in that period of time
•calculating acceleration
5.50 = 1/2at^2
5.50*2/t^2 = a
11.00/0.657 = a
16.74=a
now you got the acceleration
2. you have laws of gravitation for that

g = Gm/r^2
where g is the acceleration value
16.74 = 6.754*10^-11 × m/ 6.28*10^4
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that would be the mass of the planet ...
5 0
3 years ago
Read 2 more answers
Consider a rectangular ice floe 5.00 m high, 4.00 m long, and 3.00 m wide. a) What percentage of the ice floe is below the water
artcher [175]

Answer:

(a) 92 %

(b) 6.76 %

Explanation:

length, l = 4 m, height, h = 5 m, width, w = 3 m, density of water = 1000 kg/m^3

density of ice = 920 kg/m^3, density of mercury = 13600 kg/m^3

(a) Let v be the volume of ice below water surface.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of water x g = Total volume of ice block x density

                                                                      of ice x g

v x 1000 x g = V x 920 x g

v / V = 0.92

% of volume immersed in water = v/V x 100 = 0.92 x 100 = 92 %

(b) Let v be the volume of ice below the mercury.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of mercury x g = Total volume of ice block x  

                                                                      density of ice x g

v x 13600 x g = V x 920 x g

v / V = 0.0676

% of volume immersed in water = v/V x 100 = 0.0676 x 100 = 6.76 %

4 0
3 years ago
What is the next step if the data from an investigation do not support the original hypothesis? A. The data are revised to suppo
Leno4ka [110]
I believe the answer is D. <span>The hypothesis is revised and another experiment is conducted.</span>
5 0
3 years ago
Read 2 more answers
An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su
s2008m [1.1K]

Answer: A. 23.59 minutes.

              B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

  • mass of ice, m= 0.555 kg
  • temperature of ice, T= -16.6°C
  • rate of heat transfer, q=820 J.min^{-1}
  • specific heat of ice, c_{i}= 2100 J.kg^{-1}.K^{-1}
  • latent heat of fusion of ice, L_{i}=334\times10^{3}J.kg^{-1}

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

Q=mc\Delta T

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

Q= 0.555\times2100\times16.6

Q= 19347.3 J

Now, we require 19347.3 joules of heat to bring the ice to 0°C  and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

t=\frac{Q}{q}

=\frac{19347.3}{820}

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent  heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, Q_{L}= mL_{i}

Q_{L}= 0.555\times334\times10^{3}= 185370 J

<u>Now  the time required for supply of 185370 J:</u>

t=\frac{Q_{L}}{q}

t=\frac{185370}{820}

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0
3 years ago
Technician A says when diagnosing an overheating hydraulic system, be sure that the oil cooler is not plugged and the cooler’s f
mel-nik [20]

Answer:

Tech B is correct and Tech A is incorrect.

Explanation:

Here Tech A is wrong because when diagnosing an overheating hydraulic system, it is not necessary to un plugg the oil cooler rather it should be plugged to for proper diagnosis.

Technician B says running the hydraulic system at a lower operating temperature will reduce the possibly of oil oxidation is correct statement because at temperature oil's physical as well as chemical property tend to change.

Hence, Tech B is correct and Tech A is incorrect.

8 0
3 years ago
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