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Setler79 [48]
3 years ago
11

Marys airplane trip took 5.8 hours for one-half of that time, the airplane flew at a constant speed of 640 miles per hour and fo

r the rest of the time, it flew at a speed of 580 mph. what distance did mary travel?
Physics
1 answer:
Olenka [21]3 years ago
7 0
Distance is speed x time.  Half of the trip is 5.8/2 = 2.9hrs.
640 x 2.9 = 1856mi
580 x 2.9 = 1682mi
1856mi+1682mi=3538mi.

You could also calculate her average speed.  This is easy since it was divided in two equal time slices.  Average Speed = (640+580)/2 = 610mi/hr
Now 610mi/hr x 5.8hrs = 3538mi
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Energy is always needed to preform anything, and chemical energy is one form of energy. Chemical energy is a part of chemical reactions because it involves teh chemicals. 

5 0
3 years ago
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Need help with this question. Thirty points.
Anon25 [30]

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2 years ago
I REALLY NEED HELP PLEASE
iogann1982 [59]
The correct answer is C
6 0
4 years ago
A severe storm on January 10, 1992, caused a cargo ship near the Aleutian Islands to spill 29,000 rubber ducks and other bath to
Anna35 [415]
The distance that the rubber ducks traveled = 1600 miles.
The time taken to travel this distance = 10 months.

By definition, the approximate average speed of the ocean is
v = (1600 miles)/(10 months)

Note that
1 mile = 1609 m 
10 months = (10 months)*(30 days/month) *(24 hours/day)
                  = 7200 hours (approx.)
                  = 7200*3600 = 2.592 x 10⁷ s

Therefore
v = \frac{(1600 \, m)*(1609 \, \frac{m}{mi}) }{(10 \, months)*(2.592 \times 10^{7} \, \frac{s}{10 \, months}) } = 0.0993 \, \frac{m}{s}

Also,
v = \frac{1600 \, m}{7200 \, h} =0.222 \, \frac{mi}{h}

Answer:
The average velocity is approximately 0.1 m/s, or 0.222 m/h. 

6 0
3 years ago
An electron initially has a speed 16 km/s along the x-direction and enters an electric field of strength 27 mV/m that points in
weqwewe [10]

Answer:

a)t=1.4\times 10^{-5}\ s

b)S= 46.4 cm

Explanation:

Given that

Velocity = 16 Km/s

V= 16,000 m/s

E= 27 mV/m

E=0.027 V/m

d= 22.5 cm

d= 0.225 m

a)

lets time taken by electron is t

d = V x t

0.225 = 16,000 t

t=1.4\times 10^{-5}\ s

b)

We know that

F = m a = E q                    ------------1

Mass of electron ,m

m=9.1\times 10^{-31}\ kg

Charge on electron

q=1.6\times 10^{-19}\ C

So now by putting the values in equation 1

a=\dfrac{E q}{m}

a=\dfrac{1.6\times 10^{-19}\times 0.027}{9.1\times 10^{-31}}\ m/s^2

a=4.74\times 10^{9}\ m/s^2

S= ut+\dfrac{1}{2}at^2

Here initial velocity u= 0 m/s

S= \dfrac{1}{2}\times 4.74\times 10^{9}\times (1.4\times 10^{-5})^2\ m

S=0.464 m

S= 46.4 cm

S is the deflection of electron.

4 0
4 years ago
Read 2 more answers
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