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swat32
3 years ago
12

Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.8 km/h due east. Runner B is initia

lly 4.2 km east of the flagpole and is running with a constant velocity of 7.6 km/h due west. How far are the runners from the flagpole when their paths cross? Answer in units of km.
Physics
1 answer:
vampirchik [111]3 years ago
4 0

Answer:

0.39 km west

Explanation:

The position of Runner A is:

x = -5.7 + 8.8 t

The position of Runner B is:

x = 4.2 − 7.6 t

When the positions are equal:

-5.7 + 8.8 t = 4.2 − 7.6 t

16.4 t = 9.9

t = 0.604

Plug into either equation to find the position at this time:

x = -5.7 + 8.8 (0.604)

x = -0.39

The runners are 0.39 km west of the flagpole when they meet.

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The current through two identical light bulbs connected in series is 0.25 A. The voltage across both bulbs is 110 V. The resista
Rus_ich [418]

The resistance of a single light bulb is 220 ohms per bulb.

<h3>What is Ohm's Law?</h3>

Ohm's Law is a formula used to determine how voltage, current, and resistance in an electrical circuit relate to one another.

Ohm's Law (E = IR) is as basic to students of electronics as Einstein's Relativity equation (E = mc2) is to physicists.

E = I x R

The formula reads voltage = current x resistance, or V = A xΩ., or volts = amps x ohms.

110volts divided by .25amps = 440 ohms. 440 divided by 2 =220 ohms per bulb.

R = 110/(2*0.25) = 220 ohms

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4 0
2 years ago
A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0
3 years ago
A manganese atom is pictured below.
Rufina [12.5K]
Manganese   has  2 (two) electron  that   would  free  floating   and   able  to  form  a  metallic  bond.
  The    electronic  configuration  of  manganese  is  (Ar)  3d5 4s2.  The   two   electron  in  4s  orbital  are  the  valence    electron  which  can  freely  move  from  one  place  to  another.
5 0
3 years ago
Read 2 more answers
two thermometers, calibrated in celsius and fahrenheit respectively, are put into a liquid. the reading on the fahrenheit scale
Viefleur [7K]
Two thermometers, calibrated in celsius and fahrenheit respectively, are put into a liquid. the reading on the fahrenheit scale is four times the reading on the celsius scale. the temperature of the liquid is:

6 0
3 years ago
Ten high-technology batteries are tested for 200 hours each. One failed at 20 hours; another failed at 140 hours; all others com
Bas_tet [7]

Answer:

Failure rate   = 20%

MTBF = 880 hours

Explanation:

given data

batteries = 10

tested = 200 hours

one failed = 20 hours

another fail at =  140 hours

solution

we know that Mean Time between Failures is express as = (Total up time) ÷  (number of breakdowns)    ....................1

so here Total up time will be

Total up time = 200 × 10

Total up time = 2000

and here

Number of breakdown = 1 at 20 hour and another at 140 hour = 2

so it will be  = (Total up time) ÷ (number of breakdowns)      .......2

=  \frac{2000}{2}   =  1000  

so here gap between occurrences is

gap between occurrences=  140 - 20

gap between occurrences = 120 hour

and

MTBF  will be

MTBF = 1000 - 120

MTBF = 880 hours  

and

Failure rate (FR)  will be

Failure rate (FR) =  1 ÷ MTBF    ................3

Failure rate (FR) = R÷T     ......................4

as here R is the number of failures and T is total time

so Failure rate (FR)  = 20%

4 0
3 years ago
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