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iren2701 [21]
3 years ago
15

If she runs 3,223 meters each day and keeps this up for two weeks, how many miles will she have run

Physics
1 answer:
seropon [69]3 years ago
5 0

Answer:

28 miles

Explanation:

3,223 meters/day x 14 days = 45,122 meters

1 meter = 3.28 feet

45,122 meters = 148,000 feet

5280 feet = 1 mile

148,000 feet = 28 miles

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A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

f = f_0 (\frac{v_0}{v_0-v})

Here,

f_0 = Frequency of Source

v_s = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v  = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

f = f_0 (\frac{v_0}{v_0-v})

300 = f_0 (\frac{343}{343-v})

(300*343) - 300v = 343f_0

Now the second expression will be,

f' = f_0 (\frac{v_0}{v_0-v/2})

290 = (343)(\frac{v_0}{343-v/2})

290*343-145v = 343f_0

Dividing the two expression we have,

\frac{(300*343) - 300v}{290*343-145v} = 1

Solving for v, we have,

v = 22.12m/s

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0
3 years ago
The distance from Abdullah's house to his school is 2.4km. Abdulla takes 0.6h to go to school on his cycle but takes only 0.4h t
vladimir1956 [14]

Answer:

The average speed can be calculated as the quotient between the distance travelled and the time needed to travel that distance.

To go to the school, he travels 2.4 km in 0.6 hours, then here the average speed is:

s = (2.4km)/(0.6 hours) = 4 km/h

To return to his home, he travels 2.4km again, this time in only 0.4 hours, then here the average speed is:

s' = (2.4 km)/(0.4 hours) = 6 km/h.

Now, if we want the total average speed (of going and returning) we have that the total distance traveled is two times the distance between his home and school, and the total time is 0.6 hours plus 0.4 hours, then the average speed is:

S = (2*2.4 km)/(0.6 hours + 0.4 hours)

S = (4.8km)/(1 h) = 4.8 km/h

5 0
2 years ago
A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

8 0
3 years ago
An object of mass m moving at a speed of v1 possesses kinetic energy that is equal to KE1. When the object's speed is doubled, t
Vera_Pavlovna [14]

Answer:

KE₂ = 4KE₁

Explanation:

KE₁ = ½mv₁²

KE₂ = ½mv₂²

KE₂ = ½m(2v₁)²

KE₂ = 4(½mv₁²)

KE₂ = 4KE₁

7 0
3 years ago
in a circuit, a 10 resistor is connected in series to a parallel group containing a 60 resistor and a 5 resistor. what is the to
natka813 [3]
The 60 and the 5 in parallel have an equivalent resistance of 4.615 ohms. (rounded). The 10 in series makes it 14.615...
5 0
3 years ago
Read 2 more answers
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