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7nadin3 [17]
2 years ago
4

Walt ran 5 kilometers in 25 minutes, going eastward. What was his average velocity?

Physics
1 answer:
snow_lady [41]2 years ago
4 0
.5 kilometer a minute  
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The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:
ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

5 0
2 years ago
Conductance is directly proportional to the length of a conductor. true false user: resistance is inversely proportional to the
zlopas [31]
1) false

2) area of the conductor
7 0
2 years ago
Read 2 more answers
Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is
Harman [31]
Given:
Ca = 3Cb                      (1)
where
Ca =  heat capacity of object A
Cb =  heat capacity f object B

Also,
Ta = 2Tb                     (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf =  final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb)           (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}
where
k= \frac{M_{a}}{M_{b}} 
7 0
2 years ago
A camera gives a proper exposure when set to a shutter speed of 1/250 s at f-number F8.0. The photographer wants to change the s
Oksana_A [137]

Answer:

F4.0

Explanation:

To obtain a shutter speed of 1/1000 s to avoid any blur motion the f-number should be changed to F4.0 because the light intensity goes up by a factor of 2 when the f-number is decreased by the square root of 2.

5 0
2 years ago
Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Which student’s measureme
Lisa [10]

On comparing values , we see that student which has the largest percent error is <u>A. Student 4: 9.61 m/s2 .</u>

<u>Explanation:</u>

Here, we have Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Let's calculate which student’s measurement has the largest percent error :

<u>A. Student 4: 9.61 m/s2 </u>

Percentage of error  = \frac{9.78-9.61}{9.78} (100) = 1.738% %.

<u>B. Student 3: 9.88 m/s2 </u>

Percentage of error  = \frac{9.88-9.78}{9.78} (100) = 1.022% %.

<u>C. Student 2: 9.79 m/s2 </u>

Percentage of error  = \frac{9.79-9.78}{9.78} (100) = 0.1022%% .

<u>D. Student 1: 9.78 m/s2</u>

Percentage of error  = \frac{9.78-9.61}{9.78} (100) = 0%% .

On comparing values , we see that student which has the largest percent error is <u>A. Student 4: 9.61 m/s2 .</u>

4 0
2 years ago
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