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disa [49]
3 years ago
8

A bird flies from the south pole to the north pole. Part of the journey is 1000 miles that takes two weeks. What is the birds Ve

locity in that time?
Please show your work an include the final units,
Physics
1 answer:
alexira [117]3 years ago
8 0
Formula for Velocity = DISTANCE traveled/TIME to travel distance + direction

For this one, we can use mph(miles per hour) as unit. 

v = 1,000 miles / 336 hours   (2 weeks = 24 hours  x 14 days = 336 hrs)
   = 2.98 mph North

or we can use kph (kilometers per hour)

v = 1609.34 km / 336 hours    (1 mile = 1.60934 km)
  = 4.79 kph North
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What was your favorite thing to learn in physics
Zepler [3.9K]

Answer:

<h2>Three laws of motion haha.....</h2>
5 0
3 years ago
A mouse is walking at 2.45 m/s. He sees a cat and accelerates at 4.92 m/s/s. How much time does it take him to reach a speed of
alexandr1967 [171]
Speed= (4.92-2.45) 2.47 m/s
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5 0
3 years ago
which of the following is a chemical change A.ice melting B.ice being carved C.water boiling D.water breaking down into hydrogen
QveST [7]
The chemical change is are to water breaking down into hydrogen and oxygen.

Answer is D.
6 0
3 years ago
Read 2 more answers
A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near sid
igor_vitrenko [27]

Answer:

a) attractiva, b) dF = k \frac{Q_1 \ dQ_2}{dx}, c)  F = k Q_1 \frac{Q_2}{d \ (d+L)}, d) F = -1.09 N

Explanation:

a) q1 is negative and the charge of the bar is positive therefore the force is attractive

b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x

           dF = k \frac{Q_1 \ dQ_2}{dx}

where k is a constant, Q₁ the charge at the origin, x the distance

c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L

         ∫ dF = k \ Q_1 \int\limits^{d+L}_d     {\frac{1}{x^2} } \, dQ_2

as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density

          λ = dQ₂ / dx

          DQ₂ = λ dx

we substitute

         F = k \ Q_1 \lambda \int\limits^{d+L}_d  \, \frac{dx}{x^2}

         F = k Q1 λ (-\frac{1}{x})  

we evaluate the integral

        F = k Q₁ λ (- \frac{1}{d+L} + \frac{1}{d} )

        F = k Q₁ λ  ( \frac{L}{d \ (d+L)})

we change the linear density by its value

      λ = Q2 / L

       F = k Q_1 \frac{Q_2}{d \ (d+L)}

d) we calculate the magnitude of F

       F =9 10⁹ (-4.2 10⁻⁶)   \frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}

       F = -1.09 N

the sign indicates that the force is attractive

3 0
3 years ago
Two parallel wires carry a current in the same direction. there is ___________ between the wires.
Arte-miy333 [17]
The correct answer is "an attractive force" between the wires.

Let's see why. Assume we have wire A on the left and wire B on the right, and that the current in both wires go upward.  First, let's find the direction of the magnetic field produced by wire A at wire B: by using the right-hand rule, we see that since the current (the thumb) goes upward, the magnetic field (given by the other fingers) at wire B is directed inside the paper.
Then we can apply again the right-hand rule to see what is the force on wire B. The index gives the direction of the current (upward), the middle finger the direction of the magnetic field (inside the paper), and the thumb gives the direction of the force: to the left, so toward wire A. This means the force is attractive. (you can re-do the procedure on wire A, and you will find the force on wire A is directed toward wire B)
7 0
3 years ago
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