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diamong [38]
3 years ago
14

A person travels by car from one city to another with different constant speeds between pair of cities. She drives for 36 min at

80.5 km/h, 10 min at 92.7 km/h, and 54.3 min at 35.6 km/h, and spends 20.9 min eating lunch and buying gas. Find the distance between the initial and final cities along this route. Answer in units of km
Physics
1 answer:
Softa [21]3 years ago
3 0

 Change minutes to hrs, divide by 60:
30 min = .50 hrs
45 min = .75 hrs
12 min = .20 hrs
----------------
total + 1.45 hrs, total travel time
:

let a = average speed for the trip
:
Write a dist equation, dist = speed * time
:
80(.5) + 100(.20) + 40(.75) = 1.45a
40 + 20 + 30 = 1.45a
90 = 1.45a
a =
a = 62.069 km/h, for the entire trip
and
90 km is the total distance 

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In the following diagrams the larger vector has a magnitude of 10, and the smaller
Anton [14]

the sum of vectors with the Pythagorean theorem allows us to find that the maximum magnitude occurs for the case:

d) the two vectors are parallel

Vectors are physical quantities that have modulus and direction, for example: force, velocity, acceleration, etc.

Vector algebra has defined the sum, the product by a scalar and by a vector.  The modulus and the direction of the resulting vector must be encoded.

The sum of two quantities is done using the Pythagorean theorem

                  c² = a² + b²

where c is the resultant called hypotenuse, a and b are the summing vectors called legs; trigonometry is used for the direction.

Let's apply this expression to the present case

a, b) perpendicular vectors

              c² = a² + b²

              c = \sqrt{6^2+10^2}

              c = 11.7

the magnitude is the same in both cases, changing the direction of the vector

c) Antiparallel vectors

             

For this case the vectors are collinear, so the sum reduces to the algebraic addition

             c = a-b

             c = 6 -10

             c = -4

d) parallel vectors

             c = a + b

             c = 4 + 10

             c = 14

We can see that the vectors addition gives their maximum and minimum values ​​when the vectors are collinear.

In conclusion using the vectors addition we find that the correct answer is

d) the two vectors are parallel

learn more about vector addition here:

brainly.com/question/15074838

6 0
2 years ago
Explain the difference between displacement and distance
alisha [4.7K]
<span>Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.</span>
6 0
3 years ago
A small metal sphere has a mass of 0.19 gg and a charge of -23.0 nCnC. It is 10.0 cmcm directly above an identical sphere that h
RoseWind [281]

Answer:

  a = -7.29 m / s²

Explanation:

For this exercise we must use Newton's second law,

          F -W = m a

Force is electrical force

         F = k q₁ q₂ / r²

         k q₁ q₂ / r² -mg = m a

indicate that the charge of the two spheres is equal

         q₁ = q₂ = q

         a = (k q² / r² - m g) / m

         a = k q² / m r² - g

Let's reduce the magnitudes to the SI system

        m = 0.19 g (1kg / 1000 g) = 1.9 10⁻⁴ kg

        q1 = q2 = q = -23.0 nC (1C / 10⁹ nC) = -23.0 10⁻⁹ C

        r = 10.0 cm (1m / 100cm) = 0.1000 m

let's calculate

        a = 9 10⁹ (23.0 10⁻⁹)² / (0.1000² 1.9 10⁻⁴) - 9.8

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The negative sign indicates that the direction of this acceleration is downward

6 0
2 years ago
Calculate the acceleration of a mobile that at 4s is 32m from the origin, knowing that its initial speed is 10m / s.
ehidna [41]

Answer:

5.5 m/s^2

Explanation:

I believe this is the answer > using the formula a= v-v0/t

Hope this helps!

7 0
2 years ago
Read 2 more answers
A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3
4vir4ik [10]

Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

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I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT

8 0
3 years ago
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