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statuscvo [17]
3 years ago
5

A student decides to spend spring break by driving 50 miles due east, then 50 miles 30 degrees south of east, then 50 miles 30 d

egrees south of that direction, and to continue to drive 50 miles deviating by 30 degrees each time until he returns to his original position. How far will he drive, and how many vectors must he sum to calculate his displacement?
a. 600 ml, 12
b. 0, 0
c. 400 ml. 8
d.0,12
e. 0,8
Physics
1 answer:
PolarNik [594]3 years ago
5 0

Answer:

a. 600 ml, 12

Explanation:

The movement described in the question exhibits that of a polygon. Exhibiting a constant distance and angle with only varying direction until the starting point is reached.

The sum of exterior angles of a polygon = 360 degrees.

Exterior angle of a polygon = (360 ÷ number of sides)

Therefore,

Number of sides = 360 ÷ exterior angle

Exterior angle = 30 degrees

Hence,

Number of sides = 360 ÷ 30 = 12 sides

Since distance traveled of 50 miles is the same for each displacement ;

Total displacement = distance traveled * number of sides

Total displacement = 50 * 12 = 600 miles.

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"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e
notka56 [123]

Answer:

The Magnifying power of a telescope is M = 109.26

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective f_{objective} = \frac{R}{2}

⇒ f_{objective} = \frac{590}{2}

⇒ f_{objective} = 295 cm

Focal length of eyepiece f_{eyepiece} = 2.7 cm

Magnifying power of a telescope is given by,

M = \frac{f_{objective} }{f_{eyepiece} }

M = \frac{295}{2.7}

M = 109.26

therefore the Magnifying power of a telescope is M = 109.26

4 0
3 years ago
Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co
patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of ^{3890}Sr\, t_{\frac{1}{2}} = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

where

N = No. of nuclei left after time t

N_{o} = No. of nuclei initially started with

\frac{N}{N_{o}} = 1\times 10^{- 4}

(Since, 100% - 99.99% = 0.01%)

Thus

1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}

Taking log on both the sides:

- 4 = \frac{t}{28.5}log\frac{1}{2}

t = \frac{-4\times 28.5}{log\frac{1}{2}}

t = 387.69 yrs

5 0
3 years ago
Help!!! I need it today <br> Thank you in advance
svlad2 [7]

Answer:

 F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.

Explanation:

The student wants to prove hooke's law which has the form

          F = - k (x-xo)

To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.

Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,

we must be careful when hanging the weights so as not to create oscillations in the spring

we look for the mass of each weight

         W = mg

          m = W / g

and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.

The fact of obtaining a line already proves Hooke's law.

5 0
2 years ago
A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magni
irinina [24]

The magnitude of the magnetic field inside the solenoid is 6.46 \times 10^{-3} \ T.

The given parameters;

  • <em>length of the solenoid, L = 91 cm = 0.91 m</em>
  • <em>radius of the solenoid, r = 1.5 cm = 0.015 m</em>
  • <em>number of turns of the solenoid, N = 1300 </em>
  • <em>current in the solenoid, I = 3.6 A</em>

The magnitude of the magnetic field inside the solenoid is calculated as;

B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\

where;

\mu_o is the permeability of frees space = 4π x 10⁻⁷ T.m/A

B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T

Thus, the magnitude of the magnetic field inside the solenoid is 6.46 \times 10^{-3} \ T.

Learn more here:brainly.com/question/17137684

7 0
2 years ago
When Mendeleev organized elements in his periodic table in order of increasing mass, similar elements with similar properties we
Temka [501]
Similar elements with similar properties were in the same groups and periods for instance lithium(Li) and sodium(Na) are alkaline metals and so belong to the same group (that is group 1).Also Hydrogen(H) and Helium(He) both have only one shell or energy level and so belong to the same period.
8 0
3 years ago
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