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ollegr [7]
3 years ago
11

Utility poles are to be set every 30 meters. How many poles will be set in one mile if one was set at the beginning of the mile)

?
Physics
2 answers:
maksim [4K]3 years ago
7 0

Answer:

54 poles

Explanation:

1600 / 30 = 53.333 (only 53 poles)

note: there was one set at the beginning, so you would have to add 1  = 54

Zepler [3.9K]3 years ago
4 0

Answer:

53

Explanation:

Because there are 1609.34 meters in a mile. 1609.34÷30=53.64 but because you put one at the beginning of the mile it will stay 53 and not round up to 54

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Vectors a and b have scalar product â6.00, and their vector product has magnitude +9.00. what is the angle between these two vec
salantis [7]

Answer:

Value of angle between vector a and b is 56.30^{\circ}.

Explanation:

Vectors a and b have scalar product 6.00

Let \theta be the angle between a and b.

\vec{a}.\vec{b} = 6

ab cos \theta = 6 ...(1)

Vectors a and b have magnitude of vector product 9.00

\vec{a} \times\vec{b} = 9

ab sin \theta = 9 ...(2)

Dividing equation (2) by (1) we get

\frac{ab sin \theta}{ab cos \theta}  = \frac{9}{6}

tan \theta = 1.5

\theta = tan ^{-1} (1.5)

\theta = 56.30^{\circ}

Thus, value of angle between vector a and b is 56.30^{\circ}.

3 0
3 years ago
A laser beam is incident at an angle of 30.2° to the vertical onto a solution of corn syrup in water. (a) If the beam is refract
mote1985 [20]

Answer:

a) n2 = 1.55

b) 408.25 nm

c) 4.74*10^14 Hz

d) 1.93*10^8 m/s

Explanation:

a) To find the index of refraction of the syrup solution you use the Snell's law:

n_1sin\theta_1=n_2sin\theta_2   (1)

n1: index of refraction of air

n2: index of syrup solution

angle1: incidence angle

angle2: refraction angle

You replace the values of the parameter in (1) and calculate n2:

n_2=\frac{n_1sin\theta_1}{sin\theta_2}=\frac{(1)(sin30.2\°)}{sin18.82\°}=1.55

b) To fond the wavelength in the solution you use:

\frac{\lambda_2}{\lambda_1}=\frac{n_1}{n_2}\\\\\lambda_2=\lambda_1\frac{n_1}{n_2}=(632.8nm)\frac{1.00}{1.55}=408.25nm

c) The frequency of the wave in the solution is:

v=\lambda_2 f_2\\\\f_2=\frac{v}{\lambda_2}=\frac{c}{n_2\lambda_2}=\frac{3*10^8m/s}{(1.55)(408.25*10^{-9}m)}=4.74*10^{14}\ Hz

d) The speed in the solution is given by:

v=\frac{c}{n_2}=\frac{3*10^8m/s}{1.55}=1.93*10^8m/s

8 0
3 years ago
An 80-kg clown sits on a 20-kg bike on a tightrope attached between two trees. The center of mass of the clown is 1.6 m above th
Eduardwww [97]

Answer:

Explanation:

the center of mass formula

Ycm= [(m₁y₁) + (m₂y₂) + (m₃y₃)] / (m₁+m₂+m₃)

Rope forms the x axis and position of centre of different massses are above or below it so they represent their location on y - axis.

y₁ = 1.6 , y₂ = .7 and y₃ = - 2.1

Ycm ( given ) = - .5

Putting the values of masses and positions

- .5 = 80 x 1.6 + 20 x .7 + m₃ x - 2.1 / ( 80 + 20 + m₃ )

- .5 = 128  + 14  + m₃ x - 2.1 / ( 100+ m₃ )

- 50 - .5 m₃ = 142 - 2.1 m₃

1.6 m₃ = 192

m₃ = 120 kg .

B )

Total downward force is weight of  total mass  = 80 + 20 + 120

= 220 kg

weight = 220  x 9.8 = 2156 N .

component of weight perpendicular to rope

= 2156 cos 15 = 2082.53 N

This force will be equally distributed over each tree , so force on each tree =  2082.53 / 2 = 1041.26 N .

6 0
3 years ago
You're driving your new sports car at 85 mph over the top of a hill that has a radius of curvature of 525 m.
Bumek [7]

Explanation:

It is given that,

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The radius of curvature, r = 525 m

Let W_N is the normal weight and W_A is the apparent weight of the person. Its apparent weight is given by :

W_A=mg-\dfrac{mv^2}{r}

So, \dfrac{W_A}{W_N}=\dfrac{mg-\dfrac{mv^2}{r}}{mg}

\dfrac{W_A}{W_N}=\dfrac{g-\dfrac{v^2}{r}}{g}

\dfrac{W_A}{W_N}=\dfrac{9.8-\dfrac{(37.99)^2}{525}}{9.8}

\dfrac{W_A}{W_N}=0.719

or

\dfrac{W_A}{W_N}=71.9\%

Hence, this is the required solution.

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3 years ago
Determina la velocidad de un avión que recorre 459 km en 67 min
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Answer:

122397.7

Explanation:

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2 years ago
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