Ask question

Ask question

All categories

3 years ago

15

Three deer, a, b, and c, are grazing in a field. deer b is located 62.1 m from deer a at an angle of 54.3 ° north of west. deer

c is located 79.1 ° north of east relative to deer
a. the distance between deer b and c is 93.8 m. what is the distance between deer a and c? (hint: consider the laws of sines and cosines given in appendix
e.)

1 answer:

sveticcg [70]3 years ago

4 0

by cosine law we know that

$c^2 = a^2 + b^2 - 2 abcos\theta$

$\theta = 180 - 54.3 - 79.1 = 46.6 degree$

now using above equation

$93.8^2 = 62.1^2 + b^2 - 2*62.1*b * cos46.6$

$4942.03 = b^2 - 85.34 b$

$b^2 - 85.34b - 4942.03 = 0$

by solving above quadratic equation we have

$b = 124.9 m$

so it is at distance 124.9 m from deer a

You might be interested in

3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a

serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as $\mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}$ .

Where, ${V}_{i}$ is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

${V}_{i}$ =3 m/s, V = 0 m/s, and ∆s = 2 m

Substitute the values in the above formula,

$0=3^{2}-2 \times 2 \times a$

0 = 9 - 4a

4a = 9

$a=2.25 \mathrm{m} / \mathrm{s}^{2}$

$a=2.25 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration.

Calculating Coefficient of friction:

$\mathrm{F}=\mathrm{m} \times \mathrm{a}$

$\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}$

Compare the above equation

$\mu \times m \times g=m \times a$

Cancel "m" common term in both L.H.S and R.H.S

$\text { Equation becomes, } \mu \times g=a$

$\text { Coefficient of friction } \mu=\frac{a}{g}$

$\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}$

$\mu=\frac{2.25}{9.8}$

$\mu=0.229$

Hence coefficient of friction is 0.229.

calculating force:

$\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}$

$\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })$

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

7 0

3 years ago

Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of

FrozenT [24]

Gravity obeys the inverse square law. At 6400 km above the center of the Earth (Earth's surface) you weigh x. Twice that reduces your weight to 1/4th. Four times that height reduces your weight to 1/16th. 4 times 6400 km is 25,600 km. But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer: 19,200 km.

Incidentally, it doesn't exactly work the opposite way. At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.

6 0

3 years ago

El espectro visible en el aire está comprendido entre la longitud de onda 450 nm del color azul, Determina la velocidad de propa

TiliK225 [7]

Answer:

v = 2,99913 10⁸ m / s

Explanation:

The velocity of propagation of a wave is

v = λ f

in the case of an electromagnetic wave in a vacuum the speed that speed of light

v = c

When the wave reaches a material medium, it is transmitted through a resonant type process, whereby the molecules of the medium vibrate at the same frequency as the wave, as the speed of the wave decreases the only way that they remain the relationship is that the donut length changes in the material medium

λ = λ₀ / n

where n is the index of refraction of the material medium.

Therefore the expression is

v = $\frac{\lambda_o}{n} f$

Let's look for the frequency of blue light in a vacuum

f = $\frac{c }{\lambda_o}$

f = $\frac{3 \ 10^8}{450 \ 10^{-9}}$

f = 6.667 10¹⁴ Hz

the refractive index of air is tabulated

n = 1,00029

let's calculate

v = $\frac{450 \ 10^{-9} }{1.00029} \ 6.667 \ 10^{14}$ 450 10-9 / 1,00029 6,667 1014

v = 2,99913 10⁸ m / s

we can see that the decrease in speed is very small

8 0

2 years ago

A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +46 N·m is ap

Rom4ik [11]

Answer:

Explanation:

Let Torque due to friction be

F

Net torque

= 46 - F

Angular impulse = change in angular momentum

=( 46 - F ) x 17 = I X 580

When external torque is removed , only friction creates torque reducing its speed to zero in 120 s so

Angular impulse = change in angular momentum

F x 120 = I X 580

( 46 - F ) x 17 = F x 120

137 F = 46 x 17

F = 5.7 Nm

b )

Putting this value in first equation

5.7 x 120 = I x 580

I = 1.18 kg m²

8 0

3 years ago

Select the correct answer.

anyanavicka [17]

<h2>MULTIPLE CHOICES</h2>

$\large\mathcal{QUESTION:}$

What is the basic unit of all matter?

$\large\mathcal{CHOICES:}$

OA neutron
OB. atom
OC. electron
OD. proton
OE. nucleus

$\large\mathcal{ANSWER:}$

B. Atom

_____

<u>Atom</u> is the smallest, <u>indivisible unit of a chemical</u> <u>element and the fundamental unit of matter</u>. It consists of a nucleus (neutrons + protons) surrounded by an electron cloud. The <u>atom</u> is the smallest unit of matter that cannot be divided chemically and is a building block with its own set of properties.

<h3>#ProvideUniqueAnswers</h3>

6 0

2 years ago