A reflecting telescope is built with a 20-cm-diameter mirror having a 1.00 m focal length. it is used with a 10× eyepiece.

A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flie

cluponka [151]

Answer:

a) v₃ = 19.54 km, b) 70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

cos 45 = x₁ / V₁

sin 45 = y₁ / V₁

x₁ = v₁ cos 45

y₁ = v₁ sin 45

x₁ = 26 cos 45

y₁ = 26 sin 45

x₁ = 18.38 km

y₁ = 18.38 km

Vector 2 moves 45 km north

y₂ = 45 km

Unknown 3 vector

x3 =?

y3 =?

Vector Resulting 70 km north of the starting point

R_y = 70 km

we make the sum on each axis

X axis

Rₓ = x₁ + x₃

x₃ = Rₓ -x₁

x₃ = 0 - 18.38

x₃ = -18.38 km

Y Axis

R_y = y₁ + y₂ + y₃

y₃ = R_y - y₁ -y₂

y₃ = 70 -18.38 - 45

y₃ = 6.62 km

the vector of the third leg of the journey is

v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

v₃ = √ (18.38² + 6.62²)

v₃ = 19.54 km

to find the angle let's use trigonometry

tan θ = y₃ / x₃

θ = tan⁻¹ (y₃ / x₃)

θ = tan⁻¹ (6.62 / (- 18.38))

θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

θ’= 180 -19.8

θ’= 160.19º

I mean the address is

θ’’ = 90-19.8

θ = 70.2º

70.2º north-west

3 0

2 years ago

a torch bulb is rated 2.5V and 750mA. Calculate its power,its resistance and the energy consumed if this bulb lighted for 4 hour

Hatshy [7]

Using Ohm's Law, we can derived from this the value of resistance. If I=V/R, therefore, R = V/I
Substituting the values to the given,
P = Power = ?
R = Resistance = ?
V = Voltage = 2.5 V
I = Current = 750 mA

R = V/I = 2.5/ (750 x 10^-3)
R = 3.33 ohms

Calculating the power, we have P = IV

P = (750 x 10^-3)(2.5)
P = 1.875 W

The power consumption is the power consumed multiply by the number of hours. In here, we have;
1.875W x 4 hours = 7.5 watt-hours

3 0

1 year ago

A cyclist is riding his bike up a mountain trail. When he starts up the trail, he is going 8 m/s. As the trail gets steeper, he

taurus [48]

-3 m/s
---------
per min

oh I think 8m/s to 3m/s to 0m/s

idk probably -0.08

7 0

1 year ago

Read 2 more answers

During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

Ray Of Light [21]

Answer:

part a : The dry unit weight is 0.0616  $lb/in^3$ 

part b : The void ratio is 0.77

part c : Degree of Saturation is 0.43

part d : Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

The dry unit weight is 0.0616  $lb/in^3$ 

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

The void ratio is 0.77

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

Degree of Saturation is 0.43

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0

1 year ago

The study of alternating electric current requires the solutions of equations of the form i equals Upper I Subscript max Baselin

KiRa [710]

Answer:

Explanation:

i = Imax sin2πft

given i = 180 , Imax = 200 , f = 50 , t = ?

Put the give values in the equation above

180 = 200 sin 2πft

sin 2πft = .9

sin2π x 50t = .9

sin 360 x 50 t = sin ( 360n + 64 )

360 x 50 t = 360n + 64

360 x 50 t = 64 , ( putting n = 0 for least value of t )

18000 t = 64

t = 3.55 ms .

8 0

2 years ago