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sweet [91]
2 years ago
13

A treasure map directs you to start at palm tree

Physics
1 answer:
masya89 [10]2 years ago
6 0

The direction of the 90° turns are the possible directions used for the calculations

The four distances and directions are;

  • <u>24.2 m, 65.6° West of North</u>
  • <u>29.7 m, 42.7° East of North</u>
  • <u>24.2 m, 65.6° East of North</u>
  • <u>29.7m, 47.7° West of North</u>

Reason:

Let point A represent the motion of the treasure hunter, we have:

Turning West then West;

Walking due north to location (0, 15)

Turn 90° West and walk 22.0 m to the location (-22, 15)

Turn 90° West again and walk 5.00 m. to the location (-22, 10)

The location of point A = (-22, 10)

Direction \ of \ point \ A = arctan \left(\dfrac{10}{-22} \right) \approx -24.4^{\circ}

Direction to North = 90° - 24.4° ≈ 65.6°

Distance = √((-22)² + 10²) ≈ 24.2

Therefore, we have;

  • 24.2 m, 65.6° West of North

Turning East then West:

Turn 90° East and walk 22.0 m to the location (22, 15)

Turn 90° West again and walk 5.00 m. to the location (22, 20)

The location of point A = (22, 20)

Direction \ of \ point \ A = arctan \left(\dfrac{20}{22} \right) \approx 42.7^{\circ}

Direction to North = 90° - 42.3° ≈ 47.7° East

Distance = √((-22)² + 20²) ≈ 29.7

Therefore, we have;

  • 29.7 m, 42.7° East of North

Turning East then East:

Turn 90° East and walk 22.0 m to the location (22, 15)

Turn 90° East again and walk 5.00 m. to the location (22, 10)

The location of point A = (22, 10)

Direction \ of \ point \ A = arctan \left(\dfrac{10}{22} \right) \approx 24.4^{\circ}

Direction to North = 90° - 24.4° ≈ 65.6° East

Distance = √((-22)² + 10²) ≈ 24.2

Therefore, we have;

  • 24.2 m, 65.6° East of North

Turning West then East:

Turn 90° West and walk 22.0 m to the location (-22, 15)

Turn 90° East and walk 5.00 m. to the location (-22, 20)

The location of point A = (-22, 20)

Direction \ of \ point \ A = arctan \left(\dfrac{20}{-22} \right) \approx -42.7^{\circ}

Direction to North = 90° - 42.7° ≈ 47.7° West

Distance = √((-22)² + 20²) ≈ 29.7

Therefore, we have;

  • 29.7m, 47.7° West of North

Learn more here:

brainly.com/question/11489059

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The angular position of objects as a function of time is given, where a, b, and care constants. In which of these cases is the a
GalinKa [24]

Answer:

Explanation:

The options is not well presented

This are the options

A. θ = at³ + b

B. θ = at² + bt + c

C. θ = at² — b

D. θ = Sin(at)

So, we want to prove which of the following option have a constant angular acceleration I.e. does not depend on time

Now,

Angular acceleration can be determine using.

α = d²θ / dt²

α = θ''(t)

So, second deferential of each θ(t) will give the angular acceleration

A. θ = at³ + b

dθ/dt = 3at² + 0 = 3at²

d²θ/dt² = 6at

α = d²θ/dt² = 6at

The angular acceleration here still depend on time

B. θ = at² + bt + c

dθ/dt = 2at + b + 0 = 2at + b

d²θ/dt² = 2a + 0 = 2a

α = d²θ/dt² = 2a

Then, the angular acceleration here is constant is "a" is a constant and the angular acceleration is independent on time.

C. θ = at² —b

dθ/dt = 2at — 0 = 2at

d²θ/dt² = 2a

α = d²θ/dt² = 2a

Same as above in B. The angular acceleration here is constant is "a" is a constant and the angular acceleration is independent on time.

D. θ = Sin(at)

dθ/dt = aCos(at)

d²θ/dt² = —a²Sin(at) = —a²θ

α = d²θ/dt² = -a²θ

Since θ is not a constant, then, the angular acceleration is dependent on time and angular displacement

So,

The answer is B and C

4 0
3 years ago
A 16 g piece of Styrofoam carries a net charge of -8.6 µC and floats above the center of a large horizontal sheet of plastic tha
PilotLPTM [1.2K]

Answer:

the charge per unit area on the plastic sheet is - 3.23 x 10⁻⁷ C/m²

Explanation:

given information:

styrofoam mass, m = 16 g = 0.016 kg

net charge, q = - 8.6 μC

to calculate the charge per unit area on the plastic sheet, we can use the following equation:

F_{e} = mg

where

F_{e} = the force between the electric field

m = mass

g = gravitational force

F_{e} =qE

where

q = charge

E = electric field

and

E = σ/2ε₀

where

ε₀ = permitivity

thus

F_{e} =qE

mg = qσ/2ε₀

σ = (2mg ε₀)/q

  = 2 (0.016) (9.8)  (8.85 x 10⁻¹²)/( - 8.6 x 10⁻⁶)

  = - 3.23 x 10⁻⁷ C/m²

8 0
4 years ago
The height has a distance of 2.5 seconds to hit the ground what is gravity?
STatiana [176]

Answer:

11 times

Explanatiom:

756 fourty

5 0
3 years ago
Pilots often take advantage of the ____,which are highly-speed winds between 7km and 16 km above earths surface.
lukranit [14]
I believe it is called or referred to as the "Jet Stream". During World War II, allied pilots encountered high speed winds in the upper air. They named those winds after the fastest planes they came up against: fighters equipped with jet engines! Jet stream winds in winter time can reach up to 300 MPH as well!
6 0
3 years ago
Estimate the volume of each ball. Use the formula
Mila [183]

The given question is incomplete. The complete question is:

Estimate the volume of each ball. Use the formula v=\frac{4\pi\times r^3}{3} where v is the volume and r is the radius. record the volume in table A of your student guide. The radius of the tennis ball is 2.1 cm and the radius of thr golf ball is 2.0 cm. What is the estimated volume of the table tennis ball in cm^3&#10; What is the estimated volume of the golf ball in

Answer: Volume of the tennis ball is 38.8cm^3 and  Volume of the golf ball is 33.5cm^3

Explanation:

We have to find the Volume of tennis ball and golf ball by using the formula v=\frac{4\pi\times r^3}{3}

Radius of the tennis ball = 2.1 cm

Radius  of the golf ball =2.0 cm.

Putting the value of radius in the formula , we get:

Volume of the tennis ball =  \frac{4\times 3.14\times (2.1cm)^3}{3}&#10;=38.8cm^3

Volume of the golf ball =   \frac{4\times 3.14\times (2.0cm)^3}{3}&#10;=33.5cm^3

Volume of the tennis ball is 38.8cm^3 and  Volume of the golf ball is 33.5cm^3

6 0
3 years ago
Read 2 more answers
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