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2 years ago

HELP ASAP!!!

Physics

1 answer:

miss Akunina [59]2 years ago

4 0

Write each force in component form:

v ₁ : 50 N due east → (50 N) i

v ₂ : 80 N at N 45° E → (80 N) (cos(45°) i + sin(45°) j ) ≈ (56.5 N) (i + j )

The resultant force is the sum of these two vectors:

r = v ₁ + v ₂ ≈ (106.5 N) i + (56.5 N) j

Its magnitude is

|| r || = √[(106.5 N)² + (56.5 N)²] ≈ 121 N

and has direction θ such that

tan(θ) = (56.5 N) / (106.5 N) → θ ≈ 28.0°

i.e. a direction of about E 28.0° N. (Just to clear up any confusion, I mean 28.0° north of east, or 28.0° relative to the positive x-axis.)

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Unscramble these words TOFO - DNPOU TENOWN - TREEM HINT: they are science words

padilas [110]

It looks like they are all units of measurement:
FOOT - POUND - NEWTON - METER

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3 years ago

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At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 fi

Amanda [17]

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

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3 years ago

How many hydrogen atoms are there in all the reactants?

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Answer:

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3 years ago

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5. Principal Rodriguez has asked all students exit the building by 2:45 pm! If you want to exit

Kryger [21]

With the Pythagoras Theorem we can find that the distance traveled is:

39 m

The distance is the length of the path between two points and it is a scalar magnitude so if the path changes direction the Pythagorean theorem should be used

d = $\sqrt{(x-x_o)^2 + (y-y_o)^2 + (z-z_o)^2 }$

Where d is the distance, (x,y,z) is the interes point, (x₀,y₀,z₀) is de reference point.

In this case, let's set a reference system in the lower part of the school, take the z-axis as vertical and set the point of arrival at as the reference (0, 0, 0).

The distance that the students descend is d₁ = 30 k ^ m, when they arrive from the bottom of the school they travel d₂ = 15 j ^ m and d₃ = 20 i ^ m

let's calculate

d = $\sqrt{(20-0)^2 + (15-0)^2 + ( 30-0)^2 }$