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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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A 22.0 kg child slides down a slide that makes a 37.0° angle with the horizontal. (a) What is the magnitude of the normal force

patriot [66]

Answer:

(a) 172.185 N

(b) $53^{\circ}$

Solution:

As per the question:

Mass of the child, m = 22.0 kg

Angle, $\theta = 37.0^{\circ}$

Now,

(a) The magnitude of the normal force exerted by the slide on the child:

$F_{N} = mgcos\theta$

$F_{N} = 22\times 9.8cos37^{\circ} = 172.185\ N$

Now,

(b) The angle from the horizontal at which the force is directed is:

$90^{\circ} - 37^{\circ} = 53^{\circ}$

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3 years ago

Heat from Earth's interior causes convection currents in Earth's _________.

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3 years ago

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A bird sits on a high-voltage power line with its feet 3.87 cm apart. The wire is made from aluminum, is 2.11 cm in diameter, an

Svetlanka [38]

Answer:

ΔV=0.484mV

Explanation:

The potential difference across the end of conductor that obeys Ohms law:

ΔV=IR

Where I is current

R is resistance

The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A

R=(pL)/A

Given data

Length L=3.87 cm =0.0387m

Diameter d=2.11 cm =0.0211 m

Current I=165 A

Resistivity of aluminum p=2.65×10⁻⁸ ohms

ΔV=IR

$=I(\frac{pL}{A})\\ =I(\frac{pL}{\pi r^{2} } )\\=I(\frac{pL}{\pi (d/2)^{2} } )\\=165A((\frac{(2.65*10^{-8})(0.0387m)}{\pi (0.0211m/2)^{2} } ))\\=4.84*10^{-4}V$

ΔV=0.484mV

3 0

2 years ago

Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f

stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2$

When s = Db, t = 2t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2$

Dividing the two equations

$\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db$

Hence, Da=(1/4)Db

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3 years ago

a water bomber flying with a horizontal speed of 85m/s at a height of 3000m drops a load on a fire below. How far in front of th

Andreyy89

Answer:

2081.65 m

Explanation:

We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:

Height (h) = 3000 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

3000 = ½ × 10 × t²

3000 = 5 × t²

Divide both side by 5

t² = 3000 / 5

t² = 600

Take the square root of both side

t = √600

t = 24.49 s

Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:

Horizontal velocity (u) = 85 m/s

Time (t) = 24.49 s

Horizontal distance (s) =?

s = ut

s = 85 × 24.49

s = 2081.65 m

Thus, the load should be released from 2081.65 m.

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3 years ago