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3 years ago

The terms sanity and insanity:

Physics

2 answers:

Alex Ar [27]3 years ago

6 0

Answer:

are definitions that are used in the legal system

Explanation:

ap3x bruv :)

IgorLugansk [536]3 years ago

3 0

Whats the question exactly?

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Vector ????⃗ has a magnitude of 16.6 and is at an angle of 50.5∘ counterclockwise from the +x‑axis. Vector ????⃗ has a magnitude

natka813 [3]

Answer:

For vector u, x component = 10.558 and y component =12.808

unit vector = 0.636 i+ 0.7716 j

For vector v, x component = 23.6316 and y component = -6.464

unit vector = 0.9645 i-0.2638 j

Explanation:

Let the vector u has magnitude 16.6

u makes an angle of 50.5° from x axis

So $u_x=ucos\Theta =16.6\times cos50.5=10.558$

Vertical component $u_y=usin\Theta =16.6\times sin50.5=12.808$

So vector u will be u = 10.558 i+12.808 j

Unit vector $u=\frac{10.558i+12.808j}{\sqrt{10.558^2+12.808^2}}=0.636i+0.7716j$

Now in second case let vector v has a magnitude of 24.5

Making an angle with -15.3° from x axis

So horizontal component $v_x=vcos\Theta =24.5\times cos(-15.3)=23.6316$

Vertical component $v_y=vsin\Theta =24.5\times sin(-15.3)=-6.464$

So vector v will be 23.6316 i - 6.464 j

Unit vector of v $=\frac{23.6316i-6.464}{\sqrt{23.6316^2+6.464^2}}=0.9645i-0.2638j$

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3 years ago

Please answer the question

AfilCa [17]

The answer is Cㅤㅤㅤㅤㅤ

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3 years ago

A wave is a disturbance that carries

viktelen [127]

Answer:

(D) energy from one place to another

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2 years ago

Glucose can have a straight chain of ______________ atoms or, more commonly, form a _____________ structure.

vovikov84 [41]

Lipids and <span>Polysaccharides r the answers hoped i helped</span>

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3 years ago

A uniform, solid sphere of radius 2.50 cm and mass 4.75 kg starts with a purely translational speed of 3.00 m/s at the top of an

allsm [11]

Answer:

The final translational seed at the bottom of the ramp is approximately 4.84 m/s

Explanation:

The given parameters are;

The radius of the sphere, R = 2.50 cm

The mass of the sphere, m = 4.75 kg

The translational speed at the top of the inclined plane, v = 3.00 m/s

The length of the inclined plane, l = 2.75 m

The angle at which the plane is tilted, θ = 22.0°

We have;

$K_i$ + $U_i$ = $K_f$ + $U_f$

K = (1/2)×m×v²×(1 + I/(m·r²))

I = (2/5)·m·r²

K = (1/2)×m×v²×(1 + 2/5) = 7/10 × m×v²

U = m·g·h

h = l×sin(θ)

h = 2.75×sin(22.0°)

∴ 7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) = 7/10 × 4.75× $v_f$ ² + 0

7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) ≈ 77.93

∴ 77.93 ≈ 7/10 × 4.75× $v_f$ ²

$v_f$ ² = 77.93/(7/10 × 4.75)

$v_f$ ≈ √(77.93/(7/10 × 4.75)) ≈ 4.84

The final translational seed at the bottom of the ramp, $v_f$ ≈ 4.84 m/s.

3 0

2 years ago