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kkurt [141]
2 years ago
12

A skateboarder is moving at the speed of 3 m/s2. while traveling, the skateboard comes to a complete stop. The skateboarder howe

ver the skateboarder continues to travel at a speed of 3 m/s2 and is thrown off the skateboard. This is an example of Newton's ____?
1st Law of Motion
2nd Law of Motion
3rd Law of Motion
Physics
1 answer:
Masja [62]2 years ago
7 0

1st law of motion this is right answer

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An athletes heart beats 62 times per minute. What is the frequency of her heart beat?
PilotLPTM [1.2K]

Answer:

22

Explanation:

bc it just is

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A sample of a solid nonmetal is compared to a sample of a solid metal. Both samples have the same volume. What is the most likel
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The weight could be different, metals have a higher mass than nonmetals, so when occupying the same amount of space, the weight of the metal is far more.
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Read 2 more answers
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garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

(c) EE = ½ kx²

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

3 0
3 years ago
The photo shows falling water droplets.
levacccp [35]
I Think The answer is d I hope it helps My friend Message Me if I’m wrong and I’ll change My answer and fix it for you
4 0
3 years ago
A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also attached to a viscous damper with a damping constant of 40
11111nata11111 [884]

Quasi frequency = 4√6

Quasi period = π√6/12

t ≈ 0.4045

<u>Explanation:</u>

Given:

Mass, m = 20g

τ = 400 dyn.s/cm

k = 3920

u(0) = 2

u'(0) = 0

General differential equation:

mu" + τu' + ku = 0

Replacing the variables with the known value:

20u" + 400u' + 3920u = 0

Divide each side by 20

u" + 20u' + 196u = 0

Determining the characteristic equation by replacing y" with r², y' with r and y with 1 in the differential equation.

r² + 20r + 196 = 0

Determining the roots:

r = \frac{-20 +- \sqrt{(20)^2 - 4(1)(196)} }{2(1)}

r = -10 ± 4√6i

The general solution for two complex roots are:

y = c₁ eᵃt cosbt + c₂ eᵃt sinbt

with a the real part of the roots and b be the imaginary part of the roots.

Since, a = -10 and b = 4√6

u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6t

u(0) = 2

u'(0) = 0

(b)

Quasi frequency:

μ = \frac{\sqrt{4km - y^2} }{2m}

= \frac{\sqrt{4(3929)(20) - (400)^2} }{2(20)} \\\\= 4\sqrt{6}

(c)

Quasi period:

T = 2π / μ

T = \frac{2\pi }{4\sqrt{6} } \\\\T = \frac{\pi\sqrt{6}  }{12}

(d)

|u(t)| < 0.05 cm

u(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t < 0.05

solving for t:

τ = t ≈ 0.4045

8 0
3 years ago
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