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daser333 [38]
3 years ago
13

The Celsius temperature scale is based on which of the following

Physics
1 answer:
Lapatulllka [165]3 years ago
6 0
Freezing.................
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The captain of a boat wants to travel directly across a river that flows due exst with a speed of 100 m/sHe starts from the sout
Salsk061 [2.6K]

Answer:

86.51° North of West or 273.49°

Explanation:

Let V' = velocity of boat relative to the earth, v = velocity of boat relative to water and V = velocity of water.

Now, by vector addition V' = V + v'.

Since v' = 6.10 m/s in the north direction, v' = (6.10 m/s)j and V = 100 m/s in the east direction, V = (100 m/s)i. So that

V' = V + v'

V' =   (100 m/s)i + (6.10 m/s)j

So, we find the direction,Ф the boat must steer to from the components of V'.

So tanФ = 6.10 m/s ÷ 100 m/s

tanФ = 0.061

Ф = tan⁻¹(0.061) = 3.49°

So, the angle from the north is thus 90° - 3.49° = 86.51° North of West or 270° + 3.49° = 273.49°

7 0
3 years ago
Which of the following is the best definition of vaporization
Vinvika [58]

Change of a liquid to a gas

8 0
3 years ago
An atom of gadolinium has an atomic number of 64 and a mass number of 154. how many electrons, protons, and neutrons does it con
Svetllana [295]
No. of protons = atomic number = 64
no. of neutrons = mass no. - no. of protons= 154 -64 = 90
no. of electrons = no. of protons = 64
5 0
3 years ago
A solid circular plate has a mass of 0.25 kg and a radius of 0.30 m. It starts rolling from rest at the top of a hill 10 m long
KATRIN_1 [288]

To solve the problem it is necessary to apply the equations related to the conservation of both <em>kinetic of rolling objects</em> and potential energy and the moment of inertia.

The net height from the point where it begins to roll with an inclination of 30 degrees would be

h=Lsin30

h=10sin30

h=5m

In the case of Inertia would be given by

I = \frac{mR^2}{2}

In general, given an object of mass m, an effective radius k can be defined for an axis through its center of mass, with such a value that its moment of inertia is

I = mk^2

\frac{mR^2}{2}= mk^2

\frac{k^2}{R^2}=\frac{1}{2}

Replacing in Energy conservation Equation we have that

Potential Energy = Kinetic Energy of Rolling Object

mgh = \frac{1}{2}mv^2(1+\frac{k^2}{r^2})

9.8*5=\frac{1}{2}v^2(1+\frac{1}{2})

v^2 (1.5) = 98

v=8.0829m/s

Therefore the correct answer is C.

3 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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