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ankoles [38]
2 years ago
11

A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part

icle's speed at t=5.0s ? What is the particle's direction of motion, measured as an angle from the x-axis, at t=5.0s ?
Physics
1 answer:
mestny [16]2 years ago
4 0

Differentiate the components of position to get the corresponding components of velocity :

v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t

v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}

At <em>t</em> = 5.0 s, the particle has velocity

v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}

v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}

The speed at this time is the magnitude of the velocity :

\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}

The direction of motion at this time is the angle \theta that the velocity vector makes with the positive <em>x</em>-axis, such that

\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}

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An astronaut has a momentum of 280 kg and travels 10 m/s. what is the mass of the astronaut?
Kamila [148]

Answer:

The answer is

<h2>28 kg</h2>

Explanation:

The mass of an object given it's momentum and velocity / speed can be found by using the formula

m =  \frac{p}{v}  \\

where

m is the mass

p is the momentum

v is the speed or velocity

From the question

p = 280 kg/ms

v = 10 m/s

The mass of the object is

m =  \frac{280}{10}  = 28 \\

We have the final answer as

<h3>28 kg</h3>

Hope this helps you

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3 years ago
during conversion water to ice,forces of attraction between molecules. OPTIONS:1.)Decreases 2.)Increases 3.)Does not change 4.)C
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Dua titik yang bermuatan listrik yang sama, mula-mula berjarak 5 cm dan saling tarik menarik dengan gaya 225 N. Agar kedua titik
Serggg [28]

Answer:

15 cm

Explanation:

Dari pertanyaan yang diberikan di atas, diperoleh data sebagai berikut:

Gaya 1 (F₁) = 225 N

Jarak terpisah 1 (d) = 5 cm

Gaya 2 (F₂) = 25 N

Jarak terpisah 2 (d₂) =?

Kita dapat memperoleh persamaan yang berkaitan dengan gaya dan jarak muatan dua titik dengan menggunakan rumus berikut:

F = Kq₁q₂ / d²

Perbanyak silang

Fd² = Kq₁q₂

Menjaga Kq₁q₂ konstan, kita memiliki:

F₁d₁² = F₂d₂²

Dengan rumus di atas maka diperoleh jarak sebagai berikut:

Gaya 1 (F₁) = 225 N

Jarak terpisah 1 (d) = 5 cm

Gaya 2 (F₂) = 25 N

Jarak terpisah 2 (d₂) =?

F₁d₁² = F₂d₂²

225 × 5² = 25 × d₂²

225 × 25 = 25 × d₂²

5625 = 25 × d₂²

Bagilah kedua sisinya dengan 25

d₂² = 5625/25

d₂² = 225

Hitung akar kuadrat dari kedua sisi

d₂ = √225

d₂ = 15 cm

Oleh karena itu, muatan dua titik harus berjarak 15 cm untuk memiliki gaya tarik 25 N

4 0
2 years ago
An object is dropped from an altitude of one Earth radiusabove Earth's surface. If M is the mass of Earth and R is itsradius the
Montano1993 [528]

Answer:

The speed of the object just before it hits Earth is  \sqrt{\dfrac{GM}{R}}

(A) is correct option.

Explanation:

Given that,

M = mass of earth

R = radius of earth

The potential energy at height above the surface of the earth

P.E=-\dfrac{GmM}{R+h}

The kinetic energy at height above the surface of the earth

K.E = 0

The total energy at height above the surface of the earth

E = K.E+P.E

E = -\dfrac{GmM}{R+h}....(I)

The total energy at the surface of the earth

E'=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}....(II)

We need to calculate the speed of the object  just before it hits Earth

From equation (I) and (II)

-\dfrac{GmM}{R+h}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}

Here, h = R

-\dfrac{GmM}{2R}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}

v= \sqrt{\dfrac{GM}{R}}

Hence, The speed of the object just before it hits Earth is  \sqrt{\dfrac{GM}{R}}.

4 0
3 years ago
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