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4 years ago

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Only one of the following is possible for a moving object. Which one? The velocity is directed eastward and is decreasing, while

the acceleration is also eastward. The velocity is directed northward and is increasing, while the acceleration is zero. The velocity is constant at all times, and the acceleration is not zero. The velocity is zero at one instant, and the acceleration is not zero at that instant. The velocity is directed eastward and is increasing, while the acceleration is directed westward.

1 answer:

andreev551 [17]4 years ago

4 0

Answer:

The velocity is zero at one instant, and the acceleration is not zero at that instant.

Explanation:

We know that

$a=\dfrac{dV}{dt}$

Where

a is the acceleration of moving object

V is the velocity.

When velocity of object is increasing with time then it means that object is accelerated.

When velocity of object is decreasing with time then it means that object is de-accelerated.

When velocity is constant it means that acceleration of object is zero.

So at particular position the instant velocity of block is zero but acceleration is not zero at that instant.

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Quantum and state physics

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1 year ago

(b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1510 m/s? (c) With wh

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Answers:

(a) $2509.98 m/s$

(b) 397042.215 m

(c) 1917.76 m/s

Explanation:

The question is incomplete, please remember to write the whole question :) However, part (a) is written below:

(a) What is the escape speed on a spherical asteroid whose radius is $700 km$ and whose gravitational acceleration at the surface is $a_{g}=4.5 m/s^{2}$

Knowing this, let's begin:

a) In this part we need to find the escape speed $V_{e}$ on the asteroid:

$V_{e}=\sqrt{\frac{2GM}{R}}$ (1)

Where:

$G$ is the universal gravitational constant

$M$ is the mass of the asteroid

$R=700 km=700(10)^{3} m$ is the radius of the asteroid

On the other hand we know the gravitational acceleration is $a_{g}=4.5 m/s^{2}$ , which is given by:

$a_{g}=\frac{GM}{R^{2}}$ (2)

Isolating $GM$ :

$GM=a_{g}R^{2}$ (3)

Substituting (3) in (1):

$V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R}$ (4)

$V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)}$ (5)

$V_{e}=2509.98 m/s$ (6) This is the escape velocity

b) In this part we will use the Conservation of mechanical energy principle:

$E_{o}=E_{f}$ (7)

Being:

$E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R}$ (8)

$E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h}$ (9)

Where:

$E_{o}$ is the initial mechanical energy

$E_{f}$ is the final mechanical energy

$K_{o}$ is the initial kinetic energy

$K_{f}=0$ is the final kinetic energy

$U_{o}$ is the initial gravitational potential energy

$U_{f}$ is the final gravitational potential energy

$m$ is the mass of the object

$V=1510 m/s$ is the radial speed of the object

$h$ is the distance above the surface of the object

Then:

$\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h}$ (10)

Isolating $h$ :

$h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R$ (11)

$h=\frac{2 (4.5 m/s^{2}) (700(10)^{3} m)^{2}}{2(4.5 m/s^{2})(700(10)^{3} m)-(1510 m/s)^{2}}-700(10)^{3} m$ (11)

$h=397042.215 m$ (12) This is the distance above the asteroid's surface

c) We will use the Conservation of mechanical energy principle again, but now the condition is that the object is dropped at a distance $h=981.8 km=981.8(10)^{3} m$ . This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy:

$\frac{-GMm}{R+h}=\frac{-GMm}{R}+\frac{1}{2}mV^{2}$ (13)

Isolating $V$ :

$V=\sqrt{2a_{g} R(1-\frac{R}{R+h})}$ (14)

$V=\sqrt{2(4.5 m/s^{2}) (700(10)^{3} m)(1-\frac{700(10)^{3} m}{700(10)^{3} m+981.8(10)^{3} m})}$ (15)

Finally:

$V=1917.76 m/s$

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Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth

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Answer:

$\lambda=6.83\times 10^{-5}\ m$

Explanation:

Given that,

An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 4.39 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

We know that,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m$

So, the wavelength of the infrared radiation is $6.83\times 10^{-5}\ m$ .

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3 years ago