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3 years ago

Convert 6.32 sec to days

Physics

1 answer:

Sav [38]3 years ago

4 0

Answer:

7.315x10^-5 days

Explanation:

6.32 seconds = x days

1 day = 24 hours

1 hour = 60 minutes

1 day = 24 x 60 minutes

1 minute = 60 seconds

1 day = 24 x 60 x 60 seconds

1 day = 86400 seconds

6.32 seconds = 6.32/86400

6.32 seconds = 7.315x10^-5 days

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Answer:

Part a)

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

$v = 2.18 \times 10^5 m/s$

Explanation:

Time period of sun is given as

$T = 2.60 \times 10^8 years$

$T = 2.60 \times 10^8 (365 \times 24 \times 3600) s$

$T = 8.2 \times 10^{15} s$

Now the radius of the orbit of sun is given as

$R = 3.00 \times 10^4 Ly$

$R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m$

$R = 2.84 \times 10^20 m$

Part a)

centripetal acceleration is given as

$a_c = \omega^2 R$

$a_c = \frac{4\pi^2}{T^2} R$

$a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})$

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

orbital speed is given as

$v = \frac{2\pi R}{T}$

$v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}$

$v = 2.18 \times 10^5 m/s$

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3 years ago

A 500g object falls off a cliff and losers 100 J from its gravitational potential energy store. if the gravitational field stren

ludmilkaskok [199]

Answer : Height, h = 20.4 m

Explanation :

It is given that,

Mass of an object, m = 500 g = 0.5 kg

Gravitational potential energy, PE = 100 J

The Gravitational potential energy is the energy which is possessed due to the height and gravity of an object. It is given as :

PE = m g h

where,

h is the height of the cliff.

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h = 20.40 m

So, the height of the cliff is 20.4 m.

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3 years ago

A centrifuge in a forensics laboratory rotates at an angular speed of 3,700 rev/min. When switched off, it rotates 46.0 times be

ra1l [238]

Answer:

Explanation:

Given,

initial angular speed, ω = 3,700 rev/min

= $3700\times \dfrac{2\pi}{60}=387.27\ rad/s$

final angular speed = 0 rad/s

Number of time it rotates= 46 times

angular displacement, θ = 2π x 46 = 92 π

Angular acceleration

$\alpha = \dfrac{\omega_f^2 - \omega^2}{2\theta}$

$\alpha = \dfrac{0 - 387.27^2}{2\times 92\ pi}$

$\alpha = -259.28 rad/s^2$

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3 years ago

bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. what is the veloc

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Answer:

The final velocity of the car A is -1.053 m/s.

Explanation:

For an elastic collision both the kinetic energy and the momentum of the system are conserved.

Let us call

$m_A$ = mass of car A;

$v_{A1}$ = the initial velocity of car A;

$v_{A2}$ = the final velocity of car A;

and

$m_B$ = mass of car B;

$v_{B1}$ = the initial velocity of car B;

$v_{B2}$ = the final velocity of car B.

Then, the law of conservation of momentum demands that

$m_Av_{A1}+m_Bv_{B1} =m_Av_{A2}+m_Bv_{B2}$

And the conservation of kinetic energy says that

$\dfrac{1}{2} m_Av_{A1}^2+\dfrac{1}{2}m_Bv_{B1}^2=\dfrac{1}{2}m_Av_{A2}^2+\dfrac{1}{2}m_Bv_{B2}^2$

These two equations are solved for final velocities $v_{A2}$ and $v_{B2}$ to give

$v_{A2} =\frac{m_A-m_B}{m_A+m_B} v_{A1}+\frac{2m_B}{m_A+m_B} v_{B1}$

$v_{B2} =\frac{2m_A}{m_A+m_B} v_{A1}+\frac{m_B-m_A}{m_A+m_B} v_{B1}$

by putting in the numerical values of the variables we get

$v_{A2} =\frac{281-209}{281+209} (2.82)+\frac{2*209}{281+209} (-1.72)$

$\boxed{v_{A2} = -1.05m/s}$

and

$v_{B2} =\frac{2*281}{281+209} (2.82)+\frac{209-281}{281+209} (-1.72)$

$\boxed{v_{B2} = 3.49m/s}$

Thus, the final velocity of the car A is -1.053 m/s and of car B is 3.49 m/s.

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3 years ago

Which diagram represents an open circuit?

Kruka [31]

Answer : D) Circuit A

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3 years ago

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