Ask question

Ask question

All categories

3 years ago

15

A satellite is in orbit 36000km above the surface of the earth. Its angular velocity is 7.27*10^-5 rad/s. What is the velocity o

f the satellite? (The radius of the earth is 6400km.)

1 answer:

bagirrra123 [75]3 years ago

6 0

Answer:

v = 3.08 km/s

Explanation:

Given that,

The angular velocity of the satellite = $\omega=7.27\times 10^{-5} rad/s$

A satellite is in orbit 36000km above the surface of the earth.

The radius of the earth is 6400 km

Let v is the velocity of the satellite. It can be calculated as :

$v=r\omega\\\\v=(36000\times 10^3+6400\times 10^3)\times 7.27\times 10^{-5}\\\\v=3082.48\ m/s\\\\v=3.08\ km/s$

So, the velocity of the satellite is 3.08 km/s.

You might be interested in

Which statement correctly describes the current in a circuit that is made up of any two resistors connected in parallel with a b

AysviL [449]

B is the answer I think

8 0

3 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

<h3>a)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

<h3>b)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

7 0

3 years ago

Luigi twirls a round piece of pizza dough overhead with a frequency of

viva [34]

The linear speed of the pepperoni is 0.628 m/s. Its direction is tangential to the circle.

We know that;

v = rω

r = radius of the piece = 10 cm or 0.1 m

ω = angular velocity

We have to convert 60 revolutions per minute to radians per second

1 rev/min = 0.10472 rad/s

60 revolutions per minute = 60 rev/min × 0.10472 rad/s/1 rev/min

= 6.28 rad/s

v = 0.1 m × 6.28 rad/s

v = 0.628 m/s

The direction of this velocity is tangential to the circle.

Learn more: brainly.com/question/4612545

5 0

2 years ago

A(n) 82.7 kg boxer has his first match in the Canal Zone with gravitational acceleration 9.782 m/s 2 and his second match at the

Annette [7]

Answer:

82.7 kg

Explanation:

the mass of the boxer remains unchanged, this is because mass is a measure of the quantity of matter in an object irrespective of its location and the gravitational force acting at its location. this means mass is independent of the gravitational acceleration hence it remains the same 82.7 kg. its unit is in kilograms (Kg).

6 0

3 years ago

A block of mass 0.08 kg is pushed against a spring with spring constant k=31 N/m. The spring is compressed 0.15 meters from its

ELEN [110]

Answer:

1.11 meters

Explanation:

As the spring is compressed, elastic potential energy is built up in the spring. The total elastic potential energy can be found using the following formula

Ep = 1/2 x k x s²

where k = 31 N/m (spring constant)

s = 0.15 m (compression)

Ep = 3.4875 J

When the block of mass is released, the elastic potential energy (Ep) is converted to kinetic energy (Ek). From this we can find the initial velocity of the mass of block after release

Ek = 1/2 x m x u²

where Ek = Ep = 3.4875J

m = 0.08 kg (mass of block)

u = unknown (initial velocity)

u = 2.9526 m/s

Now that we know the initial velocity we need to find the deceleration of the mass of block due to friction. We will first find the force of friction from the following formula

F = ∪ x m x g

where F = unknown (frictional force)

∪ = 0.4 (coefficient of friction)

m = 0.08 kg (mass of block)

g = 9.81 m/s² (acceleration due to gravity)

F = 0.31392 N

From this force we calculate the deceleration based on the following formula

F = m x a

where F = 0.31392 (frictional force)

m = 0.08 kg (mass of block)

a = unknown (acceleration)

a = -3.924 m/s² -

*the negative sign is due to this value being deceleration

Now to find the total distance traveled we use the equation for motion

v² = u² + 2as

where v = 0 (final velocity)

u = 2.9526 m/s (initial velocity

a = -3.924 m/s² (deceleration due to friction)

s = unknown (distance traveled)

s = 1.11 meters

3 0

3 years ago