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4 years ago

7

An Earth satellite needs to have its orbit changed so the new orbit will be twice as far from the center of Earth as the origina

l orbit. The new orbital period will be twice as long as the original period. O true O false

1 answer:

horsena [70]4 years ago

5 0

Answer:

False.

Explanation:

From Kepler's Third Law of plenetary motion, we know that:

<em>"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit."</em>

Or, as expressed in mathematical terms:

$\frac{a^3}{T^2}=constant$ , where <em>a</em> is the semi-major axis of the orbit (the distance from the center), and <em>T </em>is the orbital period of the satellite.

From this expression we can clearly see that if the orbit's semi-major axis is doubled, orbital period will be $\sqrt{8}$ times longer to compensate the variation.

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Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above

poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

$U_{g,1} = U_{g,2} + W_{dis}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$W_{dis}$ - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

$U = m \cdot g \cdot y$

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$y$ - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

$W_{dis} = f \cdot \Delta s$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

$m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s$

$f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}$

If $m = 1000\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 40\,m$ , $y_{2} = 25\,m$ and $\Delta s = 400\,m$ , then:

$f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}$

$f = 367.763\,N$

The magnitude of the frictional force between the car and the track is 367.763 N.

7 0

4 years ago

In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at the end. the sharp upward

ladessa [460]

<u>Answer:</u>

The skier’s launch speed = 17.08 m/s

<u>Explanation:</u>

The motion of skiing is projectile motion.

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Vertical motion (Maximum height reached, H) :

We have equation of motion, $v^2=u^2+2as$ , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

$0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}$

We have θ = 63° and H = 11.8 meter.

Substituting

$11.8=\frac{u^2sin^263}{2*9.81}\\ \\ u^2=\frac{11.8*2*9.81}{sin^263}=\frac{231.516}{0.794} =291.58\\ \\ u=17.08m/s$

The skier’s launch speed = 17.08 m/s

3 0

4 years ago

Read 2 more answers

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 5.00 s, it rotates 13.9 rad. Du

alisha [4.7K]

Answer:

(a) Angular acceleration is 1.112 rad/s².

(b) Average angular velocity is 2.78 rad/s .

Explanation:

The equation of motion in Rotational kinematics is:

θ = θ₀ + 0.5αt²

Here θ is angular displacement at time t, θ₀ is angular displacement at time t=0, t is time and α is constant angular acceleration.

(a) According to the problem, θ is 13.9 rad, θ₀ is zero as it is at rest and t is 5 s. Put these values in the above equation:

13.9 = 0 + 0.5α(5)²

α = 1.112 rad/s²

(b) The equation of average angular velocity is:

ω = Δθ/Δt

ω = $\frac{13.9}{5}$

ω = 2.78 rad/s

3 0

4 years ago

An object with a mass of 78 kg is lifted through a height of 6 meters how much work is done

il63 [147K]

We got the following:
m = 78kg
h = 6m
g = 9.8 m/ $sec^{2}$ ≈ 10 m/ $sec^{2}$
----------------------------------------------------------------------------
A = ?

As we know A = Ep = mgh -> potential energy
so the answer would be A = 78 * 6 * 10 = 4680 Joule

5 0

3 years ago

If2.0J of work is done in raising a 180g apple how far is it lifted?

Alex

We know, W = F * s
W = mg * s

Here, w = 2 J
m = 180 g = 0.180 Kg
g = 9.8 m/s

Substitute their values into the expression:
2 = 0.180*9.8 * s
1.764s = 2
s = 2 / 1.764
s = 1.13 meter

In short, Your Final Answer is 1.13 m

Hope this helps!

7 0

3 years ago