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3 years ago

5

A rocket blasts off and moves straight upward from the launch pad with constant acceleration. after 3.6 s the rocket is at a hei

ght of 70.0 m. (a) what are the magnitude and direction of the rocket's acceleration?

1 answer:

const2013 [10]3 years ago

6 0

X=.5a(t^2), x=70m, t=3.6s, a=?. a=140/(3.6^2)=10.8m/s^2.

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A satellite of mass m circles a planet of mass M and radius R in an orbit at a height 2R above the surface of the planet. What m

Goryan [66]

Answer:

ΔE = GMm/24R

Explanation:

centripetal acceleration a = V^2 / R = 2T/mr

T= kinetic energy

m= mass of satellite, r= radius of earth

= gravitational acceleration = GM / r^2

Now, solving for the kinetic energy:

T = GMm / 2r = -1/2 U,

where U is the potential energy

So the total energy is:

E = T+U = -GMm / 2r

Now we want to find the energy difference as r goes from one orbital radius to another:

ΔE = GMm/2 (1/R_1 - 1/R_2)

So in this case, R_1 is 3R (planet's radius + orbital altitude) and R_2 is 4R

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Consider a spherical planet of uniform density rho. The distance from the planet's center to its surface (i.e., the planet's rad

alisha [4.7K]

Answer:

a) $g(r) = 4\pi \cdot G \cdot \rho\cdot r$ , b) $g = 4\pi \cdot G \cdot \rho\cdot r_{P}$

Explanation:

a) The acceleration due to gravity inside the planet is:

$dg = G\cdot \frac{\rho \cdot dV}{r^{2}}$

$dg = G\cdot \frac{\rho \cdot dV}{r^{2}}$

$dg = G\cdot \frac{4\pi\cdot \rho \cdot r^{2}\,dr}{r^{2}}$

$dg = 4\pi\cdot G\cdot \rho \,dr$

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b) The acceleration at the surface of the planet is:

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Answer:

29.2 ft/s

Explanation:

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y = 13 tan θ

where θ is the light's angle from perpendicular to the wall.

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w = (θ/t)

θ = wt = (2πt/3)

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y = 13 tan θ

(dy/dt) = 13 sec² θ (dθ/dt)

(dy/dt) = 13 sec² θ (2π/3)

(dy/dt) = (26π/3) sec² θ

when θ = 15°

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3 years ago

Series circuit when you had one bulb and battery voltage was at 9 volts, what was current into battery?

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Answer:

incomplete question, resistor must be there

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3 years ago

From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up

NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

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where,

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v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

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3 years ago