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Scrat [10]
3 years ago
15

A projectile has a speed of √ gm/3r directed away from a planet, when it is a distance of 4R from the center of the planet. the

planet has mass M and radius R. will this projectile be able to escape from the gravitational attraction of the planet?
Physics
1 answer:
Nookie1986 [14]3 years ago
3 0

Answer:

hat the speed of the rocket is lower than the minimum escape velocity whereby the ship cannot escape the planet

Explanation:

To find out if the projectile can escape the planet, let's find the minimum escape velocity using the concepts of energy.

Starting point. On the planet's surface

            Em∠₀ = k + U = ½ m v² - G mM / R²

End point far away

          Emf = U = - g m M / r²

          Em₀ = Emf

          ½ m v² - G m M / R² = -G m M / r²

          v² = G M (1 / R² -1 / r²)

Let's find the velocity for the height of the rocket r = 4R

          v =√GM (1 / R² - 1/16 R²) = √a GM / R²      0.968

This is the speed to escape planet

 

Let's compare this minimum escape velocity with the given value

             

            v = √GM /R²    1 /√ 3

            v = √GM / R 0.577

We can see that the speed of the rocket is lower than the minimum escape velocity whereby the ship cannot escape the planet

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Anastaziya [24]

Answer:

to much to small

Explanation:

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5 0
3 years ago
A 2.0-kg object moving 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost
densk [106]

Answer:

20J

Explanation:

In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;

m₁u₁ + m₂u₂ = (m₁ + m₂)v          ---------------(i)

Where;

m₁ and m₂ are the masses of first and second objects respectively

u₁ and u₂ are the initial velocities of the first and second objects respectively

v  is the final velocity of the two objects after collision;

From the question;

m₁ = 2.0kg

m₂ = 8.0kg

u₁ = 5.0m/s

u₂ = 0        (since the object is initially at rest)

<em>Substitute these values into equation (i) as follows;</em>

(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v

(10.0) + (0) = (10.0)v

10.0 = 10.0v

v = 1m/s

The two bodies stick together and move off with a velocity of 1m/s after collision.

The kinetic energy(KE₁) of the objects before collision is given by

KE₁ = \frac{1}{2}m₁u₁² +  \frac{1}{2}m₂u₂²       ---------------(ii)

Substitute the appropriate values into equation (ii)

KE₁ = (\frac{1}{2} x 2.0 x 5.0²) +  (\frac{1}{2} x 8.0 x 0²)

KE₁ = 25.0J

Also, the kinetic energy(KE₂) of the objects after collision is given by

KE₂ = \frac{1}{2}(m₁ + m₂)v²      ---------------(iii)

Substitute the appropriate values into equation (iii)

KE₂ = \frac{1}{2} ( 2.0 + 8.0) x 1²

KE₂ = 5J

The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision

K = KE₂ - KE₁

K = 5 - 25

K = -20J

The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J

3 0
4 years ago
A 200g ball is accelerated from 5m/s to 25m/s in 0.3sec. What force was applied?​
fomenos

Answer:

30N

Explanation:

30N of force was applied

quizlet

5 0
2 years ago
A parallel-plate vacuum capacitor has 8.60 J of energy stored in it. The separation between the plates is 3.80 mm . If the separ
creativ13 [48]

Answer:

Part a)

E = \frac{8.60}{2.62} = 3.28 J

Part b)

E = 2.62(8.60) = 22.5 J

Explanation:

As we know that the energy of capacitor when it is not connected to potential source is given as

U = \frac{Q^2}{2C}

As we know that initial energy is given as

8.60 = \frac{Q^2}{2C}

now we know that capacitance of parallel plate capacitor is given as

C = \frac{\epsilon_0A}{d}

now the new capacitance when distance is changed from 3.80 mm to 1.45 mm

C' = \frac{Cd}{d'}

C' = \frac{C(3.80)}{1.45}

C' = 2.62 C

Now the new energy of the capacitor is given as

E = \frac{Q^2}{2(2.62C)}

E = \frac{8.60}{2.62} = 3.28 J

Part b)

Now if the voltage difference between the plates of capacitor is given constant

now the energy energy of capacitor is

U = \frac{1}{2}CV^2

8.60 = \frac{1}{2}CV^2

now when capacitance is changed to new value then new energy is given as

E = \frac{1}{2}C'V^2

E = \frac{1}{2}(2.62C)V^2

E = 2.62(8.60) = 22.5 J

6 0
3 years ago
1. A block of aluminum occupies
Anvisha [2.4K]

Answer:

2700g/L

Explanation:

D=m/v

40.5g/0.015L

7 0
3 years ago
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