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Scrat [10]
3 years ago
7

The key to making a concise mathematical definition of escape velocity is to consider the energy. If an object is launched at it

s escape velocity, what is the total mechanical energy Etotal of the object at a very large (i.e., infinite) distance from the planet? Follow the usual convention and take the gravitational potential energy to be zero at very large distances?
Physics
1 answer:
aleksklad [387]3 years ago
3 0

Answer:

The total Mechanical energy will be zero

Explanation: Escape velocity is the velocity required by a free object in order to overcome the impact of the force of gravity. The total mechanical energy of an object is the total energy possessed by an object which includes its kinectic and potential energy.

since the object is moving at an escape velocity which is 11.2m/s the object will be assumed to be weightless

Etotal = kinetic energy + potential energy

kinetic energy= 1/2*M*V*V

Potential energy=MGH

Etotal=1/2*0*11.2*11.2+0*0*0

Etotal=0+0

Etotal=0.

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Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius
Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2

The centripetal acceleration for the second radius; 4.0 m

a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2

The centripetal acceleration for the third radius; 6.0 m

a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2

The centripetal acceleration for the fourth radius; 8.0 m

a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2

The centripetal acceleration for the fifth radius; 10.0 m

a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2

6 0
3 years ago
5. A box weighs 196 N. A rope is tied to the box. What is the
natta225 [31]

Answer:

296 N

Explanation:

Draw a free body diagram.  The box has two forces on it: tension up and weight down.

Apply Newton's second law:

∑F = ma

T − mg = ma

T = m (g + a)

Given m = 196 N / 9.8 m/s² = 20 kg, and a = +5 m/s²:

T = (20 kg) (9.8 m/s² + 5 m/s²)

T = 296 N

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3 years ago
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professor190 [17]

Answer:

sorry i throght i had the answer

Explanation:

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2 years ago
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