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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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A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is
valina [46]

Answer:

4.75 m/s

Explanation:

The computation of the velocity of the existing water is shown below:

Data provided in the question

Tall = 2 m

Inside diameter tank = 2m

Hole opened = 10 cm

Bottom of the tank = 0.75 m

Based on the above information, first we have to determine the height which is

= 2 - 0.75 - 0.10

= 2 - 0.85

= 1.15 m

We assume the following things

1. Compressible flow

2. Stream line followed

Now applied the Bernoulli equation to section 1 and 2

So we get

\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2

where,

P_1 = P_2 = hydrostatic

z_1 = 0

z_2 = h

Now

\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\  \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}

= 4.7476 m/sec

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6 0
3 years ago
Does a light bulb with a greater wattage have a greater brightness
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The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
Mama L [17]

A) Time needed: 6.24 s

B) Time needed: 2.86 s

Explanation:

A)

In this part, we are told that the power if the engine is constant. The power of the engine is given by

P=\frac{W}{t}

where

W is the work done

t is the time

This means that the power of the engine is proportional to the work done, and therefore, to the kinetic energy of the car:

P=\frac{\frac{1}{2}mv^2}{t}=const.

where m is the mass of the car and v its velocity.

SInce power is constant, we can write:

\frac{\frac{1}{2}mv_1^2}{t_1^2}=\frac{\frac{1}{2}mv_2^2}{t_2}

where:

t_1=1.40 s is the time the car needs to accelerates to v_1=28.0 mph

t_2 is the time the car needs to accelerate to v_2=57.0 mph

Therefore, solving for t_2,

t_2 = \frac{v^2}{u^2}t_1=\frac{57^2}{28^2}(1.40)=6.24 s

B)

First of all, we have to calculate the acceleration of the car. We can do it using the following equation:

a=\frac{v-u}{t}

where:

u = 0 is the initial velocity

v=28.0 mph \cdot \frac{1609 m/mi}{3600 s/h}=12.5 m/s is the final velocity

t = 1.40 s is the time elapsed

Substituting, we find the acceleration:

a=\frac{12.5-0}{1.40}=8.9 m/s^2

In this part, we are told that the force exerted by the engine is constant: according to Newton's second law, acceleration is proportional to the force,

F=ma

This means that the acceleration is also constant.

Now we want to find how long the car takes to accelerate to a final velocity of

v=57.0 mph \cdot \frac{1609}{3600}=25.5 m/s

From an initial velocity of

u = 0

Using again the same suvat equation, and using the acceleration we found previously, we find:

t=\frac{v-u}{a}=\frac{25.5-0}{8.9}=2.87 s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

About power:

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#LearnwithBrainly

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marishachu [46]

Answer:

The box will experience an acceleration.

Explanation:

Here, 2 N and 3 N forces are acting opposite to each other. In this case, the net force experience by the box would be (3-2)N = 1 N towards right. Since acceleration is directly proportional to the net force, therefore the box will experience an acceleration.

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