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Bess [88]
2 years ago
9

How much energy is required to accelerate a golf ball of mass 0.046 kg initially at rest to a speed of 0.75c?

Physics
1 answer:
Andrew [12]2 years ago
3 0

The required energy to accelerate the golf ball is  0.232875X10^{16} kg-m/sec^2.

<h2>What is Energy?</h2>
  • Energy, which is observable in the execution of labor as well as in the form of light and heat, is the quantitative quality that is transmitted to an object or to a physical phenomenon in physics.
  • Energy is a preserved resource; according to the rule of conservation of energy, energy can only be transformed from one form to another and cannot be created or destroyed.
  • The joule, which is defined as "the energy transmitted to an object by the effort of moving it a length of one metre against a force of one newton," is the standard measure for power in the International System of Units (SI).

Given that,

E=mc^2

=0.046X0.75^2Xc^2

=0.232875X10^{16} kg-m/sec^2 .

Learn more about energy here:

brainly.com/question/13881533

#SPJ4

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It depends on speed because it is a part of a car
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What is the main cause of global convection currents
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Read 2 more answers
A ball on the end of a string is whirled around in a horizontal circle of radius 0.478 m. The plane of the circle is 1.02 m abov
m_a_m_a [10]

Answer:

Explanation:

We shall find first the velocity of ball at the time when string breaks. Let it be v . During its fall on the ground , 1.02 m below, we use the formula

h = 1/2 gt² where t is time of fall .

1.02 = 1/2 x 9.8 x t²

t²= .2081

t = .456

During this time it travels horizontally at distance of 2.5 m with uniform velocity of v

v x .456 = 2.5

v  = 5.48 m /s

centripetal acceleration

= v² / r where r is radius of the circular path

= 5.48² / .478

= 62.82 m /s²

3 0
3 years ago
If A &gt; B, under what condition is |A-BI=|A|- IB|? a. Vectors A and B are in opposite directions b. Vectors A and B are in the
kupik [55]

Answer:

b) Vectors A and B are in the same direction.

Explanation:

To understand this problem we will say that vector A has a magnitude of 5 units and vector B a magnitude of 3 units. In the subtraction of vectors the initial parts of vectors always bind together. And the vector resulting from the subtraction is traced from the end of the second vector (B) to the end of the first vector (A).

The length of the resultant vector will be 5 - 3 = 2

In the attached image, we analyze case a), b), and d)

For a)

As we can see in the attached image the resultant vector has a length of 8 units.

For d)

As we can see in the attached image the resultant vector has a length of 5.83 units.

For b)

The resultant vector has a length of 2 units.

Therefore the case given in b) is true

5 0
3 years ago
A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and
irina [24]

The eccentricity of its orbit is $$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

<h3>What is mass?</h3>
  • Mass is a physical body's total amount of matter. It also serves as a gauge for the body's inertia or resistance to acceleration (change in velocity) in the presence of a net force. The strength of an object's gravitational pull to other bodies is also influenced by its mass.
  • The kilogram is the SI unit of mass (kg). In science and technology, a body's weight in a given reference frame is the force that causes it to accelerate at a rate equal to the local acceleration of free fall in that frame.
  • For instance, a kilogram mass weighs around 2.2 pounds at the surface of the planet. However, the same kilogram mass would weigh just about 0.8 pounds on Mars and about 5.5 pounds on Jupiter.
  • An object's mass is a crucial indicator of how much stuff it contains. Weight is a measurement of an object's gravitational pull. It is influenced by the object's location in addition to its mass. As a result, weight is a measurement of force.

The length of the semi-major axis is calculated as follows:

where, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$

$M=1.99 \times 10^{30} \mathrm{~kg}=$mass of sur

$m=1.20 \times 10^{10} \mathrm{~kg}$ - a mass of the comet

$$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

$$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

To learn more about mass, refer to:

brainly.com/question/3187640

#SPJ4

4 0
2 years ago
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