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alexandr1967 [171]
3 years ago
5

Objects that are not moving have NO _____. A. mass (only) B. mass and momentum (only) C. velocity and momentum (only) D. mass, v

elocity, and momentum
Physics
1 answer:
shepuryov [24]3 years ago
6 0

Answer:

Objects that are not moving have NO velocity and momentum (only)

Explanation:

This best explains the potential energy of an object, when an object is at rest. it is said to have possessed a potential energy but without velocity and momentum i.e energy at rest

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Which of the following are examples of natural forces? A. optical B. phase C. gravity D. nuclear
denpristay [2]
<span>The forces of nature are phase forces.</span>
6 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
1 year ago
Richard is driving home to visit his parents. 135{\rm mi} of the trip are on the interstate highway where the speed limit is 65{
nydimaria [60]

Answer:

time spent = 0.2276

Explanation:

given data

distance = 135 mi

usual speed = 65 mph

today speed = 73 mph

solution

we get here time that is express as

time = \frac{distance}{speed}    ...................1

usual time = \frac{135}{65}  = 2.0769 h

today time = \frac{135}{73} = 1.8493 h

so we get here time spent as

time spent = 2.0769 h - 1.8493 h

time spent = 0.2276

6 0
3 years ago
13. How much work does a 35 N force do to lift a potted plant from the floor to a shelf 6 m high?
Nataly_w [17]

Answer:

W= 210 N

Explanation:

Just use work = Fparallel*d

W= 35*6

W= 210 N

8 0
2 years ago
A container is filled with an ideal diatomic gas to a pressure and volume of P1 and V1, respectively. The gas is then warmed in
lilavasa [31]

Answer:

Explanation:

The  change is as follows

P₁ V₁ to 3P₁, V₁ ( constt volume )  --- first process

3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process

In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁

P₁V₁ = n R T₁ , n is no of moles of gas enclosed.

nRT₁ = P₁V₁

Heat added at constant volume  = n Cv ( 3T₁ - T₁)

= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)

= 10/3 x nRT₁

= 10/3x P₁V₁

In the second process,  Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁

Heat added at constant pressure in second case  

= n Cp ( 15T₁ - 3T₁)

= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)

= 28 x nRT₁

= 28 P₁V₁

6 0
3 years ago
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