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Ronch [10]
3 years ago
6

6. The potential energy of a freely falling object decreases

Physics
1 answer:
lilavasa [31]3 years ago
7 0

Answer:

Explanation:

As the potential energy of the freely falling object decreases, its kinetic energy increases on account of an increase in its velocity. ... Thus, the law of conservation of energy is not violated.

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the book has a mass of 2.5 kg. What net force must act on the book to mak it accelerate to the left at a rate of 7.0m/s2?
Reil [10]

Answer:

17.5 N

Explanation:

<h2>Given :</h2>

  • Mass (m) = 2.5 kg
  • Acceleration (a) = 7.0 m/s²

<h2>To calculate :</h2>

  • Force exerted (F)

<h2>Calculation :</h2>

<h3>• F = ma</h3>

→ F = (2.5 × 7.0) N

→ F = 25/10 × 7 N

→ F = 5/2 × 7 N

→ F = (5 × 7)/2 N

→ F = 35/2 N

→ <u>F</u><u> </u><u>=</u><u> </u><u>1</u><u>7</u><u>.</u><u>5</u><u> </u><u>N</u><u> </u><u>towards</u><u> </u><u>left</u>

Hence, 17.5 N of net force must act on the book to make it accelerate to the left.

5 0
3 years ago
Acceleration is caused by a force acting on a mass.​
Mumz [18]

Answer:

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

Explanation:

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4 0
4 years ago
g Let the orbital radius of a planet be R and let the orbital period of the planet be T. What quantity is constant for all plane
Wittaler [7]

Explanation:

Kepler's third law gives the relationship between the orbital radius and the orbital period of the planet. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

Here,

G is gravitational constant

M is mass of sun

It means that the mass of Sun is constant for all planets orbiting the sun, assuming circular orbits.

7 0
3 years ago
Through the process of transpiration, plants do what?
Evgesh-ka [11]
Well, the process of TRANSPIRATION deals with the release of water to the environment by plants, it is mostly in the form of vapour and is released into the atmosphere. Hope i helped, Have a nice day
8 0
3 years ago
The perihelion of the comet TOTAS is 1.69 AU and the aphelion is 4.40 AU. Given that its speed at perihelion is 28 km/s, what is
dybincka [34]

Answer:

The speed at the aphelion is 10.75 km/s.

Explanation:

The angular momentum is defined as:

L = mrv (1)

Since there is no torque acting on the system, it can be expressed in the following way:

t = \frac{\Delta L}{\Delta t}

t \Delta t = \Delta L

\Delta L = 0

L_{a} - L_{p} = 0

L_{a} = L_{p}   (2)

Replacing equation 1 in equation 2 it is gotten:

mr_{a}v_{a} =mr_{p}v_{p} (3)

Where m is the mass of the comet, r_{a} is the orbital radius at the aphelion, v_{a} is the speed at the aphelion, r_{p} is the orbital radius at the perihelion and v_{p} is the speed at the perihelion.          

From equation 3 v_{a} will be isolated:    

v_{a} = \frac{mr_{p}v_{p}}{mr_{a}}

v_{a} = \frac{r_{p}v_{p}}{r_{a}}   (4)    

Before replacing all the values in equation 4 it is necessary to express the orbital radius for the perihelion and the aphelion from AU (astronomical units) to meters, and then from meters to kilometers:

r_{p} = 1.69 AU x \frac{1.496x10^{11} m}{1 AU} ⇒ 2.528x10^{11} m

r_{p} = 2.528x10^{11} m x \frac{1km}{1000m} ⇒ 252800000 km

r_{a} = 4.40 AU x \frac{1.496x10^{11} m}{1 AU} ⇒ 6.582x10^{11} m

r_{p} = 6.582x10^{11} m x \frac{1km}{1000m} ⇒ 658200000 km  

     

Then, finally equation 4 can be used:

v_{a} = \frac{(252800000 km)(28 km/s)}{(658200000 km)}

v_{a} = 10.75 km/s

Hence, the speed at the aphelion is 10.75 km/s.

       

8 0
3 years ago
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