Answer: acceleration due to gravity of planet a would be twice that of planet b. Given that the radius are thesame.
Explanation:
Acceleration due to gravity is as a result of the gravitational force of attraction of a planet to its centre.
g = GM/r^2
Where;
g = acceleration due to gravity
G = gravitational constant
M = mass of planet
r = radius of planet
Given that the two planet have the same radius, if the mass of planet a is twice the mass of planet b the the acceleration due to gravity of planet a would be twice that of planet b, because acceleration due to gravity is directly proportional to the mass of the planet.
Answer:
PE = 44.1 J
Explanation:
Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)
We have:
- 1 kilograms = 1000 grams.
We convert it using a rule of 3, replacing, simplifying units and solving:
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Earth's gravity is known to be 9.8 m/s², so we have:
Data:
- m = 0.3 kg
- g = 9.8 m/s²
- h = 15 m
- PE = ?
Use formula of potencial energy:
Replace and solve:
Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.
The potential energy of the volleyball is <u>44.1 Joules.</u>
Greetings.
Answer:
Option (2)
Explanation:
From the figure attached,
Horizontal component,
= 7.22 m
Vertical component,
= 9.58 m
Similarly, Horizontal component of vector C,
= C[Cos(60)]
= 6[Cos(60)]
=
= 3 m
= 5.20 m
Resultant Horizontal component of the vectors A + C,
m
= 4.38 m
Now magnitude of the resultant will be,
From ΔOBC,
=
=
= 6.1 m
Direction of the resultant will be towards vector A.
tan(∠COB) =
=
=
m∠COB =
= 46°
Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.
Option (2) will be the answer.
Answer:
Potential difference and charge will also increase.
Explanation:
Asking that :
What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unchanged?
The charge is directly proportional to area of the plate. That is, increase in area of the plate of a capacitor will lead to the increase in the charges between the plates.
And since charge is also proportional to the magnitude of potential difference between the plates from the definition of capacitance of a capacitor which says that:
Q = CV
Therefore, increase in the area of the plate will also lead to increase in potential difference between the plates.
Therefore, if the plate area were increased while the plate separation remains unchanged, the charge and potential difference between them will also increase.