Answer:
Explanation:
= Mass of the Earth = 5.972 × 10²⁴ kg
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
r = Radius of Earth = 6371000 m
m = Mass of person
The force on the person will balance the gravitational force
The acceleration that the Earth will feel is
Recall the formula,
∆<em>θ</em> = <em>ω</em>₀ <em>t</em> + 1/2 <em>α</em> <em>t</em> ²
where ∆<em>θ</em> = angular displacement, <em>ω</em>₀ = initial angular speed (which is zero because the disk starts at rest), <em>α</em> = angular acceleration, and <em>t</em> = time. Solve for the acceleration with the given information:
50 rad = 1/2 <em>α</em> (5 s)²
<em>α</em> = (100 rad) / (25 s²)
<em>α</em> = 4 rad/s²
Now find the angular speed <em>ω </em>after 3 s using the formula,
<em>ω</em> = <em>ω</em>₀ + <em>α</em> <em>t</em>
<em>ω</em> = (4 rad/s²) (3 s)
<em>ω</em> = 12 rad/s
Answer:
The right solution is "0.50 m/s²". A further explanation is provided below.
Explanation:
The given values are:
Mass,
m = 50 kg
Speed,
v = 10 m/s
Rolling friction acting backward (south),
f = 10 N
Air resistance acting backward (south),
= 15 N
The total force acting will be:
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
Now,
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
The horizontal acceleration will be "0.50 m/s²" because the (-)ve sign indicates it in south direction.
Answer:
F=(-4.8*10^22,0,0) N
Explanation:
<u>Given :</u>
We are given the magnitude of the momentum of the planet and let us call this momentum (p_now) and it is given by p_now = 2.60 × 10^29 kg·m/s. Also, we are given the force exerted on the planet F = 8.5 × 10^22 N. and the angle between the planet and the star is Ф = 138°
Solution :
We are asked to find the parallel component of the force F The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force F to the momentum and a perpendicular force F' to the momentum.
The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next
F=|F|cosФp (1)
Where the parallel force F is in the opposite direction of p as the angle between them is larger than 90°. Now we can plug our values for 0 and I F I into equation (1) to get the parallel force to the planet
F=|F|cosФp
=-4.8*10^22 N*p
<em>As this force is in one direction, we could get its vector as next </em>
F=(-4.8*10^22,0,0) N
F=(0,-4.8*10^22,0) N
F=(0,0-4.8*10^22) N
The cosine of 138°, the angle between F and p is, is a negative number, so F is opposite to p. The magnitude of the planet's momentum will decrease.
Explanation:
In brief, electrons are negative charges and protons are positive charges. An electron is considered the smallest quantity of negative charge and a proton the smallest quantity of positive charge.
Two negative charges repel. Also, two positive charges repel. A positive charge and a negative charge attract each other (all experimentally verified.)
Point Charge: An accumulation of electric charges at a point (a tiny volume in space) is called a point charge.
Note: When an atom loses an electron, the separated electron forms a negative charge, but the remaining that contains one less electron or consequently one more proton becomes a positive charge. A positive charge is not necessarily a single proton. In most cases, a positive charge is an atom that has lost one or more electron(s).
