To answer the following questions for this specific problem:
a. 11.48 secs
b. Vp = a*t*3.6 =
3*11.48*3.6 = 124.0 km/h
<span>c. 9.1 secs. </span>
I am hoping that this answer has satisfied your query about
and it will be able to help you.
Answer:
1. Elastic collision
2. Inelastic collision
Explanation:
Elastic collision: collision is said to be elastic if total kinetic energy is not conserved and if there is a rebound after collision
the collision is described by the equation bellow
Inelastic collision: this type of collision occurs when the total kinetic energy of a body is conserved or when the bodies sticks together and move with a common velocity
the collision is described by the equation bellow
Wow ! I understand your shock. I shook and vibrated a little
when I looked at this one too.
The reason for our shock is all the extra junk in the question,
put there just to shock and distract us.
"Neutron star", "5.5 solar masses", "condensed burned-out star".
That's all very picturesque, and it excites cosmic fantasies in
out brains when we read it, but it's just malicious decoration.
It only gets in the way, and doesn't help a bit.
The real question is:
What is the acceleration of gravity 2000 m from
the center of a mass of 1.1 x 10³¹ kg ?
Acceleration of gravity is
G · M / R²
= (6.67 x 10⁻¹¹ N·m²/kg²) · (1.1 x 10³¹ kg) / (2000 m)²
= (6.67 x 10⁻¹¹ · 1.1 x 10³¹ / 4 x 10⁶) (N) · m² · kg / kg² · m²
= 1.83 x 10¹⁴ (kg · m / s²) · m² · kg / kg² · m²
= 1.83 x 10¹⁴ m / s²
That's about 1.87 x 10¹³ times the acceleration of gravity on
Earth's surface.
In other words, if I were standing on the surface of that neutron star,
I would weigh 1.82 x 10¹² tons, give or take.
Answer:
the center of the universe is in your mom's stomach. why do I keep seeing people wanting to look for a date. the f. u. c. k.?
Answer:
The net Electric field at the mid point is 289.19 N/C
Given:
Q = + 71 nC = 
Q' = + 42 nC = 
Separation distance, d = 1.9 m
Solution:
To find the magnitude of electric field at the mid point,
Electric field at the mid-point due to charge Q is given by:
Now,
Electric field at the mid-point due to charge Q' is given by:
Now,
The net Electric field is given by:
