Answer:
La distancia entre los dos cuerpos es aproximadamente 1.269 milímetros.
Explanation:
Asumamos que ambos cuerpos son partículas, la fuerza de atracción (
), en newtons, entre ambos cuerpos se define mediante la Ley de Newton de la Gravitación Universal, cuya ecuación es:
(1)
Donde:
- Constante de la gravitación universal, en metros cúbicos por kilogramo-segundo cuadrado.
- Masas de los cuerpos, en kilogramos.
- Distancia entre los cuerpos, en metros.
Si sabemos que 
, 
, 
y 
, entonces la distancia entre los dos cuerpos es:
Es decir, la distancia entre los dos cuerpos es aproximadamente 1.269 milímetros.
Answer:
<em>The number 0.0217 has 3 significant digits</em>
Explanation:
<u>Significant Digits
</u>
These are digits that contribute to the significance of the number. Some rules apply to discard the non-significant digits like:
- Leading zeros
- Trailing zeros (with exceptions)
Our number is 0.0217 has two leading zeros before the 2 because they only occupy space to indicate the order of magnitude of the number. Only the 2,1,7 are significant digits, thus
The number 0.0217 has 3 significant digits
3s
Explanation:
Given parameters:
Mass of car = 1000kg
Force applied = 8000N
speed = 24m/s
Unknown:
time taken for the car to stop = ?
Solution:
According to newton's second law of motion; "the force on a body is the product of its mass and acceleration".
Force = mass x acceleration
let us find the acceleration of the car;
a = 
= 
= 8m/s²
since the car is accelerating at a rate of 8m/s², when the brakes are applied, it will start decelerating at the constant rate, - 8m/s²
Applying the appropriate equation of motion;
V = U + at
V is the final velocity
U is the initial velocity
a is the acceleration
t is the time taken
final velocity = 0
0 = U + at
-U = at
-24 = -8t
t = 3s
learn more:
Newton's laws brainly.com/question/11411375
#learnwithBrainly
Answer:
568.16J
Explanation:
V = 1.4 m/s
h = 2.15m
H = 12.4m
M = 57.0kg
Fr = 41.0N
Vb = 6.80m/s
g = 9.8m/s²
K.E at the top + P.E at the top + Work done = K.E at the bottom + frictional force.
½mv² + mgh + W = ½mVb + Fr.H
(½* 57 * 1.4²) + (57 * 9.8 * 2.15) + W = (½ * 57 * 6.8²) + (41 * 12.4)
55.86 + 1202.22 + w = 1317.84 + 508.4
1258.08 + w = 1828.24
W = 1826.24 - 1258.08
W = 568.16J
