Answer:
The magnitude of the force required to bring the mass to rest is 15 N.
Explanation:
Given;
mass, m = 3 .00 kg
initial speed of the mass, u = 25 m/s
distance traveled by the mass, d = 62.5 m
The acceleration of the mass is given as;
v² = u² + 2ad
at the maximum distance of 62.5 m, the final velocity of the mass = 0
0 = u² + 2ad
-2ad = u²
-a = u²/2d
-a = (25)² / (2 x 62.5)
-a = 5
a = -5 m/s²
the magnitude of the acceleration = 5 m/s²
Apply Newton's second law of motion;
F = ma
F = 3 x 5
F = 15 N
Therefore, the magnitude of the force required to bring the mass to rest is 15 N.
Answer:
R = m⁴/kg . s
Explanation:
In this case, the best way to solve this is working with the units in the expression.
The units of velocity (V) are m/s
The units of density (d) are kg/m³
And R is a constant
If the expression is:
V = R * d
Replacing the units and solving for R we have
m/s = kg/m³ * R
m * m³ / s = kg * R
R = m * m³ / kg . s
<h2>
R = m⁴ / kg . s</h2>
This should be the units of R
Hope this helps
Speed
= (distance covered) / (time to cover the distance)
= (25 m) / (5.0 sec) = 5.0 m/s .
Gravitational force = G · (mass₁) · (mass₂) / (distance)
(distance²) = G · (mass₁) · (mass₂) / (Gravitational force)
G = 6.67 x 10⁻¹¹ n-m² / kg² (the "gravitational constant")
Distance² = (6.67 x 10⁻¹¹ n-m² / kg²) (28,500 kg) (2.2 x 10⁸ kg) / (39 N)
Distance² = (6.67 · 28,500 · 2.2 x 10⁻³ N-m²) / (39N)
Distance² = (418.209 N-m²) / (39N)
Distance² = 10.72 m²
<em>Distance = 3.275 meters</em>
An absurd scenario, but that's by golly what the math says with the numbers provided. I guess it's a teeny tiny planet orbiting 3.275 meters outside a teeny tiny black hole.
