To solve this problem it is necessary to apply the kinematic equations of motion.
By definition we know that the position of a body is given by
Where
Initial position
Initial velocity
a = Acceleration
t= time
And the velocity can be expressed as,
Where,
For our case we have that there is neither initial position nor initial velocity, then
With our values we have 
, rearranging to find a,
Therefore the final velocity would be
Therefore the final velocity is 81.14m/s
Answer:
(a) m = 33.3 kg
(b) d = 150 m
(c) vf = 30 m/s
Explanation:
Newton's second law to the block:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
Data
F= 100 N
a= 3.0 m/s²
(a) Calculating of the mass of the block:
We replace dta in the formula (1)
F = m*a
100 = m*3
m = 100 / 3
m = 33.3 kg
Kinematic analysis
Because the block moves with uniformly accelerated movement we apply the following formulas:
d= v₀t+ (1/2)*a*t² Formula (2)
vf= v₀+a*t Formula (3)
Where:
d:displacement in meters (m)
t : time interval in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
a= 3.0 m/s²
v₀= 0
t = 10 s
(b) Distance the block will travel if the force is applied for 10 s
We replace dta in the formula (2):
d= v₀t+ (1/2)*a*t²
d = 0+ (1/2)*(3)*(10)²
d =150 m
(c) Calculate the speed of the block after the force has been applied for 10 s
We replace dta in the formula (3):
vf= v₀+a*t
vf= 0+(3*(10)
vf= 30 m/s
Answer:
R = 98304.75 m = 98.3 km
Explanation:
The density of an object is given as the ratio between the mass of that object and the volume occupied by that object.
Density = Mass/Volume
Now, it is given that the density of Earth has become:
Density = 1 x 10⁹ kg/m³
Mass = Mass of Earth (Constant) = 5.97 x 10²⁴ kg
Volume = 4/3πR³ (Volume of Sphere)
R = Radius of Earth = ?
Therefore,
1 x 10⁹ kg/m³ = (5.97 x 10²⁴ kg)/[4/3πR³]
4/3πR³ = (5.97 x 10²⁴ kg)/(1 x 10⁹ kg/m³)
R³ = (3/4)(5.97 x 10¹⁵ m³)/π
R = ∛[0.95 x 10¹⁵ m³]
<u>R = 98304.75 m = 98.3 km</u>
