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:
The temperature rises by 0.52K
Explanation:
Detailed explanation and calculation is shown in the image below
For t1:
t1 = square root of 2h1 / g = square root of 2 * 0.5 / 9.8 = 0.319 sec
For t2:
t2 = sqaure root of 2h2 / g = square root of 2 * 1.0 / 9.8 = 0.451 sec
Wherein:
t = time(s) for the vertical movement
h= height
g = gravity (using the standard 9.8 m/sec measurement)
d1 = 1*0.319 = 0.319 m
d2 = 0.5 * 0.451 = 0.225 m
Where:
d = hor. distance
ratio = d1:d2
= 0.319 : 0.225
=3.19 : 2.25
The answer is 3.19 : 2.25
Either no forces or a balanced group of forces
(not a group of "balanced forces"; there's no such thing)
