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
The initial curvature and other settings of the mirrors on the Hubble did not produce good images. Space shuttle missions replaced/corrected the mirrors to give clear images.
So to find R (resistance) you do ohm's law E (volts) over I (current) to find R
So this is 12/30 = 0.4 so the resistance is 0.4
Answer:
the intensity will be 4 times that of the earth.
Explanation:
let us assume the following:
intensity of light on earth =J
distance of earth from sun = d
intensity of light on other planet = K
distance of other planet from sun = 
(from the question, the planet is half as far from the sun as earth)
from the question the intensity is inversely proportional to the square of the distance, hence
- intensity on earth : J =
J
= 1 ... equation 1
- intensity on other planet : K =
(the planet is half as far from the sun as earth)
K
= 1 ....equation 2
- equating both equation 1 and 2 we have
J = K
J = K
J = 
K = 4J
intensity of light on other planet (K) = 4 times intensity of light on earth (J)
Answer:
0.025 m
Explanation:
From the question,
Applying Hook's law
F = ke................... Equation 1
Where F = Force, k = spring constant of the scale, e = maximum distance at which the spring will compress.
make e the subject of the equation
e = F/k....................... Equation 2
Given: F = 10 N, e = 395 N/m
Substitute these values into equation 2
e = 10/395
e = 0.025 m
