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:
Explanation:
Given that,
Assume number of turn is
N= 1
Radius of coil is.
r = 5cm = 0.05m
Then, Area of the surface is given as
A = πr² = π × 0.05²
A = 7.85 × 10^-3 m²
Resistance of
R = 0.20 Ω
The magnetic field is a function of time
B = 0.50exp(-20t) T
Magnitude of induce current at
t = 2s
We need to find the induced emf
This induced voltage, ε can be quantified by:
ε = −NdΦ/dt
Φ = BAcosθ, but θ = 90°, they are perpendicular
So, Φ = BA
ε = −NdΦ/dt = −N d(BA) / dt
A is a constant
ε = −NA dB/dt
Then, B = 0.50exp(-20t)
So, dB/dt = 0.5 × -20 exp(-20t)
dB/dt = -10exp(-20t)
So,
ε = −NA dB/dt
ε = −NA × -10exp(-20t)
ε = 10 × NA exp(-20t)
Now from ohms law, ε = iR
So, I = ε / R
I = 10 × NA exp(-20t) / R
Substituting the values given
I = 10×1× 7.85 ×10^-3×exp(-20×2)/0.2
I = 1.67 × 10^-18 A
All of the orbitals in a given subshell have the same value of the "<span>magnetic and principal" quantum number
Hope this helps!</span>
Theories have both an explanatory an a predictive function. True
