Answer:
Explanation:
Given
Two masses M and 2 M with velocity v in opposite direction
After collision they stick together
Initial momentum

final momentum

Conserving momentum



i.e. system moves towards the direction of 2M mass
Initial kinetic Energy 

Final Kinetic Energy 
loss of Energy


Wouldn’t it be 1. resistance is high compared to the voltage and with less resistance, higher current
Answer:
167.354 m
Explanation:
We are given;
The mass of the car with bad shock;
m = 1500 kg
The distance at which the car sinks; x =
6 cm = 6 × 10^(−2) m
The total mass of 4 people; m_t = 11 kg
The total speed in the highway; V = 65
mph = 29.058 m/s
The spring's constant can be calculated from the formula;
F = Kx
F is also equal to mg.
Thus;
m_t × g = Kx
K = (m_t × g)/x
K = (11 × 9.81)/(6 × 10^(−2))
K = 1798.5 N/m
Mass of car and four people;m_(c+t) = 1500 + 11 = 1511 kg
Thus, the period cam be calculated from the formula;
T = 2π√((m_c+t)/k)
T = 2π√(1511/1798.5)
T = 5.759 s
the distance between adjacent bumps is calculated from;
Velocity = distance/time
Distance = velocity x time
Distance = 29.058 × 5.759
Distance = 167.354 m
Answer:
i =
, i = 0.92 A
Explanation:
This exercise asks for the electromotive force, which can be calculated with Faraday's law
fem =
where the magnetic flux
Ф = B. A
bold letters indicate vectors. We can write this equation
Ф = B A cos θ
In this case the magnetic field is perpendicular to the page and the normal to the loop of the loop is also parallel to the page, therefore the angle is zero and the cosine is 1
the loop is
A = π r²
we substitute in the first equation
fem = - π r²
we substitute the values
fem = -π r² 1
fem = - π r²
to calculate the current let's use ohm's law
V = i R
R = ρ L / A'
where A 'is the area of the wire and L is the length of the loop
L = 2π r
V = i (ρ 2π r / A ')
I =
In this case
V = fem
I = fem / R
i =
In order to complete the calculation, you need the radius of the loop and / or the wire cutter.
if we assume that the loop has a radius of r = 1 cm = 0.01 m and an area of the wire A'= π 10⁻⁶ m² a radius of the wire 1 mm
i = - 10⁻² π 10⁻⁶ / ( 2 1.7 10-8)
i = 0.92 A