Answer:
The divergence on the sensor shows the magnitude of the charges
Explanation:
This will increase as length increases since it is said to be proportional to the length. note that test charge is always positive and charge on the grid is positive as indicated (1 nC)
#1
for the block of mass 5 kg normal force is given as


friction force is given as


Net force is given as


now we know that



#2
Normal force is given as



now we know that


as object moves with constant velocity

now for coefficient of friction we can use



#3
net force upwards is given as

mass is given as

now as per newton's law we can say



#4
As we know that when block is sliding on rough surface
part a)
net force = applied force - frictional force




part b)
for coefficient of friction we can use


here normal force is given as

now we have

#5
if an object is initially at rest and moves 20 m in 5 s
so we can use kinematics to find out the acceleration



now net force is given as


#6
an object travelling with speed 25 m/s comes to stop in 1.5 s
so here acceleration of object is given as


now the force is gievn as


Answer:
a= 17.877 m/s² : Magnitude of the acceleration of the flea
β = 88.21° : Direction of the acceleration of the flea
Explanation:
Conceptual analysis
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Problem development
Look at the flea free body diagram in the attached graphic
The acceleration is presented in the direction of the resultant force (R) applied over the flea .


R= 10.905*10⁻⁶ N
We apply the formula (1) to calculate the magnitude of the acceleration of the flea
∑F = m*a m = 6.1 * 10⁻⁷ kg
R = m*a
a= R/m
a= (10.905*10⁻⁶) / (6.1 * 10⁻⁷ )
a= 17.877 m/s²
β: Direction and magnitude of the acceleration of the flea


β = 88.21°
Answer:
8.33*10^-16 Watt
Explanation:
Given that
Length of the rod, l = 2 m,
Area of the rod, A = 2 x 2 mm² = 4*10^-6 m²
resistivity of the rod, p = 6*10^-8 ohm metre,
Potential difference of the rod, V = 0.5 V
Let R be the resistance of the rod, then
R = p * l / A
R = (6*10^-8 * 2) / (4*10^-6)
R = 3*10^14 ohm
Heat generated per second = V² / R Heat = (0.5)² / (3*10^14)
Heat = 0.25 / 3*10^14
Heat = 8.33*10^-16 Watt
Therefore, the rate at which heat is generated is 8.33*10^-16 Watt