Consider the motion of the car before brakes are applied:
v₀ = maximum initial velocity of the car before the brakes are applied
t = reaction time = 0.50 s
x₀ = distance traveled by the car before brakes are applied
since car moves at constant speed before brakes are applied
Using the equation
x₀ = v₀ t
x₀ = v₀ (0.50)
Consider the motion after brakes are applied :
v₀ = initial velocity of the car before the brakes are applied
a = acceleration = - 10 m/s²
v = final velocity of the car after it comes to stop = 0 m/s
x = stopping distance = initial distance - distance traveled before applying the brakes = 38 - x₀ = 38 - v₀ (0.50)
Using the equation
v² = v²₀ + 2 a x
inserting the values
0² = v²₀ + 2 (- 10) (38 - v₀ (0.50))
v²₀ = 20 (38 - v₀ (0.50))
v₀ = 23 m/s
Mrnmrnamrnamrnamrnamrnamrnamrnamrnamrnamrna
4 times as much distance to stop
Answer:
1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Explanation:
According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.
As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :
Q₁ = ∫ ρ dV
Here dV is the volume element of sphere of radius r.
Q₁ = ρ x 4π x ∫ r² dr
The limit of integration is from 0 to r as r is less than R.
Q₁ = (4π x ρ x r³ )/3
But volume charge density, ρ = 
So, 
Applying Gauss law of electrostatics ;
∫ E ds = Q₁/ε₀
Here E is electric field inside the sphere and ds is surface element of sphere of radius r.
Substitute the value of Q₁ in the above equation. Hence,
E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
By Newton's second law, the net force on the object is
∑ <em>F</em> = <em>m</em> <em>a</em>
∑ <em>F</em> = (2.00 kg) (8 <em>i</em> + 6 <em>j</em> ) m/s^2 = (16.0 <em>i</em> + 12.0 <em>j</em> ) N
Let <em>f</em> be the unknown force. Then
∑ <em>F</em> = (30.0 <em>i</em> + 16 <em>j</em> ) N + (-12.0 <em>i</em> + 8.0 <em>j</em> ) N + <em>f</em>
=> <em>f</em> = (-2.0 <em>i</em> - 12.0 <em>j</em> ) N
