Answer:
(a) The speed of the target proton after the collision is:
, and (b) the speed of the projectile proton after the collision is: 
.
Explanation:
We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle:
, and y axle:
. Now replacing the value given as: 
, 
for the projectile proton and according to the problem 
are perpendicular so 
, and assuming that 
, we get for x axle:
and y axle: 
, then solving for 
, we get:
and replacing at the first equation we get:
, now solving for 
, we can find the speed of the projectile proton after the collision as: and 
, that is the speed of the target proton after the collision.
Answer:
Explanation:
Given that,
Capacitance 1, 
Capacitance 2, 
Capacitance 3, 
C₁ and C₂ are connected in series. Their equivalent is given by :
Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :
So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.
From the above reaction the temperature of the surroundings will increase.
Complete question is;
A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 m/s² rightward. After 3s, what will be the velocity of the rocket ship?
Answer:
v = 12 m/s
Explanation:
We are given;
Initial velocity; u = 0 m/s (because ship starts from rest)
Acceleration; a = 4 m/s²
Time; t = 3 s
To find velocity after 3 s, we will use Newton's first equation of motion;
v = u + at
v = 0 + (4 × 3)
v = 12 m/s
m = mass of the box
N = normal force on the box
f = kinetic frictional force on the box
a = acceleration of the box
μ = coefficient of kinetic friction
perpendicular to incline , force equation is given as
N = mg Cos30 eq-1
kinetic frictional force is given as
f = μ N
using eq-1
f = μ mg Cos30
parallel to incline , force equation is given as
mg Sin30 - f = ma
mg Sin30 - μ mg Cos30 = ma
"m" cancel out
a = g Sin30 - μ g Cos30
inserting the values
1.20 = (9.8) Sin30 - (9.8) Cos30 μ
μ = 0.44
