Answer:
The magnitude of the acceleration is 
The direction is
i.e the negative direction of the z-axis
Explanation:
From the question we are that
The mass of the particle 
The charge on the particle is 
The velocity is 
The the magnetic field is 
The charge experienced a force which is mathematically represented as

Substituting value



Note :

Now force is also mathematically represented as

Making a the subject

Substituting values



Answer:
<h2>1139.5 J</h2>
Explanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
work done = 265 × 4.3 = 1139.5
We have the final answer as
<h3>1139.5 J</h3>
Hope this helps you
Answer:
a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J
Explanation:
a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s
The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s
So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s
b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.
p₂ = (1.3 + 39.0)v = 40.3v
From the principle of conservation of momentum,
p₁ = p₂
37.7 kgm/s = 40.3v
v = 37.7/40.3 = 0.94 m/s
So the final velocity of the two-block system is 0.94 m/s
c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²
So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J
Answer:
The speed of Susan is 2.37 m/s
Explanation:
To visualize better this problem, we need to draw a free body diagram.
the work is defined as:

here we have the work done by Paul and the friction force, so:


Now the change of energy is:

By v = u - at
<span>=>8 = 12 - a x 0.25 </span>
<span>=>a = 4/0.25 km/hr/sec </span>
<span>=>a = 16km/hr/sec
I hope this helped!</span>