Suppose a particle is accelerated through space by a constant 10-N force. Suddenly the particle encounters a second force of 10-
1 answer:
Answer:
The particle will continue moving at constant velocity
Explanation:
When the particle encounters the second force of 10 N, the net force acting on the particle becomes zero, because the two forces are equal in magnitude but opposite in direction:
For Newton's second Law, the acceleration of the particle is proportional to the net force acting on it:
Therefore, since , the acceleration is zero (a=0) and so the particle will keep a constant velocity.
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Answer:
a) r = (0.6 i- 2039 j ^ + 0.102 k⁾ m and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s c) v_{z} = 1.02 10⁻¹m / s
Explanation:
a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration
Fe + Fm = m a
a = (Fe + Fm) / m
the electric force is
Fe = q E k ^
Fe = 4.25 10-4 60 k ^
Fe = 2.55 10-2 k ^
the magnetic force is
Fm = q v x B
Fm = 4.25 10⁻⁴
fm = 4.25 10⁻⁴ (-j ^ 30 4)
fm 0 = ^ -5,10 10⁻² j
We look for every component of acceleration
X axis
aₓ = 0
there is no force
Axis y
ay = -5.10 10²/5 10⁻⁵ j ^
ay = -1.02 107 j ^ / s2
z axis
az = 2.55 10⁻² / 5 10⁻⁵ k ^
az = 5.1 10² k ^ m / s²
Having the acceleration in each axis we can encocoar the position using kinematics
X axis
the initial velocity is vo = 30 m / s and an initial position xo = 0
x = vo t + ½ aₓ t₂2
x = 30 0.02 + 0
x = 0.6m
Axis y
acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m
y = I + go t + ½ ay t²
y = 1 + 0 + ½ (-1.02 10⁷) 0.02²
y = 1 - 2.04 10³
y = -2039 m j ^
z axis
acceleration is aza = 5.1 10² m / s², the position and initial speed are zero
z = zo + v₀ t + ½ az t²
z = 0 + 0 + ½ 5.1 10² 0.02²
z = 1.02 10⁻¹ m k ^
therefore the position of the bodies is
r = (0.6 i- 2039 j ^ + 0.102 k⁾ m
b) x axis
since there is no acceleration the speed remains constant
vₓ = 30.0 m / s
Axis y
let's use the equation v = v₀ + t
= 0 + -1.02 10⁷ 0.02
v_{y} = 2.04 10⁵ m / s
z axis
v_{z} = vo + az t
v_{z} = 0 + 5.1 10² 0.02
v_{z} = 1.02 10⁻¹m / s
Answer:
The final velocity of the bullet is 9 m/s.
Explanation:
We have,
Mass of a bullet is, m = 0.05 kg
Mass of wooden block is, M = 5 kg
Initial speed of bullet, v = 909 m/s
The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,
So, the final velocity of the bullet is 9 m/s.
Answer:
Explanation:
Aceleration is a change on the velocity of the object in a given time.
For this case: the initial velocity is
and the final velocity is :
so, the change in velocity is:
and the change in time , according to the problem:
So, the aceleration is:
You will have to use this formula:
Final Velocity (V) = 4m/s
Initial Velocity (Vo) = 8m/s
Acceleration (a) = ? m/s^2
Time (t) = 2 secs
Then:
-> 4 = 8 + a x 2
-> 4 - 8 = 2a
-> -4 = 2a
-> a = -4/2
-> a = -2 m/s^2
Ps: It's value is negative because the she was in retrograde motion.
Answer: Her acceleration is -2 m/s^2.
Final Velocity (V) = 4m/s
Initial Velocity (Vo) = 8m/s
Acceleration (a) = ? m/s^2
Time (t) = 2 secs
Then:
-> 4 = 8 + a x 2
-> 4 - 8 = 2a
-> -4 = 2a
-> a = -4/2
-> a = -2 m/s^2
Ps: It's value is negative because the she was in retrograde motion.
Answer: Her acceleration is -2 m/s^2.