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 10⁻¹ m / s
Explanation:
a) To determine the position of the particle at a specific time, it is essential to have an estimate of the object's movement and apply Newton's second law to establish 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 need to extract each component of acceleration
X axis
aₓ = 0
no force is present
Axis yay = -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 determined the acceleration for each axis, the position can be derived using kinematics
X axis
The initial speed is vo = 30 m / s, and the starting position is xo = 0 x = vo t + ½ aₓ t² x = 30 0.02 + 0 x = 0.6m Axis yThe acceleration is ay = -1.02 10⁷ m / s², starting from a position of y = 1m
y = I + go t + ½ ay t²
y = 1 + 0 + ½ (-1.02 10⁷) 0.02²y = 1 - 2.04 10³
y = -2039 m j ^ z axisThe acceleration az = 5.1 10² m / s², with a position and initial velocity of zero
z = zo + v₀ t + ½ az t²z = 0 + 0 + ½ 5.1 10² 0.02²
z = 1.02 10⁻¹ m k ^Thus, the position of the particle is
r = (0.6 i - 2039 j ^ + 0.102 k⁾ m
b) x axis
Since there's no acceleration, the speed stays unchangedvₓ = 30.0 m / s
Axis yUsing the equation v = v₀ +
t= 0 + -1.02 10⁷ 0.02
v_{y} = 2.04 10⁵ m / s
z axisv_{z} = vo + az t
v_{z} = 0 + 5.1 10² 0.02v_{z} = 1.02 10⁻¹m / s
To address this question, we will utilize concepts linked to centripetal force, aligning it with the static frictional force acting on the object. Using this relationship, we can derive the velocity and input the known values. The defined values are:
The maximum velocity can be determined using centripetal force,
Should be equal to,
As a result, the highest speed achievable through the arc without slipping is 9.93m/s
Answer:
The required energy remains identical in both scenarios since the specific heat capacity (Cp) does not change with varying pressure.
Explanation:
Given;
initial temperature, t₁ = 50 °C
final temperature, t₂ = 80 °C
Temperature change, ΔT = 80 °C - 50 °C = 30 °C
Pressure for scenario one = 1 atm
Pressure for scenario two = 3 atm
The energy needed in both scenarios is expressed as;
Where;
Cp denotes specific heat capacity, which only varies with temperature and remains unaffected by pressure.
Hence, the energy required remains the same for both scenarios since specific heat capacity (Cp) is pressure-independent.
Answer:
Explanation:
Data provided
initial velocity v₀=20 cm/s at time t=3s
final velocity vf=0 at time t=8 s
Required
Average Acceleration for the interval from 3s to 8s
Solution
Acceleration can be defined as the first derivative of velocity concerning time