Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
Explanation:
Large electrical shifting magnets have concentrated retaining strength to lift dense, ferric objects and a deep-reaching magnetization. An immensely useful materials management technique is these electromagnetic rises.
Answer:
The particle’s velocity is -16.9 m/s.
Explanation:
Given that,
Initial velocity of particle in negative x direction= 4.91 m/s
Time = 12.9 s
Final velocity of particle in positive x direction= 7.12 m/s
Before 12.4 sec,
Velocity of particle in negative x direction= 5.32 m/s
We need to calculate the acceleration
Using equation of motion
Where, v = final velocity
u = initial velocity
t = time
Put the value into the equation
We need to calculate the initial speed of the particle
Using equation of motion again
Put the value into the formula
Hence, The particle’s velocity is -16.9 m/s.
When a river flows into an ocean, it slows down and deposits materials in its delta
(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.
(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.
<h3>
Magnitude of electric field </h3>
The magnitude of electric field is given by the following equation.
F = qE
But F = mg
mg = qE
E = mg/q
where;
- E is the electric field
- m is mass of the particle
- g is acceleration due to gravity
- q is charge of the particle
<h3>For an electron</h3>
E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)
E = 5.57 x 10⁻¹¹ N/C
<h3>For proton</h3>
E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)
E = 1.02 x 10⁻⁷ N/C
Thus, the required vertical electric field is greater when the charge is proton.
Learn more about electric field here: brainly.com/question/14372859
#SPJ1
