C. is the answer because acceleration is the change in velocity in time while velocity is speed with a direction
Answer:
If the driver is wearing a seat belt, the seat belt rather than the windshield applies the unbalanced force that stops the driver's forward motion. The force from the seat belt is applied over a longer time, so the force causes less damage.
Explanation:
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.
Answer:
Momentum, p = 23250 kg m/s
Explanation:
Given that
Mass of a car, m = 1550 kg
Speed pf car, v = 15 m/s
We need to find the momentum of the car. The formula for the momentum of an object is given by :
p = mv
Substituting all the values in the above formula
p = 1550 kg × 15 m/s
p = 23250 kg m/s
So, the momentum of the car is 23250 kg m/s.
Answer:
The actual angle is 30°
Explanation:
<h2>Equation of projectile:</h2><h2>y axis:</h2>
the velocity is Zero when the projectile reach in the maximum altitude:
When the time is vo/g the projectile are in the middle of the range.
<h2>x axis:</h2>
R=Range
**sin(2A)=2sin(A)cos(A)
<h2>The maximum range occurs when A=45°
(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>
Let B the actual angle of projectile
2B=60°
B=30°
