Point charge A with a charge of +4.00 μC is located at the origin. Point charge B with a charge of +7.00 μC is located on the x
2 answers:
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The result is -15.625 m/s².
Acceleration signifies the alteration of velocity over a specified duration. It can be calculated with this formula:
Where:
vf = final velocity
vi = initial velocity
t = time
Let’s examine the information provided in your query:
Initially, the vehicle was traveling at 25 m/s before coming to a halt. Thus, it was in motion and subsequently ceased moving, indicating that the final velocity is 0 m/s.
However, we notice that the problem does not provide a time value. We need to determine the time taken from when it was in motion to when it reached the traffic light located 20 m away.
The time can be calculated using the kinematics equation:
We derive the equation by substituting the known values first.
The duration from when it was in motion until it stopped is 1.6s. Now we can utilize this in our acceleration calculation.
It is important to note that the acceleration is negative, indicating the vehicle slowed down.
Answer:
Explanation:
a) La fuerza neta que actúa sobre la caja en la dirección vertical es:
Fnet=Fg−f−Fp *sin45 °
aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.
Fnet=ma
ma=Fg−f−Fp *sin45 °
a=
=0.24 m/s²
Vf =Vi +at
=0.48+0.24*2
Vf=2.98 m/s
b)
Fnet=Fg−f−Fp *sin45 °
=Fg−0.516Fp−Fp *sin45 °
=30-1.273Fp
Fnet=0 (Ya que la velocidad es constante)
Fp=30/1.273
=23.56 N
Answer:
Jari
Explanation:
To determine who is traveling faster, we need to evaluate their gradients. A steeper slope indicates a higher speed.
For Jari's path, starting point is (0, 0) and (6, 7) is another point.
The gradient is the difference in y divided by the difference in x:
Change in y=7-0=7
Change in x=6-0=6
Thus, the slope equals 7/6.
For Jade, her first point is (0, 10) and another is (6, 16).
Change in y=16-10=6
Change in x=6-0=6
Thus, the slope equals 6/6=1.
It's evident that 7/6 exceeds 6/6 or 1, proving Jari is quicker than Jade.
