The direction of the electric field would be south.
qE/m = 115
<span> E = 115*m/q </span>
<span> = 115 * 9.1 * 10^(-31) / 1.67*10^(-19) </span>
<span> = 762.87 * 10^(-12) </span>
<span> = 6.27 x 10^-10 N/C
</span>
Hope this answers the question. Have a nice day. Feel free to ask more questions.
Now the vertical velocity of the ball thrown at an angle 10° is given as
Voy(initial vertical velocity)= 30m/s x sin 10
Voy(initial vertical velocity)= 5.2m/s
Now the ball is decelerating with an acceleration due to gravity equivalent to 9.8m/s^2.
Let Vy be the final velocity and that is equal to zero in this case.
Now
Vy= Voy- tx9.8
Where t is the time at which the vertical velocity becomes 0.
Substituting the values we get
0= 5.2-tx9.8
9.8t=5.2
t=0.53 secs
This depends on the original mass of the object having its mass doubled and the the original distance before the distance was tripled.
Answer:
F=5833.3 N N
Explanation:
Newton's second law applied to the car
F= m*a Formula (1)
F: Force in Newtons (N)
m : mass in kg
a: acceleration ( m/s²)
kinematics car
vf= v₀ + a*t Formula (2)
vf : final velocity (m/s)
v₀ : final velocity (m/s)
a : acceleration ( m/s²)
t : time t
Equivalences
1 km= 1000m
1 h = 3600 s
Data
m= 1000kg
v₀ = 90 km/h = 90*1000/3600 m/s = 25 m/s
vf= 0
t= 6 s
Problem Development
We calculate the acceleration replacing the data in the formula (2) :
0 = 25 + a*6
a= -25/6 = -4.16 m/s² ( The negative sign indicates that the car is braking)
We calculate the force is required to stop the car replacing the data in the formula (1)
-F = 1400 kg*(-4.16 m/s²)
F=5833.3 N