The moment of a couple is Force × perpendicular distance from the arm of the line of action
so the arm of the couple= moment of couple/force=8.5/34=0.25m
the arm is 0.25m
Answer:
9) a = 25 [m/s^2], t = 4 [s]
10) a = 0.0875 [m/s^2], t = 34.3 [s]
11) t = 32 [s]
Explanation:
To solve this problem we must use kinematics equations. In this way we have:
9)
a)

where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = acceleration [m/s^2]
x = distance = 200 [m]
Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.
0 = (100)^2 - (2*a*200)
a = 25 [m/s^2]
b)
Now using the following equation:

0 = 100 - (25*t)
t = 4 [s]
10)
a)
To solve this problem we must use kinematics equations. In this way we have:

Note: The positive sign of the equation means that the car increases his speed.
5^2 = 2^2 + 2*a*(125 - 5)
25 - 4 = 2*a* (120)
a = 0.0875 [m/s^2]
b)
Now using the following equation:

5 = 2 + 0.0875*t
3 = 0.0875*t
t = 34.3 [s]
11)
To solve this problem we must use kinematics equations. In this way we have:

10^2 = 2^2 + 2*a*(200 - 10)
100 - 4 = 2*a* (190)
a = 0.25 [m/s^2]
Now using the following equation:

10 = 2 + 0.25*t
8 = 0.25*t
t = 32 [s]
<h2>
Time taken is 0.632 seconds</h2>
Explanation:
Impulse momentum theorem is change in momentum is impulse.
Change in momentum = Impulse
Final momentum - Initial momentum = Impulse
Mass x Final velocity - Mass x Initial Velocity = Force x Time
Mass x Final velocity - Mass x Initial Velocity =Mass x Acceleration x Time
Final velocity - Initial Velocity = Acceleration x Time
Final velocity = 9.9 m/s
Initial Velocity = 3.7 m/s
Acceleration = 9.81 m/s²
Substituting
9.9 - 3.7 = 9.81 x Time
Time = 0.632 seconds
Time taken is 0.632 seconds
Answer:
The magnitude of the electric field and direction of electric field are
and 75.36°.
Explanation:
Given that,
First charge 
Second charge
Distance between two corners r= 50 cm
We need to calculate the electric field due to other charges at one corner
For E₁
Using formula of electric field

Put the value into the formula


For E₂,
Using formula of electric field

Put the value into the formula


We need to calculate the horizontal electric field



We need to calculate the vertical electric field



We need to calculate the net electric field

Put the value into the formula



We need to calculate the direction of electric field
Using formula of direction



Hence, The magnitude of the electric field and direction of electric field are
and 75.36°.