Answer:
3.5 N
Explanation:
Let the 0-cm end be the moment point. We know that for the system to be balanced, the total moment about this point must be 0. Let's calculate the moment at each point, in order from 0 to 100cm
- Tension of the string attached at the 0cm end is 0 as moment arm is 0
- 2 N weight suspended from the 10 cm position: 2*10 = 20 Ncm clockwise
- 2 N weight suspended from the 50 cm position: 2*50 = 100 Ncm clockwise
- 1 N stick weight at its center of mass, which is 50 cm position, since the stick is uniform: 1*50 = 50 Ncm clockwise
- 3 N weight suspended from the 60 cm position: 3*60 = 180 Ncm clockwise
- Tension T (N) of the string attached at the 100-cm end: T*100 = 100T Ncm counter-clockwise.
Total Clockwise moment = 20 + 100 + 50 + 180 = 350Ncm
Total counter-clockwise moment = 100T
For this to balance, 100 T = 350
so T = 350 / 100 = 3.5 N
The braking force is -400 N
Explanation:
We can solve this problem by using the impulse theorem, which states that the impulse applied on the ferry (the product of force and time) is equal to its change in momentum:

where in this problem, we have:
F is the force applied by the brakes
is the time interval
m = 13,000 kg is the mass of the ferry
u = 2.0 m/s is the initial velocity
v = 0 is the final velocity
And solving for F, we find the force applied by the brakes:

where the negative sign indicates that the direction is backward.
Learn more about impulse:
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Answer:
a) 447.21m
b) -62.99 m/s
c)94.17 m/s
Explanation:
This situation we can divide in 2 parts:
⇒ Vertical : y =-200 m
y =1/2 at²
-200 = 1/2 *(-9.81)*t²
t= 6.388766 s
⇒Horizontal: Vx = Δx/Δt
Δx = 70 * 6.388766 = 447.21 m
b) ⇒ Horizontal
Vx = Δx/Δt ⇒ 70 = 400 /Δt
Δt= 5.7142857 s
⇒ Vertical:
y = v0t + 1/2 at²
-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²
v0= -7 m/s ⇒ it's negative because it goes down.
v= v0 +at
v= -7 + (-9.81) * 5.7142857
v= -62.99 m/s
c) √(70² + 62.99²) = 94.17 m/s
Answer:
number of electrons = 2.18*10^18 e
Explanation:
In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.
Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:
(1)
i1 = 0.40 A
i2 = 0.75 A
you solve the equation i3 from the equation (1):

Next, you take into account that 1A = 1C/s = 6.24*10^18
Then, you have:

The number of electrons that trough the wire 3 is 2.18*10^18 e/s