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aleksandr82 [10.1K]
3 years ago
6

One boat tows another boat by means of a tow line, which is under a constant tension of 465 N. The boats move at a constant spee

d of 4.60 m/s. How much work is done by the tension in 1.10 min?
Physics
1 answer:
Ludmilka [50]3 years ago
6 0

Answer:

Work done, W = 141174 Joules

Explanation:

It is given that,

Constant tension acting on the boat, T = F = 465 N

Speed of the boat, v = 4.6 m/s

Time, t = 1.1 min = 66 seconds

Let W is the work done by the tension. It is equal to the product of force and displacement. It is given by :

W=F\times d

Since, d=vt

W=F\times v\times t

W=465\ N\times 4.6\ m/s\times 66\ s

W = 141174 Joules

So, the work is done by the tension is 141174 Joules. Hence, this is the required solution.

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3 years ago
If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. Wh
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Answer:

 v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

            v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

           a = v₀² / 2y

now they tell us that the initial velocity is half

          v’² = v₀’² - 2 a y’

          v₀ ’= v₀ / 2

at the point where turn v = 0              

          0 = v₀² /4  - 2 a y '

          v₀² /4 = 2 (v₀² / 2y)  y’

          y = 4 y'

          y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

         v² = v₀² -2a y’

         v² = 0 - 2 (v₀² / 2y) y / 4

         v² = -v₀² / 4

         v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.

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Read 2 more answers
A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanba
Phantasy [73]

Answer:

31.42383 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

\mu = Coefficient of kinetic friction = 0.48

s = Displacement = 0.935 m

m_1 = Mass of bean bag = 0.354 kg

m_2 = Mass of empty crate = 3.77 kg

v_1 = Speed of the bean bag

v_2 = Speed of the crate

Acceleration

a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g

a=--9.81\times 0.48=4.7088\ m/s^2

From equation of motion

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s

In this system the momentum is conserved

m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s

The speed of the bean bag is 31.42383 m/s

8 0
3 years ago
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