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Kaylis [27]
3 years ago
11

Corresponde al conjunto de puntos por donde pasa un cuerpo al moverse

Physics
1 answer:
Anna35 [415]3 years ago
6 0

Answer:

Means?

Explanation:

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7. Un bloque de 700 N se encuentra sobre una viga uniforme de 200 N y 6 m de longitud. El bloque está a una distancia de 1 m del
GrogVix [38]

Answer:

 x =  0.176 m

Explanation:

For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.

Let's use trigonometry to decompose the tension

      sin 60 = T_{y} / T

      T_{y} = T sin 60

       cos 60 = Tₓ / T

      Tₓ = T cos 60

we apply the equation

       ∑ τ = 0

       -W L / 2 - w x + T_{y} L = 0

 

the length of the bar is L = 6m

           -Mg 6/2 - m g x + T sin 60 6 = 0

             x = (6 T sin 60 - 3 M g) / mg

let's calculate

let's use the maximum tension that resists the cable T = 900 N

             x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)

             x = (4676 - 5880) / 6860

             x = - 0.176 m

Therefore the block can be up to 0.176m to keep the system in balance.

5 0
3 years ago
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h
Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

v_{1}=\dfrac{3.0\times10^{3}}{3600}

v_{1}=0.833\ m/s

The force F₁is constant acceleration is also a constant.

F_{1}=ma_{1}

We need to calculate the acceleration

Using formula of acceleration

a_{1}=\dfrac{v}{t}

a_{1}=\dfrac{0.833}{10}

a_{1}=0.083\ m/s^2

Similarly,

F_{2}=ma_{2}

For total force,

F_{3}=F_{2}+F_{1}

ma_{3}=ma_{2}+ma_{1}

The speed of second tugboat is

v=\dfrac{11\times10^{3}}{3600}

v=3.05\ m/s

We need to calculate total acceleration

a_{3}=\dfrac{v}{t}

a_{3}=\dfrac{3.05}{10}

a_{3}=0.305\ m/s^2

We need to calculate the acceleration a₂

0.305=a_{2}+0.083

a_{2}=0.305-0.083

a_{2}=0.222\ m/s^2

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}

\dfrac{F_{1}}{F_{2}}=3.7

F_{1}=3.7F_{2}

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0
3 years ago
The threatened or actual use of physical force to get someone to have sex without consent is called
Nimfa-mama [501]
Wouldn't it be r a p e?
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3 years ago
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How magnitude, distance and velocity affects motion?
levacccp [35]

Answer:

Velocity is the rate of motion in a specific direction. ... My velocity is 30 kilometers per hour that-a-way. Average speed is described as a measure of distance divided by time. Velocity can be constant, or it can change (acceleration).

Explanation:

Velocity is the rate of motion in a specific direction. ... My velocity is 30 kilometers per hour that-a-way. Average speed is described as a measure of distance divided by time. Velocity can be constant, or it can change (acceleration).

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