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Trava [24]
3 years ago
10

Mechanical advantage is always less than validity ratio​

Physics
1 answer:
nata0808 [166]3 years ago
7 0

That statement is true.

The question wasn't that specific, so I couldn't give you an answer...

You might be interested in
Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to thenext handhold. A 9.3kggibbon has
Mumz [18]

Upward force provided by the branch is 260 N

<u>Explanation:</u>

Given -

Mass of Gibbon, m = 9.3 kg

Length of the branch, r = 0.6 m

Speed of the movement, v = 3.3 m/s

Upward force, T = ?

The tension force in the rod must be greater than the weight at the bottom of the swing in order to provide an upward centripetal acceleration.

Therefore,

F net = T - mg

F net = ma = mv²/r

Thus,

T = mv²/r + mg

T = m ( v²/r + g)

T = 9.3 [ ( 3.3)² / 0.6 + 9.8]

T = 259.9 N ≈ 260 N

Therefore, upward force provided by the branch is 260 N

5 0
3 years ago
What problems do you think might arise if the chromosomes did not align during metaphase?
arlik [135]
What problems do you think might arise if the chromosomes did not align during metaphase?
3 0
3 years ago
A 6,000 kg train car is moving to the right at 10 m/s and connects to a 4,000-kg train car that wasn't moving. What is the veloc
vredina [299]

1) The final velocity of the two trains is 6 m/s to the right

2) It is an inelastic collision

Explanation:

1)

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two trains must be conserved before and after the collision.

Therefore we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 = 6,000 kg is the mass of the first train

u_1 = 10 m/s is the initial velocity of the first train

m_2 = 4,000 kg is the mass of the second train

u_2 = 0 is the initial velocity of the second train  (initially at rest)

v is the final combined velocity of the two trains

Solving the equation for v, we find the final velocity of the two trains:

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(6000)(10)+0}{6000+4000}=6 m/s

2)

There are two types of collisions:

  • Elastic collision: in an elastic collision, both the total momentum and the total kinetic energy of the system are conserved
  • Inelastic collision: in an inelastic collision, only the total momentum is conserved, while the total kinetic energy is not (part of it is converted into thermal energy due to internal frictions)

To verify what type of collision is this, we can compare the total kinetic energy before and after the collision:

Before:

K=\frac{1}{2}m_1 u_1^2 = \frac{1}{2}(6000)(10)^2=300,000 J

After:

K=\frac{1}{2}(m_1 +m_2)v^2 = \frac{1}{2}(6000+4000)(6)^2=180,000 J

As we can see, the kinetic energy is not conserved, so this is an inelastic collision.

Learn more about momentum here:

brainly.com/question/7973509  

brainly.com/question/6573742  

brainly.com/question/2370982  

brainly.com/question/9484203  

#LearnwithBrainly

3 0
3 years ago
Considerando que você comece a caminhar em velocidade constante, inicialmente a 350 m de um ponto referencial escolhido. Você ca
ExtremeBDS [4]

Answer:

a. S(t)=350−1t

Explanation:

To determine the equation of motion you take into account the general form of motion with constant velocity:

S(t)=S_o+vt    ( 1 )

So is the initial position from a specific reference frame. In this case is 350 m.

v is the speed of the motion, in this case is 1m/s. However, the motion is forward the zero point of the reference frame, hence, the speed is - 1m/s.

You replace the values of So and v in the equation ( 1 ) and you obtain:

S(t)=350-(1m/s)t

Hence, the answer is:

a. S(t)=350−1t

- - - - - - - - - - - - - - - - - - - - -

Para determinar a equação do movimento, você leva em consideração a forma geral do movimento com velocidade constante:

             (1)

Assim é a posição inicial de um quadro de referência específico. Neste caso, é de 350 m.

v é a velocidade do movimento, neste caso é de 1m / s. No entanto, o movimento é avançar o ponto zero do quadro de referência, portanto, a velocidade é de - 1m / s.

Você substitui os valores de So ev na equação (1) e obtém:

Portanto, a resposta é:

uma. S (t) = 350-1t, movimento retrógrado

4 0
3 years ago
A woman holds a book by placing it between her hands such that she presses at right angles to the front and back covers. The boo
In-s [12.5K]

Answer:

Explanation:

In order to solve this problem we need to make a free body diagram of the book and the forces that interact on it. In the picture below you can see the free body diagram with these forces.

The person holding the book is compressing it with his hands, thus exerting a couple of forces of equal magnitude and opposite direction with value F.

Now the key to solving this problem is to analyze the equilibrium condition (Newton's third law) on the x & y axes.

To find the weight of the book we simply multiply the mass of the book by gravity.

W = m*g

W = 1.3[kg] * 9.81[m/s^2]

W = 12.75 [N]

7 0
3 years ago
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