Ask question

Ask question

All categories

3 years ago

7

The rod of the fixed hydraulic cylinder is moving to the left with a constant speed vA = 25 mm/s. Determine the corresponding ve

locity of slider B when sA = 425 mm. The length of the cord is 1050 mm, and the effects of the radius of the small pulley A may be neglected.

1 answer:

Montano1993 [528]3 years ago

5 0

Explanation:

let vertical distance from A to C be $h=250mm$ the constraint equation is:

$l_{ac} +l_{ab} =L$

we want to find $l_{bc}$ distance can be written as $l_{bc} =h_{1}+h_{2}$ which we will find by using Pythagorean theorem h 1 is a constant and can be written as h:

$h_{2} =\sqrt{l_{ab}^2- s_{A} ^2}$

$l_{ac} =\sqrt{s_{A}^2+ h^2 }$

$l_{ab} =L-\sqrt{s_{A} ^2+h^2 }$

taking derivative w.r.t time we get

$h^._{2} =-v_{B =(l_{ab} l_{ab} ^.-s_{A} v_{A} )/h_{2}$

$l_{ab} ^.=(s_{A} v_{A} )/l_{ab} -L$

given $s_{A} =425mm$ which gives us

$l_{ab} =1050-\sqrt{450^2+250^2} =556.923mm\\\\h_{2} =\sqrt{556.923^2-425^2}=359.914mm$

$l_{ab} ^.=(425.25)/(556.923-1050)=-21.548mm\\\\-v_{B} =(-556.923*21.548-425.25)/(359.914)\\\\v_{B} =62.864mm/s$

You might be interested in

ValentinkaMS [17]

Hi there!

We know that:

U (Potential energy) = mgh

We are given the potential energy, so we can rearrange to solve for h (height):

U/mg = h

g = 9.81 m/s²

m = 30 g ⇒ 0.03 kg

0.062/(0.03 · 9.81) = 0.211 m

8 0

3 years ago

Suppose a baseball pitcher throws the ball to his catcher.

amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

$\Delta v = |v-u|$

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

$\Delta v=|v-0|=|v|$

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

$\Delta v =|0-v|=|-v|$

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

$\Delta p = m \Delta v$

where

m is the mass of the object

$\Delta v$ is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

$|\Delta p|=m|\Delta v|$

- In situation 1 (pitcher throwing the ball), the change in momentum is

$\Delta p = m|\Delta v|=m|v|=mv$

- In situation 2 (catcher receiving the ball), the change in momentum is

$\Delta p = m\Delta v = m|-v|=mv$

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

$I=\Delta p$

where

I is the impulse

$\Delta p$ is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

$I=F\Delta t$

where

I is the impulse

F is the force applied

$\Delta t$ is the time of impact

This can be rewritten as

$F=\frac{I}{\Delta t}$

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- $\Delta t$ when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to $\Delta t$ , this means that the force is larger when the ball is catched.

6 0

3 years ago

Bella makes the 6.1m distance to her food bowl in 8.8 seconds what is her average velocity

storchak [24]

Bella’s average velocity is about 0.693 meters per second.

To find the average velocity, you must divide the distance by the change in time, which should look like v=d/t

Here is how you set up the equation-
v=6.1/8.8

Once you divide 6.1 meters by 8.8 seconds, you should get a number that looks like 0.69318182.... however, I just rounded it to 0.693 meters per second. You can round it to whatever you like.

Hope this helped! If you have any questions about what I mentioned in my answer or explanation, feel free to comment on my answer and I’ll try to get back to you!

8 0

3 years ago

A vertical spring with stiffness k originally is at rest with no mass attached. Then, a mass M is attached, and the spring rocks

Alina [70]

Answer:

The position of the spring in terms of g, m & k is $x = \frac{m g}{k}$

Explanation:

Stiffness of the spring = k

Mass = m

When a mass m is attached with the spring then spring stretched. in that case the force exerted on the spring is equal to weight of the mass attached.

⇒ Force exerted on the spring F = k x

⇒ m g = k x

⇒ $x = \frac{m g}{k}$

This is the position of the spring in terms of g, m & k.

8 0

3 years ago

When a planet's orbit takes it closest to the Sun, its called:________.

andriy [413]

Answer:

perihelion

Explanation:

The point at which a planet is closest to the sun is called perihelion. The farthest point is called aphelion

3 0

3 years ago