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4 years ago

Why does the weight of water pulls the central part of the surface down?

Physics

1 answer:

Brut [27]4 years ago

5 0

Answer:

Buoyancy or Upthrust

Explanation:

This is an upward force exerted by a fluid that opposes the weight of a partially or fully immersed object.

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At the beginning of a basketball game, a referee tosses the ball straight up with a speed of 4.6 m/s. a player cannot touch the

-Dominant- [34]

At the beginning of a basketball game, a referee tosses the ball straight

up with a speed of 4.6 m/s. a player cannot touch the ball until after it reaches its maximum height and begins to fall down. what is the minimum time that a player must wait before touching the ball? s

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

kinemaic equation

v=u-at

0=4.6-9.81xt

t=4.6/9.81 ... about half a second

4 0

3 years ago

If the S-P interval of seismic waves recorded at a seismometer 8 mintues. Approximately how far is the earthquake epicenter from

Yuri [45]

8/0.00146. Hope this helps.

3 0

4 years ago

Read 2 more answers

What is the potential engery of a 150 kg diver standing on a diving board that is 10m high?

krek1111 [17]

The Answer Is Going To Be14,700

5 0

3 years ago

Define torque qualitatively and quantitatively

andrezito [222]

The rotational effect of a force is called torque.it is the cause of rotation or angular deceleration

τ=rXF
where
τ = F r sin @
hope it helps

4 0

3 years ago

In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00{\rm m/s} when it first contact

OLga [1]

Answer:

v=3.73m/s, $a=3.35m/s^{2}$

Explanation:

In order to solve this problem, we must first draw a diagram that will represent the situation (See attached picture).

So there are two ways in which we can solve this proble. One by using integrls and the other by using energy analysis. I'll do it by analyzing the energy of the system.

So first, we ned to know what the constant of the spring is, so we can find it by analyzing the energy on points A and C. So we get the following:

$K_{A}+U_{Ag}=K_{C}+U_{Cs}+U_{Cg}+W_{Cf}$

where K represents kinetic energy, U represents potential energy and W represents work.

We know that the kinetic energy in C will be zero because its speed is supposed to be zero. We also know the potential energy due to the gravity in C is also zero because we are at a height of 0m. So we can simplify the equation to get:

$K_{A}+U_{Ag}=U_{Cs}+W_{Cf}$

so now we can use the respective formulas for kinetic energy and potential energy, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}ky_{C}+fy_{C}$

so we can now solve this for k, which is the constant of the spring.

$k=\frac{mV_{A}^{2}+mgh_{A}-fy_C}{y_{C}^{2}}$

and now we can substitute the respective data:

$k=\frac{(2000kg)(4m/s)^{2}+(2000kg)(9.8m/s^{2})(2m)-(17000N)(2m)}{(2m)^{2}}$

Which yields:

k= 9300N/m

Once we got the constant of the spring, we can now find the speed of the elevator at point B, which is one meter below the contact point, so we do the same analysis of energies, like this:

$K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}$

In this case ther are no zero values to eliminate so we work with all the types of energies involved in the equation, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}mV_{B}^{2}+\frac{1}{2}ky_{B}^{2}+mgh_{B}+fy_{B}$

and we can solve for the Velocity in B, so we get:

$V_{B}=\sqrt{\frac{mV_{A}^{2}-ky_{B}^{2}+2mgh_{B}-2fy_{B}}{m}}$

we can now input all the data directly, so we get:

$V_{B}=\sqrt{\frac{(2000kg)(4m/s)^{2}-(9300N/m)(1m)^{2}+2(2000kg)(9.8m/s^{2})(1m)-2(17000N)(1m)}{(2000kg)}}$

which yields:

$V_{B}=3.73m/s$

which is the first answer.

Now, for the second answer we need to find the acceleration of the elevator when it reaches point B, for which we need to build a free body diagram (also included in the attached picture) and to a summation of forces:

$\sum F=ma$