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Alex777 [14]
3 years ago
9

A boat with an anchor on board floats in a swimming pool that is somewhat wider than the boat. Does the pool water level move up

, move down, or remain the same if the anchor is
(a) dropped into the water or
(b) thrown onto the surrounding ground
(c) Does the water level in the pool move upward, move downward, or remain the same if, instead, a cork is dropped from the boat into the water, where it floats?
Physics
1 answer:
pentagon [3]3 years ago
6 0

Answer:

a) moves down

b) moves down

c) level remains same

Explanation:

Given that the anchor is initially on the floating boat.

a)

In this condition initially the the volume of water (V_w_i) displaced is to balance its weight.

Now,

W_a=W_w

V_a.\rho_a.g=V_w_i.\rho_w.g

\frac{\rho_a}{\rho_w} =\frac{V_w_i}{V_a}

We've, the density of steel = 7850\ kg.^{-3} and the density of water = 1000\ kg.^{-3}

\therefore V_w_i=7.85\times V_a

When the anchor is dropped into water:

The volume of water displaced be (V_w_f) which will be equal to the volume of anchor since it is immersed into it.

V_a=V_w_f

\therefore V_w_i>V_w_f ...................(1)

So the level of water falls when the anchor is dropped into water.

b)

Now, when the anchor is thrown on the ground the water has now less weight to balance so the water level falls down.

c)

When the cork on the from the boat is dropped into the water and it still floats then it must displace same amount of water, hence there should be no change in the water level.

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x \rightarrow 15\frac{mg}{hour}

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The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a
grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

\sin\theta=\dfrac{3}{5}

The horizontal component is

T_{x}=T\cos\theta

The vertical component is

T_{y}=T\sin\theta

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'

Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

T\sin\theta\times 4=380+1200

T=\dfrac{1580\times5}{3\times 4}

T=658.33\ N

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

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(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

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