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3 years ago

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A cork has weight mg and density 25% of water’s density. A string is tied around the cork and attached to the bottom of a water-

filled container. The cork is totally immersed. Express the tension in the string in terms of the cork weight mg.

1 answer:

yuradex [85]3 years ago

4 0

Answer:

Explanation:

Given

Density of Cork $\rho _c=0.25\rho _{water}$

Considering V be the volume of Cork

Buoyant Force will be acting Upward and Weight is acting Downward along with T

Since density of water is more than cork therefore Cork will try to escape out of water but due to tension it will not

we can write as

$\rho _{water}Vg-\rho _Cvg-T=0$

where T=tension

Thus Tension T is

$T=Vg(\rho _{water}-\rho _c)$

Taking $\rho _c$ common

$T=\rho _cVg(\frac{\rho _{water}}{\rho _c}-1)$

$T=mg(\frac{\rho _{water}}{\rho _c}-1)$

$T=mg(4-1)$

$T=3mg$

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You are at the edge of a diving board that is 9 meters above the water. If you weigh 500 Newtons, what is your potential energy?

Semenov [28]

Answer:

4500 J

Explanation:

First, let's define some equations and derivations.

Our potential energy formula is:

$\displaystyle U = mgh$

Where <em>m </em>is mass (in kg), <em>g</em> is the gravitational constant (in m/s²), and <em>h</em> is height (in m).

We also know that <em>mg</em> is equal to the weight of an object (in N), from Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration).

Therefore, we can simply substitute force into the equation:

$\displaystyle U = Fh$

Where <em>F</em> is the force (in N) and <em>h</em> is still height (in m).

Now we can calculate the amount of potential energy in our system, measured in joules.

Substitute in the given variables, F = 500 N and h = 9 m:

$\displaystyle U = (500 \ N)(9 \ m)$

Using simple Pre-Algebra rules, we find that:

$\displaystyle U = 4500 \ J$

This tells us that the we have 4500 joules of potential energy when I am 9 meters above the water on the edge of the diving board.

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A rock is thrown with a force of 500 N and an acceleration is 75 m/s^2. What is its mass?

artcher [175]

Answer:

We conclude that the mass of a rock with a force of 500 N and an acceleration of 75 m/s² is 6.7 kg.

Hence, option D is correct.

Explanation:

Given

Force F = 500 N

Acceleration a = 75 m/s²

To determine

Mass m = ?

Important Tip:

The mass of a rock can be found using the formula F = ma

Using the formula

$F = ma$

where

F is the force (N)

m is the mass (kg)

a is the acceleration (m/s²)

now substituting F = 500, and a = 75 m/s² in the formula

$F = ma$

$500 = m(75)$

switch sides

$m\left(75\right)=500$

Divide both sides by 75

$\frac{m\cdot \:75}{75}=\frac{500}{75}$

simplify

$m=\frac{20}{3}$

$m=6.7$ kg

Therefore, we conclude that the mass of a rock with a force of 500 N and an acceleration of 75 m/s² is 6.7 kg.

Hence, option D is correct.

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3 years ago

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

<u>Now the horizontal component of velocity:</u>

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

<u>The vertical component of the velocity:</u>

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

<u>Time taken by the projectile to travel the distance of 30 m:</u>

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

<u>Vertical position of the projectile at this time:</u>

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

<u>We observe that </u> $t>t'$ <u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

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3 years ago