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3 years ago

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Logs sometimes float vertically in a lake because one end has become water-logged and denser than the other. What is the average

density of a uniform-diameter log that floats with 29.1 % of its length above water?

1 answer:

Helga [31]3 years ago

4 0

Answer:

709 kg/m³

Explanation:

Applying the expression obtained from Archimedes Principle as:

$\% of\ the\ fraction\ submerged=\frac {\rho_{object}}{\rho_{liquid}}\times 100$

Given :

Percentage of the log above water = 29.1%

Percentage of the log of the water submerged = 100-29.1 % = 70.9%

Density of water = 1000 kg/m³

So,

$70.9 \% =\frac {\rho_{object}}{1000}\times 100$

<u>The average density of the log = 709 kg/m³</u>

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vichka [17]

Answer:

$s=vt-\frac{1}{2}gt^2$

Explanation:

We could use the following suvat equation:

$s=vt-\frac{1}{2}gt^2$

where

s is the vertical displacement of the coin

v is its final velocity, when it hits the water

t is the time

g is the acceleration of gravity

Taking upward as positive direction, in this problem we have:

s = -1.2 m

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t = 1.3 s

Substituting these data, we can find v:

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3 years ago

Read 2 more answers

On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he

Pachacha [2.7K]

Answer:

d) g/2

Explanation:

We need to use one of Newton's equations of motion to find the position of the stone at any time t.

x(t) = x₀(t) + ut - ¹/₂at²

Where

x₀(t) = initial position of the stone.

x(t) - x₀(t) = distance traveled by the stone at any time.

u = initial velocity of the stone

a = acceleration of the stone

t = time taken

On both planets, before the stone was thrown by the astronaut, x = 0 and t = 0.

=> 0 = x₀(t)

=> x₀(t) = 0

On earth, when the stone returns into the hand of the astronaut at time T on earth, x = 0.

=> 0 = 0 + uT - ¹/₂gT² (a = g)

=> uT = ¹/₂gT²

=> g = 2u/T

On planet X, when the stone returns into the hand of the astronaut, time = 2T , x = 0.

=> 0 = 0 + u(2T) - ¹/₂a(2T)²

=> 2uT = 2aT²

=> a = u/T

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3 years ago

To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where

BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

$\int E.dA = \frac{q}{\epcilon_0}$

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

$q = \int \rho dV$

$q = \rho * \pi r^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r}{2 \epcilon_0}$

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

$q = \int \rho dV$

$q = \rho * \pi r_0^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r_0^2}{2 \epcilon_0 r}$

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Holly puts a box into the trunk of her car. Later, she drives around an unbanked curve that has a radius of 48 m. The speed of t

LiRa [457]

Answer:

The minimum coefficient of friction is 0.544

Solution:

As per the question:

Radius of the curve, R = 48 m

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$F_{c} = \frac{mv^{2}}{R}$

$f_{s} = \mu_{s}mg$

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$\mu_{s}$ = coefficient of static friction

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$0 = f_{s} - F_{c}$

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$\mu_{s} = \frac{v^{2}}{gR}$

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3 years ago