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3 years ago

15

Oil having a density of 923 kg/m^3 floats on water. A rectangular block of wood 4.81 cm high and with a density of 961 kg/m^3 fl

oats partly in the oil and partly in the water. The oil completely covers the block.
(a) How far below the interface between the two liquids is the bottom of the block?

1 answer:

timurjin [86]3 years ago

8 0

Answer:

x = 2.37 cm

Explanation:

given,

density of oil = 923 kg/m³

height of the block = 4.81 cm

density of the block = 961 Kg/m³

density of water = 1000 Kg/m³

Let x be the depth of the block in the water.

now,

Weight of block will be acting downward.

Buoyancy force will be acting on the block because of the oil and water.

At equilibrium position both weight and the buoyant force will balance.

$\rho_{wood}gh-\rho_{oil}g(h-x)-\rho_{water}gx= 0$

$961\times 0.0481 - 923\times 0.0481 = -923\times x +1000\times x$

$1.8278 = 77 x$

x = 0.0237 m

x = 2.37 cm

Hence, the block is 2.37 cm in the water.

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3 years ago

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

aliina [53]

Answer:

9.43 m/s

Explanation:

First of all, we calculate the final kinetic energy of the car.

According to the work-energy theorem, the work done on the car is equal to its change in kinetic energy:

$W=K_f - K_i$

where

W = -36.733 J is the work done on the car (negative because the car is slowing down, so the work is done in the direction opposite to the motion of the car)

$K_f$ is the final kinetic energy

$K_i = 66,120 J$ is the initial kinetic energy

Solving,

$K_f = K_i + W = 66,120 + (-36,733)=29,387 J$

Now we can find the final speed of the car by using the formula for kinetic energy

$K_f = \frac{1}{2}mv^2$

where

m = 661 kg is the mass of the car

v is its final speed

Solving for v, we find

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.43 m/s$

3 0

3 years ago

Complete the paragraph to compare Uranus and Neptune.

kipiarov [429]

Answer:

Uranus and Neptun are outer planets od the Solar system, since they are located after the asteroid belt. All of these outer planets are much larger then the inner ones so they are called the "ice giants". The other reason for this name is that they are very far from the Sun, so their temperature is low. Another feature they have in common is their atmosphere which is composed of gases, including methane, which is responsible for their blue color, since methane absorbs red light. However Neptune is known for very fast winds and storms in its atmosphere which is responsible for its high activity and changes.

So, the blanks should be filled with:

Sun

methane

Neptune

outer planets

5 0

3 years ago

A proton is shot perpendicularly at an infinite plane of charge. The charge density of the plane is +7.65×10^−4 C/m^2. If the pr

Alexeev081 [22]

Answer:

The velocity is $2.94\times10^{6}\ m/s$ .

Explanation:

Given that,

Charge density of the plane $\sigma=7.65\times10^{−4}\ C/m^2$

Distance = 1.05 mm

We need to calculate the electric field due to plane of charge

Using formula of electric field

$E=\dfrac{\sigma}{2\epsilon}$

Put the value into the formula

$E=\dfrac{7.65\times10^{−4}}{2\times8.85\times10^{-12}}$

$E=4.322\times10^{7}\ N/C$

We need to calculate potential difference

Using formula of potential difference

$V=E\times r$

Put the value into the formula

$V=4.322\times10^{7}\times1.05\times10^{-3}$

$V=4.5381\times10^{4}\ Volt$

We need to calculate the work requires to be done to reach the surface of the plane

Using formula of work done

$W=qV$

Put the value into the formula

$W = 1.6\times10^{-19}\times4.5381\times10^{4}$

$W=7.26096\times10^{-15}\ J$

We need to calculate the velocity

Using work energy theorem

$W=\dfrac{1}{2}mv^2$

$v^2=\dfrac{2W}{m}$

$v=\sqrt{\dfrac{2\times7.26096\times10^{-15}}{1.67\times10^{-27}}}$

$v=2.94\times10^{6}\ m/s$

Hence, The velocity is $2.94\times10^{6}\ m/s$ .

8 0

3 years ago

To test the performance of its tires, a car

yarga [219]

Answer:

0.43

Explanation:

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

There are friction forces in two directions: centripetal and tangential. The centripetal acceleration is:

ac = v² / r

ac = (35 m/s)² / 564 m

ac = 2.17 m/s²

The total acceleration is:

a = √(ac² + at²)

a = √((2.17 m/s²)² + (3.62 m/s²)²)

a = 4.22 m/s²

Sum of forces:

∑F = ma

Nμ = ma

mgμ = ma

μ = a / g

μ = 4.22 m/s² / 9.8 m/s²

μ = 0.43

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3 years ago