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2 years ago

Why does an ice cube float on water?

Physics

1 answer:

Dmitry [639]2 years ago

6 0

Answer:

Ice cubes float on water because they are less dense.

Explanation:

Ice is about 9% less dense than water. Water is heavier, so ice is lighter, causing it to float. It is less dense because the density of water decreases along with a decrease in temperature. Therefore, causing ice to be less dense than water

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A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.

pshichka [43]

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

$\sum F_y=m_1*a_m_i_n = T-m_1*g$

If the package is barely lifted, that means that T=m_2*g; then:

$\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g$

Solving the equation for a_mín, we have:

$a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2$

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: $\sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a$

For the package: $\sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a$

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: $\sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a$

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

$m_1*a+m_1*g=m_2*g-m_2*a$

Solving a, we have

$(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2$

We can then replace this value of a in one for the sums of force and find the tension T:

$T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N$

5 0

3 years ago

If the force applied to an object is not greater than the starting friction, what will happen to the object? Which answer is cor

AleksandrR [38]

Answer:

The object will move in the opposite direction of the force applied. - 2.

8 0

2 years ago

If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you

Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, $Q=121\ nC=121\times 10^{-9}\ C$

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

$Q=ne$

e is the charge on electron

$n=\dfrac{Q}{e}$

$n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}$

$n=7.5625\times 10^{11}$

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

6 0

3 years ago

Which method of measurement would be accurate but lack precision?

kodGreya [7K]

Answer:

Explanation:

reading the volume of water in a graduated cylinder which can be read to the nearest mL is accurate, it lacks precision due to the bottom meniscus formed.

the bottom meniscus may cause a wrong reading due to refraction of light

7 0

1 year ago

Two ions with masses of 4.39×10−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic field is

Ilia_Sergeevich [38]

Answer:

7.2 cm

Explanation:

magnetic field, B = 0.301 T

speed, v = 7.92 x 10^5 m/s

mass, m = 4.39 x 10^-27 kg

q = 1.6 x 10^-19 C

The radius of singly changed ion is given by

$r = \frac{mv}{Bq}$

where, m is the mass of ion, v be the speed of ion, B is the magnetic field and q be the charge

$r = \frac{4.39\times 10^{-27}\times 7.92 \times 10^{5}}{0.301\times 1.6\times 10^{-19}}$

r = 0.072 m

r = 7.2 cm

5 0

2 years ago