The magnetic field of a solenoid depends upon n, which represents select one:

A hovering mosquito is hit by a raindrop that is 50 times as massive and falling at 8.4 m/s, a typical raindrop speed. How fast

kenny6666 [7]

The final velocity after the collision is 8.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, if we consider the system to be isolated (=no external unbalanced forces), the total momentum of the raindrop+mosquito must be conserved before and after the collision.

If the collision is perfectly inelastic, moreover, the raindrop and the mosquito stick together and travel at the same velocity v after the collision.

Mathematically:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1$ is the mass of the first mosquito

$u_1 = 0$ is the initial velocity of the mosquito

$m_2 = 50 m_1$ is the mass of the raindrop

$u_2 = 8.4 m/s$ is the initial velocity of the raindrop

$v$ is the final combined velocity of the raindrop+mosquito

Re-arranging the equation and substituting, we find:

$m_1 u_1 + 50 m_1 u_2 = (m_1 + 50 m_1) v\\50 m_1 u_2 = 51 m_1 v\\50 u_2 = 51 v\\v=\frac{50}{51}u_2 = \frac{50}{51}(8.4)=8.2 m/s$

Learn more about momentum here:

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4 0

3 years ago

A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t

Anastaziya [24]

Answer:

The gauge pressure is $P_g = 2058 \ P_a$

Explanation:

From the question we are told that

The height of the water contained is $h_w = 30 \ cm = 0.3 \ m$

The height of liquid in the cylinder is $h_t = 40 \ cm = 0.4 \ m$

At the bottom of the cylinder the gauge pressure is mathematically represented as

$P_g = P_w + P_o$

Where $P_w$ is the pressure of water which is mathematically represented as

$P_w = \rho_w * g * h_w$

Now $\rho_w$ is the density of water with a constant values of $\rho_w = 1000 \ kg /m^3$

substituting values

$P_w = 1000 * 9.8 * 0.3$

$P_w = 2940 \ Pa$

While $P_o$ is the pressure of oil which is mathematically represented as

$P_o = \rho_o * g * (h_t -h_w )$

Where $\rho _o$ is the density of oil with a constant value

$\rho _o = 900 \ kg / m^3$

substituting values

$P_o = 900 * 9.8 * (0.4 - 0.3)$

$P_o = 882 \ Pa$

Therefore

$P_g = 2940 - 882$

$P_g = 2058 \ P_a$

6 0

3 years ago

Atomic physicists usually ignore the effect of gravity within an atom. To see why, we may calculate and compare the magnitude of

STatiana [176]

Answer:

$2.27\cdot 10^{49}$

Explanation:

The gravitational force between the proton and the electron is given by

$F_G=G\frac{m_p m_e}{r^2}$

where

G is the gravitational constant

$m_p$ is the proton mass

$m_e$ is the electron mass

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

$F_G=(6.67259\cdot 10^{-11} m^3 kg s^{-2})\frac{(1.67262\cdot 10^{-27}kg) (9.10939\cdot 10^{-31}kg)}{(3 m)^2}=1.13\cdot 10^{-68}N$

The electrical force between the proton and the electron is given by

$F_E=k\frac{q_p q_e}{r^2}$

where

k is the Coulomb constant

$q_p = q_e = q$ is the elementary charge (charge of the proton and of the electron)

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

$F_E=(8.98755\cdot 10^9 Nm^2 C^{-2})\frac{(1.602\cdot 10^{-19}C)^2}{(3 m)^2}=2.56\cdot 10^{-19}N$

So, the ratio of the electrical force to the gravitational force is

$\frac{F_E}{F_G}=\frac{2.56\cdot 10^{-19} N}{1.13\cdot 10^{-68}N}=2.27\cdot 10^{49}$

So, we see that the electrical force is much larger than the gravitational force.

5 0

3 years ago

What might be collected to monitor the amount of excess nitrogen in the local water system

Debora [2.8K]

Answer:

If excess nitrogen is found in the crop fields, the drainage water can introduce it into ... have aquifers that can supply a lot of freshwater very near the land surface.

4 0

3 years ago

Why does rubbing a balloons on wool or hair make it attract other objects

KonstantinChe [14]

Answer:

its because static electricity my guy

Explanation:

5 0

3 years ago

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