Ask question

Ask question

All categories

3 years ago

8

An Aichi D3A bomber, with mass of 3600 kg, departs from its aircraft carrier with a velocity of 85 m/s due east. What is the pla

ne's momentum

1 answer:

Nataliya [291]3 years ago

7 0

Answer:

306000 is the answer.......

You might be interested in

A golf club with 65J of kinetic energy strikes a stationary golf ball with a mass of 46g. The energy transfer is only 20% effici

umka21 [38]

Kinetic energy of golf club = 65J,
kinetic energy supplied to golf ball = 20% of 65 = 0.2 * 65 = 13J,
kinetic energy of ball = [mass * Velocity²]/2,
mass = 46gm = 0.046Kg,
[0.046 * V²]/2 = 13, or 0.046 *V² = 26,
V² = 26/0.046 = 565.22,
V = 23.77 m/sec = initial velocity of golf ball after hitting.

4 0

3 years ago

Fill in the blanks.<br>In what way are speed and velocity different?

IrinaVladis [17]

Volocity can be the wave length of the speed like the volume.

8 0

3 years ago

Read 2 more answers

Which factor MOST directly affects the flow of ocean currents?

Kay [80]

The factor that most affects the flow of ocean currents is B. differences in temperaturethreAnswer here

7 0

3 years ago

Read 2 more answers

If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g

EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408 × 10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M_1, M_2$ are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

$\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}$

$\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2$

Since $M_2 = 2m_2$ and $r = R/3$

$\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5$

So gravity would have been decreased by a factor of 4.5

8 0

3 years ago

The Franck-Hertz experiment involved shooting electrons into a low-density gas of mercury atoms and observing discrete amounts o

Anuta_ua [19.1K]

Answer:

the final kinetic energy is 0.9eV

Explanation:

To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:

$E_n=\frac{-13.6eV}{n^2}$

you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is

$E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV$

-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:

$E_{k}=11.1eV-10.2eV=0.9eV$

6 0

3 years ago