1) A bowling has a mass of 5 kg and a speed of 8 m/s. What is its momentum?

Exercise and health

zavuch27 [327]

Are both intertwined. exercise and you will feel brand new!!!

4 0

3 years ago

What is the technology used behind scanning probe microscopes?

Wittaler [7]

hola ya sabes la respuesta si si ,me la dices porfa

5 0

3 years ago

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If earth's mass were half its actual value but its radius stayed the same, the escape velocity of earth would be:________

siniylev [52]

If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

<h3>What is an escape velocity?</h3>

The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.

The formula to calculate the escape velocity of earth is given below:-

$V_e=\sqrt{\dfrac{2GM}{r}}$

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

$V_e=\sqrt{\dfrac{2GM}{r\times 2}}$

$V_e = \sqrt{\dfrac{GM}{r}}$ .

Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

To know more about escape velocity follow

brainly.com/question/14042253

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8 0

1 year ago

Which of the following materials is attracted to a magnet?

Butoxors [25]

Answer:

A. aluminum

Explanation:

because it's metal

4 0

3 years ago

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One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

4 years ago