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3 years ago

11

A cannon at rest fires a 32.5 kg cannonball forward at 388 m/s. After firing, the cannon recoils at 7.42 m/s. What is the mass o

f the cannon?
(Unit = kg)
Remember: right is +, left is -

1 answer:

Bond [772]3 years ago

3 0

Answer:

1700 kg

Explanation:

Let’s use conservation of momentum

32.5 * 388 = 7.42 * mc

mc = 1699.46

mc = 1700 kg

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In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

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3 years ago

Consider an old-fashion bicycle with a small wheel of radius 0.17 m and a large wheel of radius 0.92 m. Suppose the rider starts

Gala2k [10]

Answer:

10259.6 m

Explanation:

We are given that

Radius of small wheel,r=0.17 m

Radius of large wheel,r'=0.92 m

Initial velocity,u=0

Time,t=2.7 minutes=162 s

1 min=60 s

Velocity,v=10m/s

Time,t'=13.7 minutes=822 s

Time,t''=4.1 minutes=246 s

$v=u+at$

Substitute the values

$10=0+162a=162a$

$a=\frac{10}{162}=0.0617m/s^2$

$s=ut+\frac{1}{2}at^2$

Substitute the values

$s=\frac{1}{2}(0.0617)(162)^2=809.6 m$

$s'=vt'=10\times 822=8220 m$

$a'=\frac{v}{t''}=\frac{10}{246}$

$s''=\frac{1}{2}a't''^2=\frac{1}{2}\times \frac{10}{246}(246)^2=1230 m$

Total distance traveled by rider=s+s'+s''=809.6+8220+1230=10259.6 m

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