Ask question

Ask question

All categories

2 days ago

11

While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the

basket. His 74-kg body decelerated from 7.6 m/s to 0 m/s in 0.16 seconds.
a. Determine the force acting upon Logan's body.
b. If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds. Determine what the force on his body would have been for such an abrupt collision.

1 answer:

Maru [2.9K]2 days ago

6 0

a.) The force acting on Logan's body is calculated as F = 3515 N. b.) The force if Logan collided with the concrete wall while traveling at the same speed would be F = 140600 N.

You might be interested in

The total conversion of 1.00 kilogram of the Sun's mass into energy yields?

Yuliya22 [2962]

The well-known equation... E = m c²... does not address the origin of the mass involved.

Converting 1 kg of any mass entirely into energy generates

(1kg) · (c²) Joules of energy.

E = (1 kg) · (c²) = (1 kg) · (299,792,458 m/s)²

E = 8.9876 x 10¹⁶ Joules

To simplify, this equates to the energy needed to keep a 100-watt light bulb illuminated for about 10,402,259,010 days.

(This is roughly 28.5 million years, based on the current understanding of days and years.)

5 0

6 days ago

True or False: Molecules in a gas resist crowding and get as far apart as possible. Free electrons also resist crowding and get

ValentinkaMS [3076]

Answer:

This assertion is inaccurate.

Explanation:

The random nature of gas molecules results in their erratic motion and occasional collisions. While it is true that they tend to avoid being tightly packed, achieving the maximum separation from each other is not always feasible due to their lack of fixed positions. Consequently, gas molecules in a container cannot consistently maintain the furthest distance from their neighboring molecules.

In contrast, the separation among electrons is primarily influenced by repulsive forces, not random movement as in gases. Electrons maintain distance as a result of repulsion between similarly charged particles. Therefore, the arrangement of electrons on a charged copper sphere occurs not from a random distribution but rather due to repulsion, establishing a set distance between them.

4 0

25 days ago

Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta

ValentinkaMS [3076]

Answer:

The amount of heat that enters the gas throughout this two-step process totals 120 cal.

Explanation:

Given that,

Moles present = 3

Heat capacity at volume held constant = 4.9 cal/mol.K

Heat capacity at pressure held constant = 6.9 cal/mol.K

Starting temperature = 300 K

Ending temperature = 320 K

We are tasked with determining the heat absorbed by the gas at constant pressure

Employing the heat formula

$\Delta H_{1}=nC_{p}\times\Delta T$

Substituting the values into the equation

$\Delta H_{1}=3\times6.9\times(320-300)$

$\Delta H_{1}=414\ cal$

Next, we calculate the heat absorbed by the gas at constant volume

Using the corresponding heat formula

$\Delta H_{1}=nC_{v}\times\Delta T$

Insert the values into the formula

$\Delta H_{1}=3\times4.9\times(300-320)$

$\Delta H_{1}=-294\ cal$

Now, it's necessary to evaluate the total heat flow into the gas during both steps

Using the total heat formula

$\Delta H_{T}=\Delta H_{1}+\Delta H_{2}$

$\Delta H_{T}=414-294$

$\Delta H_{T}=120\ cal$

Thus, the heat that transfers into the gas throughout this two-step process amounts to 120 cal.

7 0

1 month ago

A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.

serg [3222]

A) 16.1 N

The force of electricity acting between the corks can be calculated using Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where

k represents Coulomb's constant

$q_1 = 6.0 \mu C=6.0 \cdot 10^{-6} C$ denotes the charge magnitude on the first cork

$q_2 = 4.3 \mu C = 4.3 \cdot 10^{-6}C$ indicates the charge magnitude on the second cork

r = 0.12 m is the distance separating the corks

By inserting the values into the formula, we arrive at

$F=(9\cdot 10^9 N m^2 C^{-2} )\frac{(6.0\cdot 10^{-6}C)(4.3\cdot 10^{-6} C)}{(0.12 m)^2}=16.1 N$

B) Attractive

<pas per="" coulomb="" law="" the="" orientation="" of="" electric="" force="" between="" two="" charged="" entities="" relies="" on="" their="" charge="" signs.=""><pmore specifically="">

- when both are similarly charged (e.g. positive-positive or negative-negative), the force is repulsive

- when charges are of opposite signs (e.g. positive-negative), the resulting force is attractive

<pin this="" case="" we="" have="">

Cork 1 holds a positive charge

Cork 2 possesses a negative charge

<pthus the="" force="" acting="" between="" them="" is="" attractive.="">

C) $2.69\cdot 10^{13}$

The total charge of the negative cork is

$q_2 = -4.3 \cdot 10^{-6}C$

<pwe understand="" that="" a="" single="" electron="" has="" charge="" of="">

$e=-1.6\cdot 10^{-19}C$

<pthe total="" charge="" of="" the="" negative="" cork="" arises="" from="" having="" n="" extra="" electrons="" so="" we="" can="" express="" it="" as="">

$q_2 = Ne$

<pafter solving="" for="" n="" we="" can="" determine="" the="" count="" of="" excess="" electrons:="">

$N=\frac{q_2}{e}=\frac{-4.3\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=2.69\cdot 10^{13}$

D) $3.75\cdot 10^{13}$

The overall charge on the positive cork is

$q_1 = +6.0\cdot 10^{-6}C$

<pthe charge="" of="" a="" single="" electron="" is="" known="" to="" be="">

$e=-1.6\cdot 10^{-19}C$

<pthe total="" charge="" of="" the="" positive="" cork="" results="" from="" n="" excess="" electrons="" which="" can="" be="" depicted="" as="">

$q_1 = -Ne$

<pby calculating="" for="" n="" we="" derive="" the="" number="" of="" electrons="" cork="" has="" lost:="">

$N=-\frac{q_1}{e}=-\frac{+6.0\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=3.75\cdot 10^{13}$

</pby></pthe></pthe></pafter></pthe></pwe></pthus></pin></pmore></pas>

6 0

27 days ago

A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac

Keith_Richards [2877]

Answer:

The flux across the cube's surface is $2.314\ Nm^{2}/C$ .

Solution:

According to the details provided:

Cube edge length, a = 8.0 cm = $8.0\times 10^{- 2}\ m$ .

Volume charge density, $\rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}$ .

Now,

To find the electric flux:

$\phi = \frac{q}{\epsilon_{o}}$

where

$\phi$ = electric flux

$\epsilon_{o} = 8.85\times 10^{- 12}\ F/m$ = permittivity of vacuum.

The volume charge density for this scenario is described by:

$\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}$

Cube volume, $V = a^{3}$ .

Thus,

$V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}$ .

The total charge can be derived from equation (2):

$q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}$ .

$q = 2.048\times 10^{-11}\ F = 20.48\ pF$ .

Now, insert the value of 'q' into equation (1):

$\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C$ .

5 0

1 month ago