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3 years ago

14

Jerome plays middle linebacker for South's varsity football team. In a game against cross-town rival North, he delivered a hit t

o North's 82-kg running back, changing his eastward velocity of 5.6 m/s into a westward velocity of 2.5 m/s.

1 answer:

anygoal [31]3 years ago

5 0

Explanation:

We have,

Mass of the running back, m = 82 kg

Initial speed of the running back, u = +5.6 m/s

Final speed of the running back, v = -2.5 m/s

It is required to find the momentum change of the running back. For this we will subtract final momentum to the initial momentum. So,

$\Delta p=m(v-u)\\\\\Delta p=82\times (-2.5-5.6)\\\\\Delta p=-664.2\ kg-m/s$

So, the magnitude of the change in momentum of the running back is 664.2 m/s in west direction.

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Particle motion in surface waves is __________ motion.

gulaghasi [49]

<h3>Answer;</h3>

<em>B.)neither longitudinal nor transverse</em>

<h3><u>Explanation;</u></h3>

<em><u>Longitudinal waves</u></em> are waves in which the vibration of particles is parallel to the direction of the wave motion.
<em><u>Transverse waves</u></em> on the other hand are those waves in which the vibration of particles is perpendicular to the direction of the wave motion.
In <em><u>surface waves particles in the medium of transmission move in a circular motion.</u></em> Therefore, they are neither transverse waves nor longitudinal waves.

7 0

3 years ago

Read 2 more answers

A 50-kg copper block initially at 140Â°c is dropped into an insulated tank that contains 90 l of water at 10Â°c. Determine the f

xxMikexx [17]

Answer:

$T_f=24.71$

Explanation:

From the question we are told that:

Mass of block $m=50$

Temperature of block $T_b =140 \textdegree C$

Volume of water $V= 90L$

Temperature of water $T_w=10 \textdegree C$

Density of water $\rho=1000kg/m^3$

Specific heat of water $C_w=4.18KJ/kg-k$

Specific heat of copper $C_p=0.96KJ/kg-k$

Generally the equation for equilibrium stage is mathematically given by

$mC_p(T_b-T_f)=\rho*VV*c(T_f-T_w)$

$50*0.96(140-T_f)=1000*90*10^-3*c_w(T_f-10)$

$48(140-T_f)=376.2(T_f-10)$

$140-T_f=7.8375(T_f-10)$

$140-T_f=7.8375T_f-78.375$

$-8.8375T_f=-218.375$

$T_f=\frac{-218.375}{-8.8375}$

$T_f=\frac{-218.375}{-8.8375}$

$T_f=24.71$

6 0

3 years ago

A object with a mass of 1.5 kg is lifted from the ground to a height of 0.22 m what is the objects potential energy

svet-max [94.6K]

Answer:

<h2>3.3 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 1.5 × 10 × 0.22

We have the final answer as

<h3>3.3 J</h3>

Hope this helps you

5 0

2 years ago

Observe yourself breathing and count the number of times you inhale per second. During each breath you probably inhale 0.66 L of

Pavel [41]

To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

$PV=nRT$

Where,

P = Pressure

V = Volume

T = Temperature

n = amount of substance

R = Ideal gas constant

We start by calculating the volume of inhaled O_2 for it:

$V = 21\% * 0.66L$

$V = 0.1386L$

Our values are given as

P = 1atm

T=293K $R = 0.083145kJ*mol^{-1}K^{-1}$

Using the equation to find n, we have:

$PV=nRT$

$n = \frac{PV}{RT}$

$n = \frac{(1)(0.1386)}{(0.0821)(293)}$

$n = 5.761*10^{-3}mol$

Number of molecules would be found through Avogadro number, then

$\#Molecules = 5.761*10^{-3}*6.022*10^{23}$

$\#Molecules = 3.469*10^{21} molecules$

7 0

3 years ago

How can a 1kg ball have more kinetic energy than a 100kg ball? Explain both using words and by providing a numerical example

MariettaO [177]

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the $(velocity)^2$ and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

$\text { Kinetic Energy }=\frac{1}{2} m v^{2}$

Better understood from numerical example as given:

If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?

This can be solved as follows:

$\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}$

$\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}$

It shows that man A will have more K.E.

Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.

4 0

3 years ago