Ask question

Ask question

All categories

3 years ago

9

Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the

floor. The tension is a constant 29.0 N and the coefficient of friction is 0.180. Use work and energy to find Paul's speed after being pulled 2.70m .Express your answer with the appropriate units.

1 answer:

Pavlova-9 [17]3 years ago

8 0

Answer:1.71 m/s

Explanation:

Given

mass of Susan $m=12 kg$

Inclination $\theta =30^{\circ}$

Tension $T=29 N$

coefficient of Friction $\mu =0.18$

Resolving Forces Along x axis

$F_x=T\cos \theta -f_r$

where $f_r=friction\ Force$

$F_y=mg-N-T\sin \theta$

since there is no movement in Y direction therefore

$N=mg-T\sin \theta$

and $f_r=\mu N$

Thus $F_x=T\cos \theta -\mu N$

$F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))$

$F_x=25.114-18.558$

$F_x=6.556 N$

Work done by applied Force is equal to change to kinetic Energy

$F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2$

$6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2$

$v_f^2=\frac{6.556\times 2.7\times 2}{12}$

$v_f^2=2.95$

$v_f=1.717 m/s$

You might be interested in

Alexis walked 1.5 km to her house in 0.5 hours. What is Alexis’ speed?

katrin2010 [14]

Answer:

Speed = 3 [km/h]

Explanation:

To solve this problem we must use the definition of speed which relates the distance traveled for a while.

Distance = 1.5 [km] = 1500 [m].

time = 0.5 [hr] = 1800 [s]

Speed = Distance/time

Speed = 1.5/0.5

Speed = 3 [km/h] or 1500/1800 = 0.8333[m/s]

4 0

2 years ago

Given the isotope 2Fes, which has an actual mass of 55.934939 u: a) b) Determine the mass defect of the nucleus in atomic mass u

SSSSS [86.1K]

Answer:

Mass defect of each iron-56 nuclei:

The binding energy per nucleon of Iron-56 is approximately 8.6 MeV.

Explanation:

According to the physics constants table on Chemistry Libretexts:

Proton rest mass: $\rm 1.0072765\;amu$ ;
Neutron rest mass: $\rm 1.0086649\; amu$ .
Speed of light in vacuum: $\rm 2.99792458\times 10^{8}\;m\cdot s^{-1}$ .
Charge on an electron: $\rm 1.6021765\times 10^{-19}\;C$ .

<h3>a)</h3>

The mass defect of a nucleus is equal to the sum of the mass of its parts (protons and, in most cases, neutrons) minus the mass of the nucleus.

The atomic number of iron is 26. There are 26 protons in each iron-56 nucleus. The mass number 56 indicates that there are 56 nucleons (neutrons and protons) in each iron-56 nucleus. The other $56 - 26 = 30$ particles are neutrons.

The mass of protons and neutrons in each iron-56 nucleus will be:

$\rm 26 \times 1.0072765 + 30 \times 1.0086649 = 56.464736\;amu$ .

According to this question, the mass of an iron-56 nucleus is equal to 55.934939 amu. The mass defect will be

$\rm 56.464736 - 55.934939 = 0.514197\;amu$ .

<h3>b)</h3>

By the mass-energy equivalence,

$E = m\cdot c^{2}$ .

Refer to this equation, the speed of light in vacuum $c^{2}$ is the conversion factor between mass $m$ and energy $E$ . The value of $c$ is usually given only in SI units $\rm m\cdot s^{-1}$ . Accordingly, the value of $c^{2}$ will be in the SI unit $\rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}$ .

Convert million electron-volts to joules.

One electron-volt is equal to the electrical work done moving an electron across a potential difference of one volt.

$\begin{aligned}\rm 1 MeV&= \rm 10^{6}\; eV\\ &= \rm (10^{6}\times 1.6021765\times 10^{-19}\;C)\times 1\; V\\&=\rm 1.6021765\times 10^{-19}\;J\end{aligned}$ .

Convert the unit of $c^{2}$ from $\rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}$ to the desired $\rm MeV \cdot amu^{-1}$ :

$\begin{aligned}c^{2} &= \rm {\left(2.99792458\times 10^{8}\;m\cdot s^{-1}\right)}^{2}\\&=\rm 8.987551787\times 10^{16}\; m^{2}\cdot s^{-2}\\ &= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\\&= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\times \frac{1\;MeV}{1.6021765\times 10^{-13}\;J}\times \frac{1\times 10^{-3}\;kg}{6.022142\times 10^{23}\;amu}\\&\approx \rm 931.602164\;MeV\cdot amu^{-1}\end{aligned}$ .

Total binding energy in each iron-56 nucleus:

$\begin{aligned}E &= m\cdot c^{2}\\&= \rm 0.514197\;amu \times 9.31602164\;MeV\cdot amu^{-1} \\&=\rm 479.027038\; MeV \end{aligned}$ .

Again, the mass number 56 indicates that there are 56 nucleons in each iron-56 nucleus. The binding energy per nucleon of iron-56 $\mathrm{^{56}Fe}$ will be:

$\displaystyle \rm \frac{479.027038\; MeV}{56} \approx 8.6\; MeV$ .

6 0

3 years ago

(25 pts) Estimate how much collector area and storage capacity would be required for an active solar hot-water system designed t

Aneli [31]

Answer:

The required total area is 1.48 m²

Explanation:

Given that,

Latitude = 44+° N

New Mexico,

Latitude= 35+° N

Heat capacity = 4200 J/Kg°C

Temperature = 60°C

Let us assume the input temperature 22°C

Estimate volume of water 100 ltr for 4 person.

We need to calculate the heat

Using formula of heat

$H=mc_{p}\Delta T$

$H=mc_{p}(T_{f}-T_{i})$

Put the value into the formula

$H=100\times4200\times(60-22)$

$H=15960\ KJ$ ...(I)

Let solar radiation for 6 hours/day.

We need to calculate the total energy per unit area

Using formula of energy

$E=1000\times6\times3600\ J/m^2$

$E=21600\ KJ/m^2$

Let the efficiency of collector is 50 %

Then, the total energy per unit area will be

$E=21600\times\dfrac{50}{100}$

$E=10800\ KJ/m^2$ ....(II)

We need to calculate the required total area

Using equation (I) and (II)

$A=\dfrac{H}{E}$

Where, H = heat

E = total energy

Put the value into the formula

$A=\dfrac{15960}{10800}$

$A=1.48\ m^2$

Hence, The required total area is 1.48 m²

6 0

3 years ago

What happens to light when it changes speed?

Brut [27]

Hey there!

When light changes speed, it REFRACTS.
Your answer is going to be option C.

Hope this helps you.
Have a great day!

5 0

3 years ago

The de Broglie wave is produced only by sub atomic particle and photon. O True O False

Alina [70]

Answer:

True

Explanation:

Matter can be in the form of a particle or a wave. This is known as the dual nature of matter. This concept was proposed by Louis de Broglie and was named after him. This phenomenon has been observed for all the elementary particles.

The de Broglie wavelength is given by

$\lambda=\frac{h}{p}=\frac{h}{mv}$

Where

h = Planck's constant

p = Particles momentum

m = Mass of particle

v = Velocity of particle

8 0

3 years ago