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3 years ago

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a chuck wagon with an initial velocity of 4 m/s and a mass of 35 kg gets a push with 350 joules of force. what is the wagon's fi

nal velocity?

1 answer:

Kitty [74]3 years ago

8 0

Answer:

the final velocity of the wagon is 6 m/s.

Explanation:

Given;

initial velocity of the wagon, u = 4 m/s

mass of the wagon, m = 35 kg

energy applied to the wagon, E = 350 J

The final velocity of the wagon is calculated as;

E = ¹/₂m(v² - u²)

$m(v^2-u^2) = 2E\\\\v^2-u^2 = \frac{2E}{m} \\\\v^2 = \frac{2E}{m} + u^2\\\\v = \sqrt{\frac{2E}{m} + u^2} \\\\v = \sqrt{\frac{2(350)}{35} + (4)^2}\\\\v = 6 \ m/s$

Therefore, the final velocity of the wagon is 6 m/s.

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The magnetic field at the center of a 1.50-cm-diameter loop is 2.70 mT . Part A. What is the current in the loop?

elixir [45]

Explanation:

It is given that,

Diameter of the circular loop, d = 1.5 cm

Radius of the circular loop, r = 0.0075 m

Magnetic field, $B=2.7\ mT=2.7\times 10^{-3}\ T$

(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :

$B=\dfrac{\mu_o I}{2r}$

$I=\dfrac{2Br}{\mu_o}$

$I=\dfrac{2\times 2.7\times 10^{-3}\times 0.0075}{4\pi \times 10^{-7}}$

I = 32.22 A

(b) The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$r=\dfrac{\mu_o I}{2\pi B}$

$r=\dfrac{4\pi \times 10^{-7}\times 32.22}{2\pi \times 2.7\times 10^{-3}}$

r = 0.00238 m

$r=2.38\times 10^{-3}\ m$

Hence, this is the required solution.

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3 years ago

The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu

Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

$E=\phi + K$

where

$E=\frac{hc}{\lambda}$ is the energy of the incident light, with h being the Planck constant, c being the speed of light, and $\lambda$ being the wavelength

$\phi$ is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

$\lambda=190 nm=1.9\cdot 10^{-7}m$ , so the energy of the incident light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J$

Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: $\phi = 4.5 eV$

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

$E=\phi+K=4.5 eV+2.7 eV=7.2 eV$

Then the copper is replaced with sodium, which has work function of

$\phi = 2.3 eV$

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

$K=E-\phi = 7.2 eV-2.3 eV=4.9 eV$

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3 years ago

A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle

expeople1 [14]

Explanation:

Given that,

Wavelength of the light, $\lambda=691\ nm=691\times 10^{-9}\ m$

(a) Slit width, $a=3.8\times 10^{-4}\ m$

The angle that locates the first dark fringe is given by :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}$

$\theta=0.104^{\circ}$

(b) Slit width, $a=3.8\times 10^{-6}\ m$

The angle that locates the first dark fringe is given by :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}$

$\theta=10.47^{\circ}$

Hence, this is the required solution.

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3 years ago

Why is it better to use the metric system, rather than the English system, in scientific measurement?

fiasKO [112]

A. The English system uses one unit for each category of measurement.

4 0

3 years ago

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Problem 5 You are playing a game where you drop a coin into a water tank and try to land it on a target. You often find this gam

Dvinal [7]

Answer:

Option D: 1.5in in front of the target

Explanation:

The object distance is $y= 6in$ .

Because the surface is flat, the radius of curvature is infinity .

The incident index is $n_i=\frac{4}{3}$ and the transmitted index is $n_t= 1$ .

The single interface equation is $\frac{n_i}{y}+\frac{n_t}{y^i}=\frac{n_t-n_i}{r}$

Substituting the quantities given in the problem,

$\frac{\frac{4}{3}}{6in}+\frac{1}{y^i}= 0$

The image distance is then $y^i=-\frac{18}{4}in =-4.5in$

Therefore, the coin falls $1.5in$ in front of the target

6 0

3 years ago