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3 years ago

6

The static frictional force between a 95-kilogram object and the floor is 45 Newtons. The kinetic frictional force is only 22 Ne

wtons. What force must be exerted to accelerate the box at 0.5 meters per second to the south?

1 answer:

Lisa [10]3 years ago

6 0

Answer:

F = 69.5 [N]

Explanation:

We must remember that the friction force is defined as the product of the normal force by the coefficient of friction, and it can be calculated by the following expression.

$f=N*miu$

where:

N = normal force [N]

miu = friction coefficient

f = friction force = 22 [N]

Now we must calculate the force exerted by means of Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

$F - f = m*a$

where:

F = force exerted [N]

f = friction force [N]

m = mass = 95 [kg]

a = acceleration = 0.5 [m/s²]

Now replacing:

$F - 22 = 95*0.5\\F = 47.5 + 22\\F = 69.5 [N]$

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Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm

Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: <em>Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.</em>

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

$\eta\cdot \dot W = \dot K$

$\dot W = \frac{\dot K}{\eta}$ (Eq. 1)

Where:

$\eta$ - Efficiency of fan, dimensionless.

$\dot W$ - Electric power supplied fan, measured in watts.

$\dot K$ - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

$\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2}$ (Eq. 2)

Where:

$\rho_{a}$ - Density of air, measured in kilograms per cubic meter.

$\dot V$ - Volume flow, measured in cubic meters per second.

$A_{s}$ - Cross-sectional area of fan, measured in square meters.

If we know that $\rho_{a} = 1.20\,\frac{kg}{m^{3}}$ , $\dot V = 0.05\,\frac{m^{3}}{s}$ , $\eta = 0.3$ and $A_{s} = 20\times 10^{-4}\,m^{2}$ , the power needed to be supplied by the fan is:

$\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}$

$\dot K = 62.5\,W$

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

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3 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

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3 years ago

If two particles have equal charges and are placed near one another, how

navik [9.2K]

Answer: C) The two particles will move away from each other

Explanation:

When two electrically charged bodies come closer, appears a force that attracts or repels them, depending on the sign of the charges of this two bodies.

This is stated by Coulomb's Law:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

Mathematically this law is written as:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$

Where $K$ is a proportionality constant.

Now, if $q_{1}$ and $q_{2}$ have the same sign charge (both positive or both negative), a repulsive force will act on these charges.

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3 years ago

How do valence electrons relate to the chemical reactions of an element?

11111nata11111 [884]

Valence electrons are required for certain bonding processes. Without them you cannot bond with certain elements. For example, carbon bonds really well with carbon because it has the same amount of valence electrons. However carbon would not bond well with uranium due to the massive differences in valence electrons. Hope this helps!

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3 years ago

An isotope undergoes radioactive decay by emitting radiation that has a –1 charge. What other characteristic does the radiation

weeeeeb [17]

<u>Answer:</u> The radiation emitted will have negligible mass number.

<u>Explanation:</u>

Radioactive decay is defined as the process in which an unstable nuclei breaks down into stable nuclei via various methods.

An isotope undergoes a radioactive decay to attain stability.

There are three types of decay process, but the process in which the emitted radiation carries a charge of -1 is beta decay.

Beta decay is defined as the decay process in which a neutron gets converted to a proton and an electron. In this decay process, beta particle is emitted. The emitted particle carries a charge of -1 units and has a mass of 0 units. The released beta particle is also known as electron.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

Hence, the radiation emitted will have negligible mass number.

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3 years ago

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