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3 years ago

How much work is requried to uniformly accelerate a merry-go-round?

1 answer:

5 0

Do not worry if you don't recognize both parts of the problem at this point. If you recognize the dynamics problem, On the other hand, if you recognize this as a kinematics problem you will quickly see that you need to find angular acceleration before you can begin and so will need to do that pre-step first.

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A box sits on the back of a flatbed truck. If the coefficient of static friction between the box and the truck bed is μs = 0.400

slava [35]

Answer:

28,699m

Explanation:

The force to make the box move should be μs.N=μs.m.g=m.|a|

then,

|a|=μs.g

Being

μs coefficient of static friction,

N the force made by the truck on the box caused by the gravity force,

m the mass,

g the acceleration of gravity

and a the acceleration of the truck.

$x = v \times t + \frac{1}{2} \times a \times {t}^{2}$

as the truck is stopping, the acceleration is negative. then,

$x = v \times t - \frac{1}{2} \times |a| \times {t}^{2}$

$|a| = v \div t \\ t = v \div |a|$

$x = v \times (v \div |a| ) - \frac{1}{2} \times |a| \times {(v \div |a|)}^{2}$

$x = v \times (v \div μs.g ) - \frac{1}{2} \times |a| \times {(v \div μs.g)}^{2}$

$x = \frac{{v}^{2}}{μs \times g} - \frac{1}{2} \times \frac{{v}^{2}}{μs \times g} \\ x = \frac{1}{2} \times \frac{{v}^{2}}{μs \times g} \\ x = 0.5 \times \frac{{(15m/s) }^{2} }{0.4 \times 9.8m/ {s}^{2} } = 28.699m$

28,699m

7 0

3 years ago

NEED HELP!!!!

melamori03 [73]

Answer:

The correct choice is D. Concentric exercise can be hard on joints, so she should be very careful.

4 0

2 years ago

Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge

kramer

Answer:

E = 440816.32 N/C

Explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m

$K =1/4 \pi \epsilon_0 = 8.99 \times 10^9$ N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have

$E_1 + E_3 = 0$

$E =E_2 + E_4$

$E = 2 E_2$

[ $E_2 =\frac{2\times k \times q}{r^2}$

[ $r= \frac{(0.5^2 + 0.5^2)^2}{2} = 0.35 m$ ]

plugging all value

$E = 2 E_2$

$E = 2 E_2 =\frac{2\times k \times q}{r^2}$

$E = \frac{2 \times 8.99 \times 10^93\times 10^{-6}}{0.35^2}$

E = 440816.32 N/C

3 0

3 years ago

Read 2 more answers

Traveling in circle requires a net force

seraphim [82]

Answer:

Circular motion requires a net inward or "centripetal" force. Without a net centripetal force, an object cannot travel in circular motion. In fact, if the forces are balanced, then an object in motion continues in motion in a straight line at constant speed.

Explanation:

3 0

3 years ago

What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec

jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

$E = \sigma T^{4}$ (1)

Where $\sigma$ is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

$\lambda max = \frac{2.898x10^{-3} m. K}{T}$ (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

$E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}$ (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

$T = \frac{2.898x10^{-3} m. K}{\lambda max}$ (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), $\lambda max$ (450 nm) will be express in meters:

$450 nm . \frac{1m}{1x10^{9} nm}$ ⇒ $4.5x10^{-7}m$

$T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}$

$T = 6440 K$

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

$700 nm . \frac{1m}{1x10^{9} nm}$ ⇒ $7x10^{-7}m$

$T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}$

$T = 4140 K$

Comparison:

$\frac{6440 K}{4140 K} = 1.55$

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0

3 years ago