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RoseWind [281]
3 years ago
14

A person starts running from 2 m/s to 6 m/s in 2 seconds.

Physics
1 answer:
fiasKO [112]3 years ago
8 0

Answer:

2 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 2 m/s

Final velocity (v) = 6 m/s

Time (t) = 2 s

Acceleration (a) =?

The acceleration of the person can be obtained as follow:

v = u + at

6 = 2 + (a×2)

6 = 2 + 2a

Collect like terms

6 – 2 = 2a

4 = 2a

Divide both side by 2

a = 4 / 2

a = 2 m/s²

Therefore, the acceleration of the person is 2 m/s²

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A truck is traveling east at 80 km/h. At an intersection 32 km ahead, a car is traveling north at 50 km/h. How long after this m
nirvana33 [79]

The time elapsed when the vehicles are closest to each other is 20 min.

The given parameters:

  • Speed of the truck, u = 80 km/h
  • Distance, d = 32 km
  • Speed of the car, v = 50 km/h

<h3>Principles of relative speed</h3>

The time elapsed when the cars are close to each other is calculated by applying the principles of relative speed.

(V_r) t = d\\\\V_r^2 = 50^2 + 80^2\\\\V_r =\sqrt{50^2 + 80^2} \\\\V_r = 94.34 \ km/h

94.34 t = 32\\\\t = \frac{32}{94.34} \\\\t = 0.34 \ hr\\\\t \approx 20 \min

Thus, the time elapsed when the vehicles are closest to each other is 20 min.

Learn more about relative velocity here: brainly.com/question/24430414

3 0
2 years ago
A ____ is the time required for one half of the nuclei in a radio- ____ isotope to decay.
sukhopar [10]

Answer:

A half-life is the time required for one half of the nuclei in a radio- active isotope to decay.

Explanation:

A radio-active isotope is an isotope which undergoes radioactive decay.

Radioactive decay is a spontaneous process in which the nucleus of an atom changes its state (turning into a different nucleus, or de-exciting), emitting radiation, which can be of three different types: alpha, beta or gamma.

The half-life of a radio-active isotope is the time required for half of the nuclei of the initial sample to decay.

The law of radio-active decay can be expressed as follows:

N(t) = N_0 (\frac{1}{2})^{t/t_{1/2}}

where

N(t) is the number of undecayed nuclei left at time t

N0 is the initial number of nuclei

t is the time

t_{1/2} is the half-life

We see that when t=t_{1/2} (that means, when 1 half-life has passed), the number of undecayed nuclei left is

N(t) = N_0 (\frac{1}{2})^{t_{1/2}/t_{1/2}}=N_0 (\frac{1}{2})^1=\frac{N_0}{2}

So, half of the initial nuclei.

5 0
2 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
3 years ago
Complete the equation to show the radioactive decay of carbon-14 to nitrogen-14
blsea [12.9K]

Answer:

The beta decay takes place.

Explanation:

The reaction of radioactivity of carbon 14 to nitrogen 14 is

There is a beta decay.  

The reaction is

C_{6}^{14}\rightarrow N_{7}^{14}+\beta _{-1}^{0}+ energy

Here some energy is released in form of neutrino.

7 0
2 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
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