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alexandr1967 [171]

3 years ago

12

Most houses in the countries that experience winter have air heaters installed.These heaters are usually installed near the floo

r.Explain why.

1 answer:

Tatiana [17]3 years ago

6 0

Answer: because the air currents move upwards so the hot air fills up the whole room. Similarly if the air conditioner is placed neer the ceiling because it gives cool air

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The spectrum of a distant star shows that one in 2 e6 of the atoms of a particular element is in its first excited state 7.5 eV

Alchen [17]

Answer:

The temperature of star is 5473.87 K

Explanation:

Given:

Energy difference $\Delta E = 7.5$ eV

The ratio of number of particle $\frac{N_{f} }{N_{i} } = \frac{1}{2 \times 10^{6} }$

Degeneracy ratio $\frac{g_{f} }{g_{i} } = 4$

From the formula of boltzmann distribution for population levels,

$\frac{N_{f} }{N_{i} } =\frac{g_{f} }{g_{i} } e^{-\frac{\Delta E}{kT} }$

Where $k =$ boltzmann constant = $8.62 \times 10^{-5} \frac{eV}{K}$

$\frac{1}{2 \times 10^{6} } =4 e^{-\frac{7.5 eV}{8.62 \times 10^{-5} T} }$

$8 \times 10^{6} } = e^{\frac{7.5 eV}{8.62 \times 10^{-5} T} }$

$\ln(8 \times 10^{6}) = {\frac{7.5 eV}{8.62 \times 10^{-5} T} }$

$T = {\frac{7.5 eV}{8.62 \times 10^{-5} \ln(8 \times 10^{6})} }$

$T = 5473.87$ K

Therefore, the temperature of star is 5473.87 K

5 0

3 years ago

Why does the light from the touch reach upto only certain distance

ad-work [718]

It is also likely (but not certain) that the photons will be absorbed by atoms. ... Light particles( or photons) never”run out” or loose their energy, so they can go an infinite distance, or until it reaches an object, that reflects the light or obsorbs it. Ie, a planet, or a mirror.

6 0

4 years ago

Breanna is standing beside a merry-go-round pushing 19° from the tangential direction and is able to accelerate the ride and her

leva [86]

To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.

Torque in a body is defined as,

$\tau_l = F*d$

And in angular movement like

$\tau_a = I*\alpha$

Where,

F= Force

d= Distance

I = Inertia

$\alpha =$ Acceleration Angular

PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.

$\tau= F*cos(19)*d$

On the other hand we have the speed data expressed in RPM, as well

$\omega_f = 10rpm = 10\frac{1rev}{1min}(\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 1.0471rad/s$

Acceleration can be calculated by

$\alpha = \frac{\omega_f}{t}$

$\alpha = \frac{1.0471}{9}$

$\alpha = 0.11rad/s^2$

In the case of Inertia we know that it is equivalent to

$I = \frac{1}{2}mr^2 = \frac{1}{2}(750)*(2.3)^2$

$I = 1983.75kg.m^2$

Matching the two types of torque we have to,

$\tau_l=\tau_a$

$Fd=I\alpha$

$Fcos(19)*2.3=1983.75(0.11)$

$F=100.34N$

PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,

$W = \frac{1}{2}I\omega_f^2$

$W= \frac{1}{2}(1983.75)(1.0471)^2$

$W=1087.51J$

7 0

3 years ago

An object elongates from a length of 45 cm to a length of 55 cm. The percent strain is

suter [353]

Answer:

22%

Explanation:

55 - 45 = 10

10/45 simplifies to 2/9

2/9 = 0.22222... so 22.22% (rounded 22%)

5 0

3 years ago

A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t

tresset_1 [31]

Answer:

Part a)

$Reading = 2.00 kg$

Part b)

$Reading = 2.00 kg$

Part c)

$Reading = 4.04 kg$

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part b)

When elevator is decending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

$F_{net} = ma$

$T - mg = ma$

$T = mg + ma$

Reading is given as

$Reading = \frac{mg + ma}{g}$

$Reading = 2.00\frac{9.81 + 10}{9.81}$

$Reading = 4.04 kg$

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

3 0

3 years ago