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babunello [35]
3 years ago
9

You've been tired and lethargy what are two possible reasons?

Physics
1 answer:
dybincka [34]3 years ago
5 0

Explanation:

physical exertion.

lack of physical activity.

lack of sleep.

being overweight or obese.

periods of emotional stress.

boredom.

grief.

taking certain medications, such as antidepressants or sedatives.

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Need help on this please
marin [14]
The answer is D, the amount of energy stays the same.
3 0
3 years ago
As a rain storm passes through a region, there is an associated drop in atmospheric pressure. If the height of a mercury baromet
Pani-rosa [81]

Answer:

new atmospheric pressure is 0.9838 × 10^{5}  Pa

Explanation:

given data

height = 21.6 mm = 0.0216 m

Normal atmospheric pressure = 1.013 ✕ 10^5 Pa

density of mercury = 13.6 g/cm³

to find out

atmospheric pressure

solution

we find first height of mercury when normal pressure that is

pressure p = ρ×g×h

put here value

1.013 × 10^{5}  = 13.6 × 10³ × 9.81 × h

h =  0.759 m

so change in height Δh = 0.759 - 0.0216

new height H = 0.7374 m

so new pressure = ρ×g×H

put here value

new pressure = 13.6 × 10³ × 9.81 × 0.7374

atmospheric pressure = 98380.9584

so new atmospheric pressure is 0.9838 × 10^{5}  Pa

6 0
3 years ago
I)
Natalija [7]

Answer:

I think its (A)The specific heat of an object explains  how easily it changes temperatures.

Explanation:

6 0
3 years ago
A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of grav
erastovalidia [21]

I attached a Diagram for this problem.

We star considering the system is in equlibrium, so

Fm makes 90-(\theta+5) with vertical

Fm makes 70 with vertical

Applying summatory in X we have,

\sum F_x = 0

W+1.4-Fm cos(70)

We know that W is equal to

W= 0.06*100N = 6N

Substituting,

Fm cos (70) = W+1.4N

Fm cos (70) = 6N + 1.4N

Fm = \frac{7.4}{cos(70)}

Fm = 21.636N

<em>For the second part we know that the reaction force Fj on deltoid Muscle is equal to Fm, We can assume also that</em> \beta  = \theta

3 0
3 years ago
7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf
Serga [27]

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to  1 μV/m

Intrinsic impedance of air = 120 π  Ω

Intrinsic impedance of  water = \dfrac{120\pi}{\epsilon_r}

Using equation to solve the problem

  E(z) = E_0 e^{-\alpha\ z}

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}

\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}

\alpha =21.07\ Np/m

now ,

  1 \times 10^{-6} = 20 e^{-21.07\times z}

  e^{21.07\times z}= 20\times 10^{6}

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

5 0
3 years ago
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