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garri49 [273]
3 years ago
7

The chemical energy in your food is

Physics
1 answer:
ale4655 [162]3 years ago
8 0

Answer:

Electromanetic

Explanation:ESPERO TE AYUDE

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Calculate the pressure of water in a will if the deep of the water is 10 m​
Bas_tet [7]

Answer:

98,000 pa

Explanation:

The formula for water pressure is as follows:

pressure = pgh

Where <em>p </em>is the density of water (in kg/m3), <em>g </em>is the gravitational field strength, and <em>h </em>is the height of the water.

The density of water is 1000kg/m3, the gravitational field strength is 9.8, and the height is 10. Substituting in these values:

pressure = 1000 \times 9.8 \times 10

pressure = 98000

7 0
2 years ago
What do nebulae have the most in common with?
alexdok [17]
Most commonly the answer is A, the sun. A nebula is a nursery for stars. The Sun itself was created in a nebula. Also,you do not need college physics to understand this question


3 0
3 years ago
The length of the minute hand of the clock is 14cm. Calculate the speed with which the tip of the minute hand moves
Molodets [167]

Speed of the tip of the minute hand=V=0.0244 cm/s

Explanation:

The angular velocity of the minute hand is given by

\omega= \frac{2\pi}{T}

T= time period of the minute hand=60 min=3600 s

so ω= 2 π/3600 rad/s

Now linear velocity v= r ω

r= radius of minute hand=14 cm

so v= 14 (2 π/3600)

V=0.0244 cm/s

8 0
3 years ago
Efficiency is given by A. actual output divided by effective capacity. B. design capacity divided by utilization. C. effective c
jok3333 [9.3K]

Answer:

A

Explanation:

Actual output divided by the effective capacity. It is the ratio of output to effectiveness

5 0
3 years ago
Read 2 more answers
A generator with �# ' = 300 V and Zg = 50 Ω is connected to a load ZL = 75 Ω through a 50-Ω lossless line of length l = 0.15λ. (
ki77a [65]

Answer:

a. Zin = 41.25 - j 16.35 Ω

b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.  Pin = 216 w

d. PL = Pin = 216 w

e. Pg = 478.4 w , Pzg = 262.4 w

Explanation:

a.

Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]  

βl = 2π / λ * 0.15 λ = 54 °

Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]

Zin = 41.25 - j 16.35 Ω

b.

I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶

V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)

V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.

Pin = ¹/₂ * Re * [V₁ * I₁]

Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)

Pin = 216 w

d.

The power PL and Pin are the same as the line is lossless input to the line ends up in the load so

PL = Pin

PL = 216 w

e.

Pg Generator

Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)

Pg = 478.4 w

Pzg dissipated

Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50

Pzg = 262.4 w

4 0
3 years ago
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