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Aleonysh [2.5K]

3 years ago

12

The energy transfer diagram shows energy transfer in an MP3 player. Useful energy is transferred away from the MP3 player by lig

ht and what else

1 answer:

OLEGan [10]3 years ago

4 0

Answer:

heat and sound.

Explanation:

Though some would argue that the heat is not useful. I guess it depends on if your hands are cold.

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A 167-? resistor and a 114-? resistor are both connected in parallel across a power supply. If the potential drop across the 167

Troyanec [42]

19-? Is the exact p.d across the 114-?resistor.
Current will different
But p.d will same in parallel circuit .

7 0

3 years ago

Lab: Magnetic and Electric Fields lab report. Edge 2020

Lubov Fominskaja [6]

Answer:

what is the question

Explanation: plz give the brainliest

3 0

3 years ago

A rubber-wheeled 50 kg cart rolls down a 35 degree concrete incline. Coefficient of rolling friction between rubber and concrete

lapo4ka [179]

Answer:

(a) 5.62 m/s²

(b) 5.46 m/s²

Explanation:

The given values are:

mass,

m = 50 kg

angle

= 35°

(a)

If friction neglected,

⇒ $F_x=mgSin \theta=ma$

⇒ $a_x=gSin \theta$

$=9.8 \ Sin35^{\circ}$

$=5.62 \ m/s^2$

(b)

If friction present,

⇒ $F_x=mgCos \theta$

⇒ $F_x=mgSin \theta-\mu_r mgCos \theta$

⇒ $a=gSin \theta-\mu_rgCos \theta$

$=9.8 \ Sin35^{\circ}-0.02\times 9.8 \ Cos30^{\circ}$

$=5.46 \ m/s^2$

8 0

3 years ago

A jogger runs at a speed of 3 m/s. How far does he run in 120 seconds?<br>

Licemer1 [7]

Answer: *360 mph*

Explanation:

I am pretty sure that it is 360 mph

3 times 120 = 360

8 0

3 years ago

The main water line enters a house on the first floor. The line has a gauge pressure of 2.10 x 105 Pa. (a) A faucet on the secon

Eva8 [605]

To answer this question is necessary to apply the concepts related to Bernoulli's equation. The Bernoulli-related concept describes the behavior of a liquid moving along a streamline. Pressure can be defined as the proportional ratio between height, density and gravity:

$P = h\rho g$

Where,

h = Height

$\rho$ = Density

g = Gravity

Our values are

$\rho = 1000kg/m^3 \rightarow$ density of water at normal conditions

h = 7.3m

$g = 9.8m/s^2$

PART A) Replacing these values to find the total pressure difference we have to

$P_1 = h_1 \rho g$

$P_1 = (7.3)(1000)(9.8)$

$P_1 = 71540Pa$

In this way the pressure change would be subject to

$\Delta = P_2-P_1$

$\Delta = 2.1*10^5Pa- 0.7154*10^5Pa$

$\Delta = 138460Pa$

$\Delta = 0.135Mpa$

PART B) Considering the pressure gauge of the group as the ideal so that at a height H the water cannot flow even if it is open we have to,

$P_2 = H\rho g$

$2.1*10^5 = H (1000)(9.8)$

$H = 21.42m$

Therefore the high which could a faucet be before no water would flow from it is 21.42m

5 0

3 years ago