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yan [13]
2 years ago
5

Pleeease help meeeeee

Physics
1 answer:
VLD [36.1K]2 years ago
5 0
The average speed is 52km
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An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage of 120 V to a voltage of 5.00 V. T
Maksim231197 [3]

Answer:

35

Explanation:

We are given that

Initial voltage,V_1=120 V

Final voltage, V_2=5 V

Number of tuns in primary coil of the transformer, N_p=840

Rms current, I_{rms}=580mA=580\times 10^{-3} A

1 mA=10^{-3} A

We have to find the number of turns  are there on the secondary coil.

We know that

\frac{N_s}{N_p}=\frac{V_2}{V_1}

Using the formula

\frac{N_s}{840}=\frac{5}{120}

N_s=\frac{5}{120}\times 840=35

Hence, there are  number of turns on the secondary coil=35

8 0
3 years ago
30. The length of mercury thread when it is at 0°C, 100°C and at an unknown temperature 0 is 25mm, 225mm and 175mm respectively.
yulyashka [42]

Answer:

75°C

Explanation:

175−25/225−25=x−0/100−0

150/200=x/100

x=150×100/200

= 75°C

3 0
3 years ago
We know today that atoms cannot be divided into smaller parts true or false
djverab [1.8K]
Hi , the answer is false ,atoms can be divided into smaller parts , electrons , protons and neutrons.
3 0
3 years ago
The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.
Vitek1552 [10]

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

  • The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, <em>C</em> represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}

The steepness of the slope is therefore;

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100

Where;

\overline{AM} = Half the width of the truck = \dfrac{2.4 \, m}{2} = 1.2 m

\overline{CM} = The elevation of the center of gravity above the ground = 2.2 m

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%

tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}

Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right)  \approx 28.6 ^{\circ}

The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.

Learn more here:

brainly.com/question/20793607

3 0
2 years ago
PLEASE HELP
Oksana_A [137]

Answer:

The answer is A

Explanation: hope this helps :)

5 0
2 years ago
Read 2 more answers
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