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Andreyy89
3 years ago
5

a resistance of 30 ohm is placed in a circuit with 90 volt battery what current Flows In the circuit​

Physics
1 answer:
guajiro [1.7K]3 years ago
5 0

Answer:

3amperes

Explanation:

current = voltage ÷ resistance = 90÷30= 3amperes

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An athlete runs 200 m by running 100 m in a straight line and then turning around and running 100m back to the start point.
siniylev [52]

Answer:

o

Explanation:

The athlete ran a total distance of zero because they ran 100m forward then turned around so they went back to their starting position

6 0
3 years ago
What are precautions in a simple pendulum experiment ​
monitta

Explanation:

1. draught

2. Parallax error

3. angle if displacement

4. air resistance or any form of obstruction

6 0
3 years ago
Find the following answer based on the image.
saul85 [17]

<u>We are given:</u>

Mass of Neptune = 1.03 * 10²⁶ kg

Distance from the center of Neptune (r) = 2.27 * 10⁷

now, computing the value of the acceleration due to gravity (g)

<u>Finding g:</u>

We know the formula:

g = G(mass of planet) / (r)²

g = [6.67 * 10⁻¹¹ * 1.03*10²⁶] / (2.27*10⁷)                      [since G is 6.67*10⁻¹¹]

g = (6.87 * 10¹⁵) / (5.15 * 10¹⁴)

which can be rewritten as:

g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15

g = (6.87 * 10¹⁵⁻¹⁴) / 5.15

g = (6.87/5.15) * 10

g = 1.34 * 10

g = 13.4 m/s² <em>(approx)</em>

5 0
3 years ago
An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
Veronika [31]

Answer:

s=5.79\ km

\theta=47^{\circ} east of south

Explanation:

Given:

  • distance of the person form the initial position, d'=8.4\ km
  • direction of the person from the initial position, 47^{\circ} north of east
  • distance supposed to travel form the initial position, d=5.3\ km
  • direction supposed to travel from the initial position, due North

<u>Now refer the schematic for visualization of situation:</u>

y=d'.\sin47^{\circ}-d

y=8.4\times \sin47-5.3 ...............(1)

x=d'.\cos47^{\circ}

x=8.4\times \cos47^{\circ} .................(2)

<u>Now the direction of the desired position with respect to south:</u>

\tan\theta=\frac{y}{x}

\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}

\theta=47^{\circ} east of south

<u>Now the distance from the current position to the desired position:</u>

s=\sqrt{x^2+y^2}

s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}

s=5.79\ km

4 0
3 years ago
A 400.0 kg storage box is held 10 m above ground by a forklift. What is its gravitational potential energy? (PE = mgh)
castortr0y [4]

Answer: D.) 39,200 J

Via the equation of potential energy PE = mgh where m is mass, g is the average gravity on earth and h is the height.  In this case m = 400 kg, g = 9.8, h = 10 m thus:

P.E.=(400kg)(9.8\frac{m}{s^2} )(10m)=39,200 J

P.E.= 39,200 Joules

7 0
3 years ago
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