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ASHA 777 [7]
3 years ago
8

Use the circuit to answer the following questions.

Physics
1 answer:
Zanzabum3 years ago
7 0

Answer:

1)ammeter

2)ised to check measure of current flow through a circuit

3)o.90 ambere

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At what point will the electric field of a charged object be strongest?
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Answer:

The answer is C because they have to be close to be able to interact

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3 years ago
Pls help it’s easy ik but I want to make sure of them both I don’t have much time
LuckyWell [14K]

Answer:

To the right

Weight

Explanation:

8 0
2 years ago
A gas occupies a volume of 20 cubic meters at 9,000 pascals. If the pressure is lowered to 5,000 pascals, what volume will the g
forsale [732]
We need to consider no change in the temperature of gas (isothermal transformation)

Volume and pressure are inversely proportional magnitudes, so we can write:

P_1.V_1=P_2.V_2\\
\\
9.20=5.V_2\\
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V_2=\frac{180}{5}=36 \ m^3
5 0
3 years ago
A boy of mass 40kg eats bananas contains of 980 joule. If this energy is used to lift him up from ground,the height to which he
Oksanka [162]

Answer:

h = 2.49 [m]

Explanation:

In order to solve this problem we must use the definition of potential energy, which tells us that energy is equal to the product of mass by gravity by height.

The potential energy can be calculated by means of this equation:

Ep = m*g*h

where:

Ep = potential energy = 980 [J]

m = mass = 40 [kg]

g = gravity acceleration = 9.81 [m/s^2]

h = elevation [m]

Now replacing:

980 = 40*9.81*h

h = 2.49 [m]

7 0
3 years ago
A planet is discovered orbiting around a star in the galaxy Andromeda at four times the distance from the star as Earth is from
Dmitriy789 [7]

Answer:

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

Explanation:

The orbital period of a planet around a star can be expressed mathematically as;

T = 2π√(r^3)/(Gm)

Where;

r = radius of orbit

G = gravitational constant

m = mass of the star

Given;

Let R represent radius of earth orbit and r the radius of planet orbit,

Let M represent the mass of sun and m the mass of the star.

r = 4R

m = 16M

For earth;

Te = 2π√(R^3)/(GM)

For planet;

Tp = 2π√(r^3)/(Gm)

Substituting the given values;

Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)

Tp = 2π√(4R^3)/(GM)

Tp = 2 × 2π√(R^3)/(GM)

So,

Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

5 0
3 years ago
Read 2 more answers
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