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1 year ago

Calculate the current through the ammeter (look at photo)

Physics

1 answer:

AleksAgata [21]1 year ago

8 0

Answer:

6 amps

Explanation:by Kirchhoff's loop rule the current at any point in the loop must be equal or charge would be building up. The current at the ammeter is equally to the total current through the sun of the paths in parallel which it is in series with

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Consider the following statements:

valentinak56 [21]

Answer:

1. A only

Explanation:

First law of thermodynamics it sates that energy of the universe is constant.It is also known as energy conservation law.

Therefore according to the first law of thermodynamics ,the thermal energy of the system is unchanged.

Second law of thermodynamics states that ,it is impossible to make a device which take heat from higher temperature energy resource and give 100 % work without heat rejecting.

Therefore only option A is correct.

1. A only

4 0

2 years ago

why is the earth more attracted to the sun gravitational pull than the gravitational pull of the moon

Ierofanga [76]

First, let's take a look at the equation for the force of gravity between two objects:

F = (GMm)/r², where,

G = gravitational constant = 6.67 x 10⁻¹¹
M = mass of one object
m = mass of the other object
r = distance between the two objects

From this equation, we can see that the force of gravity is directly proportional to the mass of the two objects and inversely proportional to the distance between them. We can then say that the Earth is more attracted to the sun than the moon because of the massive mass of the Sun (1.9891 x 10³⁰) compared to moon (7.3577 x 10²²). Although, the moon is nearer to the Earth, it has little effect to bring down the gravitational pull of the Sun.

5 0

2 years ago

When an object is placed 110 cm from a diverging thin lens, its image is found to be 55 cm from the lens. The lens is removed, a

Kazeer [188]

Explanation:

Given that,

Object distance u= -110 cm

Image distance v= 55 cm

We need to calculate the focal length for diverging lens

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-f}=\dfrac{1}{55}-\dfrac{1}{-110}$

$\dfrac{1}{f}=-\dfrac{3}{110}$

$f=-36.6\ cm$

The focal length of the diverging lens is 36.6 cm.

Now given a thin lens with same magnitude of focal length 36.6 cm is replaced.

Here, The object distance is again the same.

We need to calculate the image distance for converging lens

Using formula of lens

$\dfrac{1}{36.6}=\dfrac{1}{v}-\dfrac{1}{-110}$

Here, focal length is positive for converging lens

$\dfrac{1}{v}=\dfrac{1}{36.6}-\dfrac{1}{110}$

$\dfrac{1}{v}=\dfrac{367}{20130}$

$v=54.85\ cm$

The distance of the image is 54.85 cm from converging lens.

Hence, This is the required solution.

5 0

2 years ago

A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general

valentina_108 [34]

Answer:

Explanation:

(A)

The string has set of normal modes and the string is oscillating in one of its modes.

The resonant frequencies of a physical object depend on its material, structure and boundary conditions.

The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.

Given below are the incorrect options about the wave in the string.

• The wave is travelling in the +x direction

• The wave is travelling in the -x direction

• The wave will satisfy the given boundary conditions for any arbitrary wavelength $\lambda_i$

• The wave does not satisfy the boundary conditions $y_i(0;t)=0
$

Here, the string of length L held fixed at both ends, located at x=0 and x=L

The key constraint with normal modes is that there are two spatial boundary conditions, $y(0,1)=0
$

and $y(L,t)=0$

.The spring is fixed at its two ends.

The correct options about the wave in the string is

• The wavelength $\lambda_i$ can have only certain specific values if the boundary conditions are to be satisfied.

(B)

The key factors producing the normal mode is that there are two spatial boundary conditions, $y_i(0;t)=0$ and $y_i(L;t)=0$ , that are satisfied only for particular value of $\lambda_i$ .

Given below are the incorrect options about the wave in the string.

• $A_i$ must be chosen so that the wave fits exactly o the string.

• Any one of $A_i$ or $\lambda_i$ or $f_i$ can be chosen to make the solution a normal mode.

Hence, the correct option is that the system can resonate at only certain resonance frequencies $f_i$ and the wavelength $\lambda_i$ must be such that $y_i(0;t) = y_i(L;t)=0
$