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1 year ago

What happens to the force between two objects if the mass of one object is doubled.

Physics

1 answer:

irinina [24]1 year ago

8 0

If the mass of one object is doubled, the force between these objects will also double.

Force refers to an external agent which is capable of changing the state of a body which may be at rest or in motion.

Force F = G(mM/d^2) (where m is the mass of first object, M is the mass of the second object and G is the gravitational pull and d is the distance between the two objects)

The force between two objects (m1 and m2) (according to the universal law of gravitation) is proportional to their mass and reciprocally proportional to the square of their separation between them.

So F' = G(2mM/d^2),

which means F'=2F

Therefore, the when the mass is doubled, the force also doubles.

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3 years ago

A concave mirror with a radius of curvature of 20 cm has a focal length of

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Answer:

A concave mirror has a radius of curvature of 20 cm. What is it's focal length? If an object is placed 15 cm in front of it, where would the image be formed? What is it's magnification?

The focal length is of 10 cm, object distance is 30 cm and magnification is -2.

Explanation:

Given:

A concave mirror:

Radius of curvature of the mirror, as C = 20 cm

Object distance in-front of the mirror = 15 cm

Focal length:

Focal length is half of the radius of curvature.

Focal length of the mirror = $\frac{C}{2}$ = 10 cm

According to the sign convention we will put the mirror on (0,0) point, of the Cartesian coordinate open towards the negative x-axis.

Object and the focal length are also on the negative x-axis where focal length and image distance will be negative numerically.

We have to find the object distance:

Formula to be use:

⇒ $\frac{1}{focal\ length}= \frac{1}{image\ distance} + \frac{1}{object\ distance}$

⇒ Plugging the values.

⇒ $\frac{1}{-10} =\frac{1}{image\ distance}+\frac{1}{-15}$

⇒ $\frac{1}{-10} -\frac{1}{-15}=\frac{1}{image\ distance}$

⇒ $\frac{1}{-10} + \frac{1}{15}=\frac{1}{image\ distance}$

⇒ $\frac{-3+2}{30} =\frac{1}{image\ distance}$

⇒ $\frac{-1}{30} =\frac{1}{image\ distance}$

⇒ $-30\ cm=image\ distance$

Image will be formed towards negative x-axis 30 cm away from the pole.

Magnification (m) is the negative ratio of mage distance and object distance:

⇒ $m=-\frac{image\ distance}{object\ distance}$

⇒ $m=-\frac{(-30)}{(-15)}$

⇒ $m=-2$

The focal length of the concave mirror, is of 10 cm, object distance is 30 cm and magnification is -2.

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3 years ago

An incident ray strikes a mirror with an angle of 30 degrees to the surface of the mirror. what is the angle of the reflected ra

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3 years ago

A ball is thrown upward with an initial vertical velocity of v0 to a maximum height of hmax and falls back toward earth. on the

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Answer: $v_f=g\frac{t}{2}$

The ball was thrown at the speed of $v_o$ .

Maximum height achieved is $h_{max}$

Time of flight is t.

Now, the time the ball takes to achieve maximum height = the time taken by ball to fall back = $\frac{t}{2}$

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Using the equation of motion:

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3 years ago

Transverse waves are being generated on a rope under constant tension. By what factor is the required power increased or decreas

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Answer:

a) when the length of the rope is doubled and the angular frequency remains constant: The power increases by factor of two

b) when the amplitude is doubled and the angular frequency is halved: The power is the same

c) when both the wavelength and the amplitude are doubled: The power increases by a factor of 8

d) when both the length of the rope and the wavelength are halved: The power increases by factor of two

Explanation:

For a sinusoidal mechanical wave (Transverse wave), the time-averaged power is the energy associated with a wavelength divided by the period of the wave.

$P_{avg} =\frac{E \lambda}{T}=\frac{1}{2} \mu A^2 \omega^2 \frac{\lambda}{T}$

where;

A is the Amplitude

ω is the angular frequency

λ is the wavelength

a) when the length of the rope is doubled and the angular frequency remains constant

L = λ/2, λ = 2L

The power increases by factor of two

b) when the amplitude is doubled and the angular frequency is halved

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The power is the same

c) when both the wavelength and the amplitude are doubled

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The power increases by a factor of 8

d) when both the length of the rope and the wavelength are halved

L = λ/2

when both are halved

L/2 = λ/4, λ = 2L

The power increases by factor of two

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4 years ago