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torisob [31]
3 years ago
7

What happens to an object when you halve the mass? Hurry please !

Physics
1 answer:
Vika [28.1K]3 years ago
5 0

Answer:

It means that if force is constant, as mass is increased, acceleration decreases

Explanation:

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A 3.35 kg object initially moving in the positive x direction with a velocity of 4.90 m s collides with and sticks to a 1.88 kg
ahrayia [7]

Answer:

The final components of velocity of the composite object is 3.33 m/s.

Explanation:

Given;

mass of the first object, m₁ = 3.35 kg

initial velocity of the first object, u₁ = 4.90 m/s in positive x-direction

mass of the second object, m₂ = 1.88 kg

initial velocity of the second object, u₂ = 3.12 m/s in negative y-direction

initial momentum of the first object, P₁ = 3.35 x 4.9 = 16.415 kgm/s

initial momentum of the second object, P₂ = 1.88 x 3.12 = 5.8656 kgm/s

The resultant velocity of the two objects is given by;

R² = 16.415² + 5.8656²

R² = 303.858

R = √303.858

R = 17.432 kgm/s

Apply the principle of conservation of linear momentum for inelastic collision;

total initial momentum before = total final momentum after collision

P₁(x) + P₂(y) = Pf

R = Pf

R = v(m₁ + m₂)

17.432 = v(m₁ + m₂)

where;

v is the final components of velocity of the composite object

v = \frac{17.432}{m_1 + m_2} \\\\v = \frac{17.432}{3.35+1.88} \\\\v = 3.33 \ m/s

Therefore, the final components of velocity of the composite object is 3.33 m/s.

8 0
2 years ago
A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of
Paul [167]

Answer:

<em>The height of water in the column = 348.14 cm</em>

Explanation:

<em>Pressure:</em><em>This is defined as the ratio of the force acting normally ( perpendicular) to the area of surface in contact. The S.I unit of  pressure is N/m²</em>

<em>p = Dgh............... Equation 1</em>

<em>Where p = pressure, D = density, g = acceleration due to gravity, h = height.</em>

<em>From the question, the same pressure will support the column of mercury and water.</em>

<em>p₁ = p₂</em>

<em>Where p₁ = pressure of mercury, p₂ = pressure of water</em>

D₁gh₁ = D₂gh₂.................. Equation 2

making h₂ the subject of equation 2

h₂ = D₁gh/D₂g............... Equation 3

Where D₁ and D₂ = Density of mercury and water respectively, h₁ and h₂ = height of mercury and water respectively

Given: D₁ = 13.6 g/cm³, D₂ = 1.00 g/cm³, h₁ = 256 mm = 25.6 cm.

Constant: g = 9.8 m/s²

Substituting these values into Equation 3,

h₂ = (13.6×9.8×25.6)/1×9.8

<em>h₂ = 348.14 cm</em>

<em>The height of water in the column = 348.14 cm</em>

6 0
3 years ago
Does the refraction of light make a swimming pool seem deeper or shallower? explain your answer.
larisa [96]

The refraction of light makes a swimming pool seem <u>shallower</u>.

The swimming pool seems shallower because the rays of light coming from the bottom of the pool do not come with a straight path. The path of light is straight as long as it is in the water.

When lights come out of the water into the air it bents downwards. This bending is called refraction.

Refraction forms a virtual image of the pool and it seems shallower than it actually is to the observer. This only happens when light travels from one transparent medium into another having lower density.

If you need to learn more about why a swimming pool appears <u>shallower</u>, click here

https://brainly.in/question/7136803?referrer=searchResults

#SPJ4

6 0
1 year ago
Joe moved to a new town and was having a hard time making friends. Recently he met some people he really liked at a party. They
siniylev [52]
Joey should say no to drugs next time he is offered to smoke weed
3 0
3 years ago
Part complete Sound with frequency 1240 Hz leaves a room through a doorway with a width of 1.11 m . At what minimum angle relati
Sedbober [7]

Answer:

14.43° or 0.25184 rad

Explanation:

v = Speed of sound in air = 343 m/s

f = Frequency = 1240 Hz

d = Width in doorway = 1.11 m

Wavelength is given by

\lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{343}{1240}\\\Rightarrow \lambda=0.2766\ m

In the case of Fraunhofer diffraction we have the relation

dsin\theta=\lambda\\\Rightarrow \theta=sin^{-1}\frac{\lambda}{d}\\\Rightarrow \theta=sin^{-1}\frac{0.2766}{1.11}\\\Rightarrow \theta=14.43^{\circ}\ or\ 0.25184\ rad

The minimum angle relative to the center line perpendicular to the doorway will someone outside the room hear no sound is 14.43° or 0.25184 rad

6 0
3 years ago
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