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Aloiza [94]
3 years ago
10

The gravitational force between two objects is 2400 N. What will be the gravitational force between the objects if the mass of o

ne object is doubled
600N
1200N
4800N
9600N
Physics
1 answer:
motikmotik3 years ago
6 0

Answer:

4800N

Explanation:

Lets assume,

Mass of first object = m₁

Mass of second object = m₂

Distance between the two objects = r

Thus the force between the two objects will be

F = \frac{G\times m_{1}\times m_{2}}{r^{2}}

where, G = Universal gravitational constant

Given, F = 2400N

New mass of second object = 2m₂

Now, the force will be

F_{2} = \frac{G\times m_{1}\times 2m_{2}}{r^{2}}

F_{2}= 2\frac{G\times m_{1}\times m_{2}}{r^{2}}

F_{2}= 2F

F_{2}= 2\times2400

Thus, F₂ = 4800N

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A 1000 kg rocket carrying 25 kg of fuel and oxygen rises at a velocity of 305 m/s. If all the mass of fuel and oxygen is burned
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If all the mass of fuel and oxygen is burned to form gases of combustion, the downward velocity of these gases would be 12,505 m/s.

<h3>Conservation of linear momentum</h3>

The principle of conservation of linear momentum states that, the total momentum of an isolated system is constant.

The downward velocity of thes gases is calculated as follows;

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305(1000 + 25) = v2(25)

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Thus, If all the mass of fuel and oxygen is burned to form gases of combustion, the downward velocity of these gases would be 12,505 m/s.

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In an Atwood machine the two masses (m1 =5.0kg and m2 = 3.0 kg) are released from rest, with m1 at a height of h=0.75 m above th
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Answer:

Explanation:

m₁ is heavy so it will go down and m₂ will go up. Let common acceleration be a .

For downward acceleration of m₁

m₁g - T₁ = m₁a

For upward acceleration of m₂

T₂- m₂g = m₂a

adding

m₁g - T₁ +T₂- m₂g = (m₁+m₂)a

m₁g- m₂g  +(T₂- T₁) = (m₁+m₂)a  --------------- ( 1 )

For rotatory motion of disc of mass mp and moment of inertia I

(T₁ -T₂ )R = I α , α is angular acceleration ----------( 2 )

From  1 and 2

m₁g- m₂g -  I α / R = (m₁+m₂)a

m₁g- m₂g -  I a / R² = (m₁+m₂)a

m₁g- m₂g = (m₁+m₂)a + I a / R²

m₁g- m₂g = [(m₁+m₂) + mp k² / R² ] a   ( k is radius of gyration of disk )

a = (m₁- m₂)g / [(m₁+m₂) +mp k² / R² ]

=  (m₁- m₂)g / [(m₁+m₂) + .5 mp  ]  ( for cylinder k² / R² = .5 )

If h be the height by which m₁ falls and its velocity becomes v

v² = 2 a h

a = v² / 2h

v² / 2h = (m₁- m₂)g / [(m₁+m₂) + .5 mp  ]

v²[(m₁+m₂) + .5 mp  ] = 2(m₁- m₂) gh

- v²[(m₁+m₂) +2(m₁- m₂) gh =  .5 mp v²

- v²[(m₁+m₂) +2(m₁- m₂) gh / .5 v ² = mp .

- 1.8² ( 5 + 3 ) + 2 ( 5 - 3 ) 9.8 x .75 / .5 x 1.8² = mp

- 25.92 + 29.4 / 1.62 = mp

2.15 kg = mp .

K.E  of masses = 1/2 x 5 x 1.8² + 1/2 x 3 x 1.8²

= 12.96 J

Rotational KE of disc

= 1/2 I ω²

= 1/4 mp r²ω²

= 1/4 x mp x v²

= .25 x 2.15 x 1.8²

= 1.74 J

in percent terms

(1.74 / 12.96) x 100

= 13.5 %

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3 years ago
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