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umka2103 [35]
4 years ago
13

Two objects attract each other gravitationally with a force of 2.5×10−10 n when they are 0.25 m apart. their total mass is 4.40

kg . find their individual masses.
Physics
1 answer:
FrozenT [24]4 years ago
4 0
Let the two masses be m₁ and m₂ kg.

Because their combined mass is 4.40 kg, therefore
m₁ + m₂ = 4.4                (1)

The force of attraction between the masses, when separated by d = 0.25 m, is
F = 2.5 x 10⁻¹⁰ N.
According to Newton's Law of gravitational attraction,
F = \frac{G m_{1} m_{2}}{d^{2}}
where
G = 6.673 x 10⁻¹¹ (N-m²)/kg²

Therefore
\frac{6.673 \times 10^{-11} m_{1}m_{2}}{0.25^{2}} =2.5 \times 10^{-10}
m₁m₂ = 0.2342            (2)
That is,
m₂ = 0.2342/m₁           3)

Substitute (3) into (1).
m₁ + 0.2342/m₁ = 4.4
m₁² + 0.2342 = 4.4m₁
m₁² - 4.4m₁ + 0.2342 = 0

Solve with the quadratic formula.
m₁ = 0.5[4.4 +/- √(18.423 )]
      = 4.346 or 0.0539 kg

When m₁ = 4.346, then m₂ = 0.2342/4.346 = 0.0539 
When m₁ = 0.0539, then m₂ = 0.2342/0.0539 = 4.346

Answer: 4.346 kg and 0.054 kg

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otez555 [7]

Answer:

<em>Both vehicles move east at 3.97 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states that the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is:

P=mv.

If we have a system of two bodies, then the total momentum is the sum of both momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

Assume both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

Assuming east direction to be positive, we have an m1=1459 kg car traveling west at v1=-43 m/s. An m2=9755 kg truck is traveling east at v2=11 m/s. They collide head-on and stick together after that.

Computing the resultant velocity after the collision:

\displaystyle v'=\frac{1459*(-43)+9755*11}{1459+9755}

\displaystyle v'=\frac{44568}{11214}

v' = 3.97 m/s

Both vehicles move east at 3.97 m/s

4 0
3 years ago
Gravity can be described as?
torisob [31]
A constant? A force? Sorry... Not too sure what you're asking.
7 0
3 years ago
A different scaffold that weighs 400 N supports two painters, one 500 N and the other 400 N. The reading in the left scale is 80
fredd [130]

Answer:

500 N

Explanation:

Given that,

The upward force is 800 N and the downward forces are 400 N, 500 N, 400 N.

At equilibrium, the upward forces will become equal to the downward forces. Let the reading in the right hand scale.

x + 800 = 400 + 500 + 400

x + 800 = 1300

x = 1300 - 800

= 500 N

So, the reading in the right hand scale is 500 N.

3 0
3 years ago
If the current in the circuit decreases, what does that mean about the rate at which the charge(and voltage) change in a capacit
nasty-shy [4]

Answer:

`1. charge Q, on the capacitor increases, while the current will decrease

2. τ = t = secs

Explanation:

1. consider RC  of a circuit to be am external source

voltage across the circuit is given as

v =v₀(1 - e^{\frac{t}{τ} })

where v = voltage

v₀ = peak voltage

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as the charge across the capacitor increases, current decreases

the charge across the circuit is given as

Q= Q₀(1 - e^{\frac{t}{τ} })

charge Q is inversely proportional to the current I

hence the charge across the circuit increases

2. τ = RC

unit of time constant, τ,

= Ω × F

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5 0
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The transformation equation fom potential to kinetic energy is =

m×g×h = \frac{1}{2} mv^{2} + \frac{1}{2} (\frac{2}{5} mr^{2} )(\frac{v}{r}) ^{2}

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the first is the sphere 1 followed by the disc 2 then the hoop 3

the correct order is a, 1, 2, 3

8 0
4 years ago
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