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3 years ago

Based on Newtons law of universal gravitation, complete the following table.

Physics

2 answers:

madam [21]3 years ago

5 0

Gravity increases as mass increases.

Gravity decreases when distance decreases

I hope I help.

mart [117]3 years ago

4 0

Explanation:

According to the Newton's law of universal gravitation, the force between two bodies is given by :

$F=G\dfrac{m_1m_2}{d^2}$ ..........(1)

$m_1\ and\ m_2$ are masses

d is the distance between objects

From equation (1), the force of gravity increases when distance decreases or mass increases and the force of gravity decreases when mass decreases or distance increases.

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In order to increase the kinetic energy of a speeding train by 44%, 42MJ of work must be performed (1 MJ = 106 J). If the final

victus00 [196]

Answer:

Explanation:

100 % increase makes an amount 2 times .

44 % increase = 42 MJ

100 % increase = 95.45 MJ

final kinetic energy = 2 x 95.45 MJ

= 190.9 x 10⁶ J

1/2 m v² = 190.9 x 10⁶

.5 x m x 9² = 190.9 x 10⁶

m = 4.71 x 10⁶ kg

= 4710 metric ton .

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3 years ago

How does the movement of particles of matter change when matter goes from solid to liquid?

icang [17]

Solid to liquid: molecules spread out to have their own spaces and move energetic and lively.

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3 years ago

Express 2759 mockingbirds in kilomockingbirds. answer in units of kmockingbird

Nadya [2.5K]

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6 0

3 years ago

A basketball player throws the ball at a 47 angle above the horizontal to a hoop which is located a horizontal distance L = 5.0

FromTheMoon [43]

Answer:

$v_0 =$ 1.71

Explanation:

the parabolic movment is described by the following equation:

$y = tan(a)x-\frac{1}{2v_0^2(cos(a))^2}gx^2$

where y is the height of the ball, a is the angle of launch, $v_0$ the initial velocity, g the gravity and x is the horizontal distance of the ball.

So, if we want that the ball reach the hood, we will replace values on the equation as:

$0.8 = tan(47)(5)-\frac{1}{2v_0^2(cos(47))^2}(9.8)(5)^2$

Finally, solving for $v_0$ , we get:

$v_0=\sqrt{\frac{-9.8(5)^2}{(0.8-tan(47)(5))2cos^2(47)}}$

$v_0 =$ 1.71

4 0

4 years ago

Three liquids are at temperatures of 4 ◦C, 24◦C, and 29◦C, respectively. Equal masses of the first two liquids are mixed, and th

Scilla [17]

Answer:

T₁₃ = 24.1°C

Explanation:

Given

m = mass of each liquids (all masses are equal)

C₁ = specific heat of the first liquid

C₂ = specific heat of the second liquid

C₃ = specific heat of the third liquid

T₁ = 4°C (temperature of the first liquid)

T₂ = 24°C (temperature of the second liquid)

T₃ = 29°C (temperature of the third liquid)

Temperature of 1+2 liquids mix: T₁₂ = 21°C

Temperature of 2+3 liquids mix: T₂₃ = 26.1°C

Temperature of 1+3 liquids mix: T₁₃ = ?

We can apply the relation ∑Q = 0

We assume the system is isolated and the process is adiabatic.

Mix 1:

Q₁ + Q₂ = 0

where

Q₁ = m*C₁*(T₁-T₁₂)

and

Q₂ = m*C₂*(T₂-T₁₂)

then

m*C₁*(T₁-T₁₂) + m*C₂*(T₂-T₁₂) = 0

⇒ C₁*(T₁-T₁₂) + C₂*(T₂-T₁₂) = 0

⇒ (4°C-21°C)*C₁ + (24°C-21°C)*C₂ = 0

⇒ -17°C*C₁ + 3°C*C₂ = 0

⇒ C₁ = (3/17)*C₂ = 0.176*C₂ (i)

Mix 2:

Q₂ + Q₃ = 0

where

Q₂ = m*C₂*(T₂-T₂₃)

and

Q₃ = m*C₃*(T₃-T₂₃)

then

m*C₂*(T₂-T₂₃) + m*C₃*(T₃-T₂₃) = 0

⇒ C₂*(T₂-T₂₃) + C₃*(T₃-T₂₃) = 0

⇒ (24°C-26.1°C)*C₂ + (29°C-26.1°C)*C₃ = 0

⇒ -2.1°C*C₂ + 2.9°C*C₃ = 0

⇒ C₃ = 0.724*C₂ (ii)

Mix 3:

Q₁ + Q₃ = 0

where

Q₁ = m*C₁*(T₁-T₁₃)

and

Q₃ = m*C₃*(T₃-T₁₃)

then

m*C₁*(T₁-T₁₃) + m*C₃*(T₃-T₁₃) = 0

⇒ C₁*(T₁-T₁₃) + C₃*(T₃-T₁₃) = 0

⇒ (4°C-T₁₃)*C₁ + (29°C-T₁₃)*C₃ = 0

If we use the following relations C₁ = 0.176*C₂ and C₃ = 0.724*C₂ we obtain

(4°C-T₁₃)*0.176*C₂ + (29°C-T₁₃)*0.724*C₂ = 0

⇒ (4°C-T₁₃)*0.176 + (29°C-T₁₃)*0.724 = 0

⇒ 0.706°C - 0.176*T₁₃ + 21°C - 0.724*T₁₃ = 0

⇒ 0.9*T₁₃ = 21.706°C

⇒ T₁₃ = 24.1°C

5 0

3 years ago