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liubo4ka [24]
3 years ago
11

Why does the density of a substance remain the same for different amounts of the substance

Physics
1 answer:
Novay_Z [31]3 years ago
5 0
The molarity remains the same so the ratio does not change
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PLEASE HELP ME!!!!!!!!
Valentin [98]

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4 0
3 years ago
(Superposition, quadratic formuła) The Earth (mass M) is at origin, and the Moon (mass m ) is located at a distance d. (a) Using
jekas [21]

Answer:

a)    Fa = G m2 [M / r² - m / (d-r)²]  

b) r2 = 31 10⁶ m

Explanation:

The equation of the law of universal gravitation is

            F = G m1 m1/ r²

The value of the gravitation constant is 6.67 10-11 N m²/kg². This force is always attractive.

Let's calculate the value of that force on the spacecraft, add the strength

Earth's force to the ship

            F1 = G M m2 / r²

The moon force ship

            F2 = G m m2 / (d-r)²

Total force is

            Fa = F1 — F2

            Fa = G M M2 / r² - G m m2 / (d-r)²

            Fa = G m2 [M / r² - m / (d-r)²]

This is the force on the spaceship

b)  Let's look for the point where the force is zero, for this we can see that the value of the bracket must be zero

           Fa = 0

           [M / r² - m / (d-r)²] = 0

            M / r² = m / (d-r)²

           (d-r)² = m/M   r²

           d² -2rd + r² - m/M   r² = 0

           r² [m/M - 1] + r 2d - d² = 0

This is a second degree equation for r, we solve the to find the results.

           r = {-2d ±√[4d² - 4 [m/M -1] (-d²)]} / (2 [m/M-1])

           r = {-2d ± √ [4d² (1 + (m/M-1)]} / 2(m/M-1)

           r = {-2d ± 2d √(m/M)}  / (2(m/M-1))

           r = 2d {-1 ± √(m/M)} / 2(m/M-1)

           r = d [-1 ± √(m/M)] /  (m/M-1)

To find the explicit value we substitute the values ​​that we can find in tables

          m = 7.36 1022 kg

          M = 5.98 1024 kg

          d = 380000 km (1000m / 1 km) = 380 10⁶ m

          r = 380 10⁶ [-1 ±√(7.36 10²² / 5.98 10²⁴)] /(7.36 10²² / 5.98 / 10²⁴ -1)

          r = 380 10⁶ [-1 - √ (1.23 10²)] / (123-1)

          r = 380 106 [-1 ± 11] / 122

 

          r1 = 380 10⁶ 10/122 = 380 10⁶ 0.08197

          r1 =  31 10⁶ m

          r2 = 380 106 [-12/122] = 380 10 6 0.09836

          r2 = -37 10⁶ m

The correct distance is the positive r2 = 31 10⁶ m

c) let's use Newton's second law, to find the acceleration in the spacecraft

 

          F = m a

          a = Fa / m2 = G m2 [M / r² - m / (d-r)²] / m2

          a = G [M/r² - m/(d-r)²]

Since we have acceleration, we can use the definition of kinematics

           a = dv / dt = dv / dr dr / dt = dv / dr v

           v dv = a dr

           v dV = G [M /r² - m /(d-r)²] dr

We integrate

            ½ (V² - Vo²) = G [M (-1 /r) -m (1 / (d-r)

We evaluate between the initial point where we can assume that the initial velocity is zero for the Xo position and the final point with velocity v at the points

            V² = 2G [M (1 / Xo - 1 /X) - m (1 / (d-X) - 1 /(d-xo)]

            V² = [-M /X -m /(d-X)] 2G + constant

We now use the definition of speed

            v = dx / dt

            dx = V dt

We substitute, perform the integral and simplify, if we can make the constant zero

             dx = √([-M / X -m / (d-X)] 2G) dt

             dx / √([-M / X -m / (d-X)] 2G = dt

8 0
3 years ago
Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equ
sdas [7]

Answer:

x_{3}=2.35

Explanation:

Given x^2-2x-1=0,x_1=2

From Newton's method

x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}

f(x)=x^2-2x-1

f'(x)=2x-2

Now

x_{2}=x_1-\dfrac{f(x_1)}{f'(x_1)}

f(x)=x^2-2x-1

f(2)=2^2-2\times 2-1

f(x)=-1

f'(2)=2x-2

f'(1)=2\times 2-2

f'(1)=2

x_{2}=2+\dfrac{1}{2}

x_{2}=2.5

x_{3}=x_2-\dfrac{f(x_2)}{f'(x_2)}

f(2.5)=2.5^2-2\times 2.5-1

f(2.5)=0.45

f'(2.5)=2\times 2.5-2

f'(2.5)=3

x_{3}=2.5-\dfrac{0.45}{3}

So

x_{3}=2.35

8 0
3 years ago
The coefficient of friction on a surface is 0.25. A box requires 100N to slide it across the surface. What is the weight of the
Novosadov [1.4K]
So we know coefficient of f times normal force is friction. So do 100= .25 times x. Now solve for x. You get 400. So 400 is the normal force. And we know normal force equals weight in these types of problems so the answer is 400
5 0
3 years ago
A car travels 8km in 7 minutes. Find the speed of the car.
AleksandrR [38]

Answer:

42.6083 mi/h

Explanation:

Given: A car travels 8km in 7 minutes.

To find: Find the speed of the car.

Formula: Speed = \frac{Distance}{Time}

Solution: Since the formula for the speed of an object (which is the car) is speed = distance ÷ time, divide the distance (8km) by the time (7min)

Speed = 42.6083 miles per hour

3 0
3 years ago
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