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Stells [14]
3 years ago
11

It is estimated that uranium is relatively common in the earth's crust, occurring in amounts of 4 g / metric ton. A metric ton i

s 1000 kg. At this concentration, what mass of uranium is present in 1.2 mg of the earth's crust?
Physics
1 answer:
timama [110]3 years ago
7 0

Answer:

The mass of Uranium present in a 1.2mg sample is 4.8 \cdot 10^{-6}\,mg

Explanation:

The ration between Uranium mass and total sample mass is: \frac{4g}{1000kg} =\frac{4g}{1000000g}=\frac{1}{250000}

For a sample of mass 1.2 mg, the amount of uranium is:

1.2\, mg \cdot \frac{1}{250000}=4.8 \cdot 10^{-6}\,mg

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You are planning to make an open rectangular box from an 8-inch by 15-inch piece of cardboard by cutting congruent squares from
Phoenix [80]

Let us say that x is the cut that we will make on the sides to make a box, therefore the new dimensions are:

l = 15 – 2x

w = 8 – 2x

It is 2x since we cut on two sides.

 

We know that volume is:

V = l w x

V = (15 – 2x) (8 – 2x) x

V = 120x – 30x^2 – 16x^2 + 4x^3

V = 120x – 46x^2 + 4x^3

 

Taking the 1st derivative:

dV/dx = 120 – 92x + 12x^2

 

Set dV/dx = 0 to get maxima:

120 – 92x + 12x^2 = 0

 

Divide by 12:

x^2 – (92/12)x + 10 = 0

(x – (92/24))^2 = -10 + (92/24)^2

x - 92/24 = ±2.17

x = 1.66, 6

We cannot have x = 6 because that will make our w negative, so:

x = 1.66 inches

 

So the largest volume is:

V = 120x – 46x^2 + 4x^3

V = 120(1.66) – 46(1.66)^2 + 4(1.66)^3

V = 90.74 cubic inches

4 0
3 years ago
When 1 kg of water and 1 kg of wood absorb the same amount of heat, the change in temperature of the wood is greater than the ch
Strike441 [17]

Intermolecular forces in water are greater than those in wood. APEX

6 0
3 years ago
There are four charges, each with a magnitude of 4.25 C. Two are positive and two are negative. The charges are fixed to the cor
VMariaS [17]

Answer:

 F = 7.68 10¹¹ N,  θ = 45º

Explanation:

In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges

The net force is

          F_ {net} = F₂₁ + F₂₃ + F₂₄

bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.

let's use trigonometry

          cos 45 = F₂₄ₓ / F₂₄

          sin 45 = F_{24y) / F₂₄

          F₂₄ₓ = F₂₄ cos 45

          F_{24y} = F₂₄ sin 45

let's do the sum on each axis

X axis

          Fₓ = -F₂₁ + F₂₄ₓ

          Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis  

         F_y = - F₂₃ + F_{24y}

         F_y = -F₂₃ + F₂₄ sin 45

They indicate that the magnitude of all charges is the same, therefore

         F₂₁ = F₂₃

Let's use Coulomb's law

         F₂₁ = k q₁ q₂ / r₁₂²

       

the distance between the two charges is

         r = a

         F₂₁ = k q² / a²

we calculate F₂₄

           F₂₄ = k q₂ q₄ / r₂₄²

the distance is

           r² = a² + a²

           r² = 2 a²

         

we substitute

           F₂₄ = k  q² / 2 a²

we substitute in the components of the forces

          Fx = - k \frac{q^2}{a^2} +  k \frac{q^2}{2 a^2}  \ cos 45

          Fx = k \frac{q^2}{a^2}  ( -1 + ½ cos 45)

          F_y = k \frac{q^2}{a^2} ( -1 +  ½ sin 45)    

         

We calculate

            F₀ = 9 10⁹ 4.25² / 0.440²

            F₀ = 8.40 10¹¹ N

       

            Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

            Fₓ = -5.43 10¹¹ N

         

remember cos 45 = sin 45

             F_y = - 5.43 10¹¹  N

We can give the resultant force in two ways

a) F = Fₓ î + F_y ^j

          F = -5.43 10¹¹ (i + j)   N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

          F = \sqrt{F_x^2 + F_y^2}

          F = 5.43 10¹¹  √2

          F = 7.68 10¹¹ N

in angle is

           θ = 45º

7 0
2 years ago
A brand new corvette can go from 0 to 85 miles per hour in 4.8 seconds.
klasskru [66]

Answer:

(A) 7.9 m/s^{2}  

(B) 19 m/s

(C) 91 m

Explanation:

initial velocity (U) = 0 mph = 0 m/s

final velocity (V) = 85 mph = 85 x 0.447 = 38 m/s

initial time (ti) = 0 s

final time (t) = 4.8 s

(A) acceleration = \frac{V-U}{t}

         = \frac{38-0}{4.8} = 7.9 m/s^{2}  

(B) average velocity = \frac{V+U}{2}

     =\frac{38+0}{2} = 19 m/s

(C) distance travelled (S) = ut + 0.5at^{2}

  =  (0 x 4.8) + 0.5 x 7.9 x 4.8^{2} = 91 m

5 0
3 years ago
If a car accelerates from a speed of 10 m/s at an acceleration of 2 m/s2 for 3s.
Flauer [41]

Answer:

\boxed{\sf Final \ speed \ of \ the \ car = 16 \ m/s}

Given:

Initial speed (u) = 10 m/s

Acceleration (a) = 2 m/s²

Time taken (t) = 3s

To Find:

Final speed (v) of the car

Explanation:

From equation of motion we have:

\boxed{ \bold{v = u + at}}

By substituting value of u, a & t in the equation we get:

\sf \implies v = 10  + 2\times 3 \\  \\ \sf \implies v = 10 + 6 \\  \\ \sf \implies v = 16 \: m {s}^{ - 1}

\therefore

Final speed (v) of the car = 16 m/s

3 0
3 years ago
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