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morpeh
6 days ago
14

a crane cable can support a maximum load of 15,000 kg. if a bucket has a mass of 2,000 kg and gravel has a mass of 1,500 kg for

every cubic meter, how many cubic meters of gravel (g) can be safely lifted by the crane?
Mathematics
2 answers:
tester [10.8K]6 days ago
7 0
2000 + 1500g ≤ 15000
1500g ≤ 15000 - 2000
1500g ≤ 13000
g ≤ 13000/1500
g ≤ 8 2/3

Thus, the crane is capable of safely lifting no more than 8 2/3 cubic meters of gravel.
Inessa [11.1K]6 days ago
3 0

Response:

26/3

Detailed breakdown:

I just completed it

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Answer:

w and y are the knights

Step-by-step explanation:

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1 month ago
Jerry buys a pack of pencils that cost $3 David buys 4 sets of markers each set of marker also cost 3 what is the total cost of
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The aggregate expense for the markers is 12.
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A round cake has a diameter of 30\text{ cm}30 cm30, start text, space, c, m, end text. Angela places the cake on a circular cake
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The cake board’s circumference measures 35π cm. Initially, noting the cake's diameter of 30 cm, the board's diameter is established as an additional 5 cm, leading to circumference determination by applying the formula, yielding 35π cm.
5 0
14 days ago
The temperature, H, in degrees Celsius, of a cup of coffee placed on the kitchen counter is given by H = f(t), where t is in min
Leona [11195]

Response:

a

    f'(t) is negative

b

 The unit is degC/min

c

At 35 minutes since the coffee was placed on the counter, the temperature is 68, and it will drop by approximately 0.75 in the next 30 seconds.

Detailed explanation:

According to the question, we are informed that

   The temperature of a cup of coffee on a counter is represented by the equation

         H= f(t)

where t indicates the time in minutes since the coffee was set on the counter.

In general, f(t)' denotes the derivative of f(t), indicating the change in temperature over time for the coffee

As the coffee rests on the counter, its temperature tends to decrease after a few minutes; thus, f(t)' is negative.

The temperature is measured in degrees Celsius while time is expressed in minutes.

Therefore, the change in temperature with respect to time, f(35)', will be given in degrees Celsius per minute (i.e., degC/min).

The question states that

        |f(35)'| = 1.5

which means the rate of temperature change after 35 minutes is 1.5.

And

         |f(35)| = 68

indicating that the coffee's temperature after 35 minutes is 67 degrees Celsius.

Subsequently, after another t= 30 seconds = \frac{30}{60}= 0.5 \ minutes the rate at which the coffee's temperature changes is

          R = |f(35)'| * t

=>        R = 1.5 * 0.5

=>        R = 0.75    

Thus

At 35 minutes since the coffee was placed on the counter, its temperature is 68 and will drop by roughly 0.75 in the next 30 seconds.

     

5 0
1 month ago
The ground-state wave function for a particle confined to a one-dimensional box of length L is Ψ=(2/L)^1/2 Sin(πx/L). Suppose th
AnnZ [10979]

Respuesta:

(a) 4.98x10⁻⁵

(b) 7.89x10⁻⁶

(c) 1.89x10⁻⁴

(d) 0.5

(e) 2.9x10⁻²

Explicación paso a paso:

La probabilidad (P) de encontrar la partícula está dada por:

P=\int_{x_{1}}^{x_{2}}(\Psi\cdot \Psi) dx = \int_{x_{1}}^{x_{2}} ((2/L)^{1/2} Sin(\pi x/L))^{2}dx  

P = \int_{x_{1}}^{x_{2}} (2/L) Sin^{2}(\pi x/L)dx     (1)

La solución de la integral de la ecuación (1) es:

P=\frac{2}{L} [\frac{X}{2} - \frac{Sin(2\pi x/L)}{4\pi /L}]|_{x_{1}}^{x_{2}}  

(a) La probabilidad de encontrar la partícula entre x = 4.95 nm y 5.05 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{4.95}^{5.05} = 4.98 \cdot 10^{-5}    

(b) La probabilidad de encontrar la partícula entre x = 1.95 nm y 2.05 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{1.95}^{2.05} = 7.89 \cdot 10^{-6}  

(c) La probabilidad de encontrar la partícula entre x = 9.90 nm y 10.00 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{9.90}^{10.00} = 1.89 \cdot 10^{-4}    

(d) La probabilidad de encontrar la partícula en la mitad derecha de la caja, es decir, entre x = 0 nm y 50 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{50.00} = 0.5

(e) La probabilidad de encontrar la partícula en el tercio central de la caja, es decir, entre x = 0 nm y 100/6 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{16.7} = 2.9 \cdot 10^{-2}

Espero que te ayude.

3 0
11 days ago
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