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3 years ago

Neptune is approximately 4.5 billion kilometers from the sun. What is this distance in AU?

1 answer:

7 0

Well, one AU is 149,597,870 km. So, we would basically have to divide 4.5 billion km by 149,597,870, right?

4,500,000,000/149,597,870=30.080642 AU.

So, the correct answer would be 30 AU. Hoped this helped!

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The weight of an object on Earth is 350 newtons. On Mars, the same object would weigh 134 newtons. What is the acceleration due

sineoko [7]

W = mg = 350 newton
m = W/g = 350/9.8 = 35.71 kg
on mars
W = mg = 134 newton
g = W/m = 134/35.71 = 3.75 meters/second2

4 0

3 years ago

Read 2 more answers

The specification limits for a product are 8 cm and 10 cm. A process that produces the product has a mean of 9.5 cm and a standa

My name is Ann [436]

Answer:

The value of Cpk is 0.83.

Explanation:

Given that,

Upper specification limits = 10 cm

lower specification limits = 8 cm

Mean = 9.5

Standard deviation = 0.2 cm

We need to calculate the process capability

Using formula of Cpk

$Cpk=min(\dfrac{USL-mean}{3\times SD}, \dfrac{mean-LSL}{3\times SD})$

Put the value into the formula

$Cpk=min(\dfrac{10-9.5}{3\times0.2}, \dfrac{9.5-8}{3\times0.2})$

$Cpk=min(0.83,2.5)$

$Cpk=0.83$

Hence, The value of Cpk is 0.83.

4 0

2 years ago

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

3 years ago

AGREE OR DISAGREE: All light can be seen.

Tema [17]

Answer:

Disagree

Explanation:

We cannot see infared lights.

3 0

2 years ago

A grey hound can run about 40/mph is it velocity,acceleration, or speed

loris [4]

Answer: Speed

Explanation:

Speed is the time it takes something to travel a certain distance. Accelaration is the rate at which an object's speed changes. Velocity is a vector of the object's speed and direction.

5 0

3 years ago