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3 years ago

9. If a car travels 18,000 meters in 30 minutes, what is its average speed?

Physics

1 answer:

erica [24]3 years ago

5 0

If a car travels 18,000 meters in 30 minutes, what is its average speed?

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Novosadov [1.4K]

Explanation:

The are the same but difference

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2 years ago

What is the distance between a 900 kg compact car and a 1600 kg pickup truck if the gravitational force between them is about 0.

elena-14-01-66 [18.8K]

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

$r= \sqrt{\frac{Gm1m2}{F} }$

$r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m$

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required:

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

$r= \sqrt{\frac{Gm1m2}{F} }$

$r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m$

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2 years ago

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In this lab, you will use a magnet and a simple circuit to examine the concepts of electricity and magnetism. Think about how th

Alex73 [517]

Sample Response: How can magnetic and electric fields be demonstrated?

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3 years ago

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Ocean currents near the equator are primarily: eastward westward north and south clockwise

sladkih [1.3K]

Equatorial currents are primarily westward. This is because the dominant current in the northern hemisphere has a clockwise direction, while the southern hemisphere has a counterclockwise direction. When these two currents meet at the equator, a common westward current exists.

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3 years ago

g (12 points) The time between incoming phone calls at a call center is a random variable with exponential density p(x) = 1 r e

rusak2 [61]

Answer:

$(1)p(x)\geq 0\\(2)\int_{0}^{\infty} p(x) dx=0$

Explanation:

A function f(x) is a Probability Density Function if it satisfies the following conditions:

$(1)f(x)\geq 0\\(2)\int_{0}^{\infty} f(x) dx=0$

Given the function:

$p(x)=\dfrac{1}{r}e^{-x/r} \: on\: [0,\infty), where\:r=\dfrac{20}{ln(2)}$

(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in $[0,\infty).$

(2)

$\int_{0}^{\infty} p(x)=\int_{0}^{\infty} \dfrac{1}{r}e^{-x/r}\\=\dfrac{1}{r} \int_{0}^{\infty} e^{-x/r}\\=-\dfrac{r}{r}\left[e^{-x/r}\right]_{0}^{\infty}\\=-\left[e^{-\infty/r}-e^{-0/r}\right]\\=-e^{-\infty}+e^{-0}\\SInce \: e^{-\infty} \rightarrow 0\\e^{-0}=1\\\int_{0}^{\infty} p(x)=1$

The function p(x) satisfies the conditions for a probability density function.

6 0

3 years ago