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2 years ago

Determine the velocity of a car that travels 28km in 2.2 hours

Physics

1 answer:

OLga [1]2 years ago

5 0

Answer:

<h2>velocity = 12.73 km/hr.</h2><h2 />

Explanation:

velocity = distance / time

=<u> 28 km </u>

2.2 hr

= 12.73 km/hr.

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The logarithm of x, written log(x), tells you the power to which you would raise 10 to get x. So, if y=log(x), then x=10^y. It i

fomenos

To solve this problem it is necessary to apply the rules and concepts related to logarithmic operations.

From the definition of logarithm we know that,

$Log_{10}(10) = 1$

In this way for the given example we have that a logarithm with base 10 expressed in the problem can be represented as,

$log_{10}(1,000,000)$

We can express this also as,

$log_{10}(10^6)$

By properties of the logarithms we know that the logarithm of a power of a number is equal to the product between the exponent of the power and the logarithm of the number.

So this can be expressed as

$6*log_{10}(10)$

Since the definition of the base logarithm 10 of 10 is equal to 1 then

$6*1=6$

The value of the given logarithm is equal to 6

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3 years ago

An organism has ____ copies of each gene.

Thepotemich [5.8K]

Answer:

Explanation:

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2 years ago

A train moving at a constant speed on a surface inclined upward at 10.0° with the horizontal travels a distance of 400 meters in

Amiraneli [1.4K]

If the velocity of the train is v=s/t, where s is the distance and t is time, then v=400/5=80m/s. To get the vertical component of the velocity we need to multiply the velocity v with a sin(α): Vv=v*sin(α), where Vv is the vertical component of the velocity and α is the angle with the horizontal. So:

Vv=80*sin(10)=80*0.1736=13.888 m/s.

So the vertical component of the velocity of the train is Vv=13.888 m/s.

7 0

3 years ago

Please help meeeee!!!!!!

Vinil7 [7]

Answer:

Net force

Explanation:

Bruh, easy question

3 0

2 years ago

I'm not really sure how to go about creating the equation, can anyone help me?

AlexFokin [52]

The displacement vector (SI units) is
$\vec{r} =At\hat{i}+A[t^{3}-6t^{2}]\hat{j}$

The speed is a scalar quantity. Its magnitude is
$v= \sqrt{A^{2}t^{2}+A^{2}(t^{3}-6t^{2})^{2}} \\ v=A \sqrt{t^{2}+t^{6}-12t^{5}+36t^{4}} \\ v=At \sqrt{t^{4}-12t^{3}+36t^{2}+1}$

Answer: At√(t⁴ - 12t³ + 36t² + 1)

3 0

3 years ago