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3 years ago

Use the image below to answer the following question (ruler not to scale).

Physics

2 answers:

maks197457 [2]3 years ago

7 0

Answer:

Yes, they are correct, you need to include all parts of the question, don't just copy and paste the words. But in the diagram that was supposed to be there, the magnets' north poles are facing each other. Therefore they would push away from each other if they were to get any closer. So, the magnetic potential energy would decrease due to their movement away from each other. So your answer is (C. It will decrease.

Explanation:

Svetradugi [14.3K]3 years ago

4 0

Answer:

it depends on wether the + and - are facing eachother

or away from eachother

Explanation:

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Car 1 goes around a level curve at a constant speed of 65 km/h . The curve is a circular arc with a radius of 95 m . Car 2 goes

Arte-miy333 [17]

Answer:

The radius of the curve that Car 2 travels on is 380 meters.

Explanation:

Speed of car 1, $v_1=65\ km/h$

Radius of the circular arc, $r_1=95\ m$

Car 2 has twice the speed of Car 1, $v_2=130\ km/h$

We need to find the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration. We know that the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

According to given condition,

$\dfrac{v_1^2}{r_1}=\dfrac{v_2^2}{r_2}$

$\dfrac{65^2}{95}=\dfrac{130^2}{r_2}$

On solving we get :

$r_2=380\ m$

So, the radius of the curve that Car 2 travels on is 380 meters. Hence, this is the required solution.

4 0

3 years ago

You shake a bottle of soda and take off the cap. If the soda shoots out of the

Harman [31]

Answer:

A. 16.9 m

Explanation:

I think this is the answer i am not sure

but hope it helps

3 0

2 years ago

Read 2 more answers

Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars

zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

$T^2 \propto R^3$

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

$T^2 \propto R^3$

$T^2 = kR^3$

Substituting the values, we get the value of constant k for mars.

$687^2 = k\times (2.279 \times 10^{11})^3$

$k = 3.92 \times 10^{-29}$

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

$365^3 = 3.92\times 10^{-29} R^3$

$R^3 = 3.39 \times 10^{33}$

$R= 1.50 \times 10^{11}\;\rm m$

Hence we can conclude that the distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

6 0

2 years ago

I need help with these questions.

strojnjashka [21]

I'll go ahead and answer the ones here without an answer. For reference, the half-life formula is <em>final amount = original amount(1/2)^(time/half-life)</em>

<em />

4) 12.5g

x = 100(1/2)^(63/21)

5) 50g

3.125 = x(1/2)^(0.1/0.025)

6) 500g

x = 4000(1/2)^(525/175)

7) 0.24g

0.06 = x(1/2)^(11430/5730)

8) 125g

x = 1000(1/2)^(17100/5700)

Hope this helps! :)

4 0

3 years ago

In a solid , the atoms are tightly locked in position an do not change posistion , true or false

attashe74 [19]

the answer is true because they're so close together that they can't move they can't slide past each other or anything

6 0

3 years ago