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ivolga24 [154]
2 years ago
7

what two objects have stored energy? A. ball rolling on the ground B. a small rock sitting on top of a big rock C. a stretched r

ubber band D. a stone lying on the ground
Physics
2 answers:
Verizon [17]2 years ago
8 0

The correct statements are:

B. a small rock sitting on top of a big rock

As the rock is at a height with respect to ground it has potential Energy

and

C. a stretched rubber band

A stretched rubber band has elastic potential energy


The others are actually moving and hence would consist of Kinetic energy. Potential energy is stored in objects that do not move and are stationary.

Dmitry_Shevchenko [17]2 years ago
6 0
C and as other have expained below, B as well

Potential energy basically means something has the possibilty to change. A <em>stretched</em> rubber band could be released at any moment, which would then create kinetic energy. If the band wasn't stretched it wouldn't have any potential energy. 
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If earth increase the distance from the sun, what will happen to the period of orbi t(the time it takes to complete one revoluti
Mandarinka [93]

The period of the orbit would increase as well

Explanation:

We can answer this question by applying Kepler's third law, which states that:

"The square of the orbital period of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit"

Mathematically,

\frac{T^2}{a^3}=const.

Where

T is the orbital period

a is the semi-major axis of the orbit

In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit, a: but as we saw from the previous equation, the orbital period of the Earth is proportional to a, therefore as a increases, T increases as well.

Therefore, the period of the orbit would increase.

Learn more about Kepler's third law:

brainly.com/question/11168300

#LearnwithBrainly

5 0
3 years ago
Which type of exercise is weightlifting?
MatroZZZ [7]

The answer is option B "anaerobic." Weightlifting deals with stress to the muscles when lifting weights and due time the muscles will begin to adapt and get stronger. Other examples of anaerobic exercise are things like: weight training, sprinting, cycling, and jumping anything that has short exertion, and high-intensity movement is an anaerobic exercise.

Hope this helps!

Nonportrit

5 0
3 years ago
Read 2 more answers
I need help with questions 4 and 5 because my teacher is not good with explaining them to a better understanding (Geometry)
Helga [31]
Hello! I can help you with this! 

4. For this problem, we have to write and solve a proportion. We would set this proportion up as 12/15 = 8/x. This is because we're looking for the length of the shadow and we know the height of the items, so we line them up horizontally and x goes with 8, because we're looking for the shadow length. Let's cross multiply the values. 15 * 8 = 120. 12 * x = 12. You get 120 = 12x. Now, we must divide each side by 12 to isolate the "x". 120/12 is 10. x = 10. There. The cardboard box casts a shadow that is 10 ft long.

5. For this question, you do the same thing. This time, you're finding the height of the tower, so you would do 1.2/0.6 = x/7. Cross multiply the values in order to get 8.4 = 0.6x. Now, divide each side by 0.6x to isolate the "x". 8.4/0.6 is 14. x = 14. There. The tower is 14 m tall.

If you need more help on proportions and using proportions in real life situations, feel free to search on the internet to find more information about how you solve them.
4 0
2 years ago
An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.
Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

u -v= d

u=71+26=97\ cm

We need to calculate the focal length

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

put the value into the formula

\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}

\dfrac{1}{f}=-\dfrac{71}{2522}

f=-35.52\ cm

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value in to the formula

\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}

\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}

\dfrac{1}{v}=-\dfrac{124}{2755}

v=-22.21\ cm

We need to calculate the magnification

Using formula of magnification

m=\dfrac{v}{u}

m=\dfrac{22.21}{95}

m=0.233

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

4 0
3 years ago
Sammy squirrel is steering his boat at a heading of 327 degree at 18mph. The current is flowing at 4mph at a heading of 60 degre
Tanya [424]

Answer:

  • 59.97 º at 18.23 mph

Explanation:

To find Sammy's course you have to add the two velocities (vectors), 18 mph 327º and 4 mph 60º.

To add the two vectors analytically you decompose each vector into their vertical and horizontal components.

<u>1. 18 mph 327º</u>

  • Horizontal component: 18 mph × cos (327º) = 15.10 mph

  • Vertical component: 18 mph × sin (327º) = - 9.80 mph

  • Vector notation:

       15.10\hat i-9.80\hat j

<u>2. 4 mph 60º</u>

  • Horizontal component: 4 mph × cos (60º) = 2.00 mph

  • Vertical component: 4 mph × sin (60º) = 3.46 mph

  • Vector notation:

       2.00\hat i+3.46\hat j

<u>3. Addition:</u>

You add the corresponding components:

15.10\hat i-9.80\hat j+2.00\hat i+3.46\hat j\\ \\ 17.10\hat i-6.34\hat j

To find the magnitude use Pythagorean theorem:

  • \sqrt{17.1^2+6.34^2}= 18.23

<u>4. Direction:</u>

Use the tangent ratio:

  • tan(\alpha )=opposite/adjacent=3.46/2.00=1.73

Find the inverse:

  • arctan (1.73) ≈ 59.97º
5 0
3 years ago
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