All categories

3 years ago

5. Did objects have to touch to interact? What causes this?

Physics

2 answers:

dimulka [17.4K]3 years ago

6 0

Answer:

★ For example, a useful analogy for explaining the Earth's gravity force is that the Earth can pull on objects without touching them just like a magnet can affect other objects without touching them. In Addition, the main notion to convey here is that forces can act at a distance with no perceivable substance in between.

Explanation:

Hope you have a great day :)

djverab [1.8K]3 years ago

5 0

Answer/Explanation

A useful analogy for explaining the Earth's gravity force is that the Earth can pull on objects without touching them just like a magnet can affect other objects without touching them. The main notion to convey here is that forces can act at a distance with no perceivable substance in between.

You might be interested in

4-08 t 1 - / S P = 1 / P = S 1 S= P Def. of Speed

nordsb [41]

Answer:

what do I answer? not enough information

8 0

3 years ago

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 n frictional force. he pushes in a dire

alexandr1967 [171]

Work is calculated as the product of Force, Distance, and angular motion. In this case, the work done by gravity is perpendicular to the motion of the cart, so θ = 90°

and W=Fdcosθ

W=35.0 N x 20.0 m x cos90

W=0 J

This means that work done perpendicular to the direction of the motion is always zero.

5 0

3 years ago

A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?

vesna_86 [32]

Answer:

The time for final 15 cm of the jump equals 0.1423 seconds.

Explanation:

The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as

$v^2=u^2+2as$

where

'v' is the final velocity of the player

'u' is the initial velocity of the player

'a' is acceleration due to gravity

's' is the height the player jumps

Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

$0^2=u^2-2\times 9.81\times 0.76\\\\\therefore u=\sqrt{2\times 9.81\times 0.76}=3.86m/s$

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

$v^{2}=3.86^2-2\times 9.81\times 0.66\\\\\therefore v=\sqrt{3.86^2-2\times 9.81\times 0.66}=1.3966m/s$

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as

$v=u+at$

where symbols have the usual meaning

Applying the given values we get

$t=\frac{v-u}{g}\\\\t=\frac{0-1.3966}{-9.81}=0.1423seconds$

4 0

3 years ago

The mass of a string is 5.9 × 10-3 kg, and it is stretched so that the tension in it is 200 n. a transverse wave traveling on th

bagirrra123 [75]

The velocity of the wave on the string is given by

$v=\sqrt{\frac{T}{\frac{m}{L}}} \\ v=\sqrt{\frac{TL}{m}}$

Solving the above equation,

$v^2=\frac{TL}{m} \\ L=\frac{v^2m}{T}$

The frequency of the wave $f=300$ and wave length is $0.76$

The velocity is $v=(300)(0.76)=228$

Substituting numerical values,

$L=\frac{228^2(0.0059)}{200}\\ T=1.534$

The length of the string is $1.534 m$

4 0

3 years ago

Adita lifts a book from the floor, carries it across the room, and places it on a high shelf. When is Adita doing work on the bo

noname [10]

She does work from the moment she touches the book until she lets it go. Work is anything that requires energy. Therefore, she is working as she picks up the book, carries it, and when she is lifting it onto the shelf.

5 0

3 years ago

Read 2 more answers