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Sedaia [141]
3 years ago
6

BRAINLIEST PLZ HELP!

Physics
1 answer:
balu736 [363]3 years ago
5 0
The correct answer should be A
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A ping pong ball with a dent in it can be put into a pan of boiling water. After a short amount of time, the dent will pop out.
kifflom [539]

heat from water goes into air in ball

air expands

ping goes the dent

6 0
3 years ago
Read 2 more answers
A 6.0-kilogram cart initially traveling at 4.0 meters per second east accelerates uniformly at 0.50 meter per second squared eas
VMariaS [17]

Answer: 5.5m/s

Explanation:

vf=vi+at

vf= 4.0m/s + (0.50m/s^2)(3.0s)

3 0
3 years ago
A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to
uysha [10]

Answer: 2.94×10^8 J

Explanation:

Using the relation

T^2 = (4π^2/GMe) r^3

Where v= velocity

r = radius

T = period

Me = mass of earth= 6×10^24

G = gravitational constant= 6.67×10^-11

4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]

= 0.9865 x 10^-13

Therefore,

T^2 = (0.9865 × 10^-13) × r^3

r^3 = 1/(0.9865 × 10^-13) ×T^2

r^3 = (1.014 x 10^13) × T^2

To find r1 and r2

T1 = 120min = 120*60 = 7200s

T2 = 180min = 180*60= 10800s

Therefore,

r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m

r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m

Required Mechanical energy

= - GMem/2 [1/r2 - 1/r1]

= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]

= (2001 x 10^7)/2 * (0.1239 - 0.0945)

= (1000.5 × 10^7) × 0.0294

= 29.4147 × 10^7 J

= 2.94 x 10^8 J.

6 0
3 years ago
Networks that are designed to connect similar computers that share data and software with each other are called: client/server n
aev [14]

peer-to-peer networks

Explanation:

Networks designed to connects similar computers that shares data and software with each other are called peer-to-peer networks.

These networks are closely around and do not rely on information passing through a central data point.

  • In a peer to peer connection, the computers have direct access to one another.
  • The computers over the network becomes both client and server.

learn more:

Machines on public network brainly.com/question/10338479

#learnwithBrainly

8 0
3 years ago
A particular spiral galaxy can be approximated by a thin disk-like volume 62 Thousand Light Years in radius and 7 Hundred Light
SVETLANKA909090 [29]

Answer:

Approximate linear dimension is 2 light years.

Explanation:

Radius of the spiral galaxy r = 62000 LY

Thickness of the galaxy h = 700 LY

Volume of the galaxy = πr²h

                                   = (3.14)(62000)²(700)

                                   = (3.14)(62)²(7)(10)⁸

                                   = 84568×10⁸

                                   = 8.45\times 10^{12} (LY)³

Since galaxy contains number of stars = 1078 billion stars ≈ 1.078\times 10^{12}

Now volume covered by each star of the galaxy = \frac{\text{Total volume of the galaxy}}{\text{Number of stars}}

= \frac{8.45\times 10^{12} }{1.078\times 10^{12}}

= 7.839 Light Years

Now the linear dimension across the volume

= (\text{Average volume per star})^{\frac{1}{3}}

= (7.839)^{\frac{1}{3}}

= 1.99 LY

≈ 2 Light Years

Therefore, approximate linear dimension is 2 light years.

8 0
3 years ago
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