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3 years ago

Give an example of a reputable website

2 answers:

8 0

Answer: A reputable website is one that possesses authentic and credible information that can be easily relied upon. For instance, all official government websites are reputable.

Example- www.csirhrdg.res.in is a reputable website of CSIR ( counsil of scientific and industrial research) India. It is one of the premier national and world's largest publically funded resource and development organization.

Leni [432]3 years ago

6 0

Well, almost any website can be, but if you are meaning for money, then google. if you are meaning reliable, then look for something with .org, its history and stuff

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The mass of the particles that a river can transport is proportional to the sixth power of the speed of the river. A certain riv

kow [346]

Answer:

1.122 m/s

Explanation:

So usually a river with a speed of 1 meters per second can transport particle that weighs:

$1^6 = 1 kg$

If the particle is twice as massive as usual, then its weights would be 1 * 2 = 2kg

This means the river must be flowing at a speed of

$2^{\frac{1}{6}} = 1.122 m/s$

5 0

3 years ago

A student visits the beach and wants to explore how landforms, such as sand dunes, are the result of changes caused by wind. wha

posledela

C Camera. I think this because you can make timelapses with cameras which makes it easy to see.

3 0

3 years ago

Read 2 more answers

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1

tamaranim1 [39]

The acceleration would be 6m/sThis is because of the formula, "f/m=a" to find the acceleration; We would need to subtract the force of the friction which equals 1380, then divide that by the mass (which was 230) to get the answer 6

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3 years ago

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a lamp or bulb is marked 12 volt 240 Watts how many joules does it consume in one hour and what is the current that passes throu

valkas [14]

Answer:

1 watt is 1 Joule per second. 240 watt would then be 240 joules per second. How many seconds are there in an hour? Current formula is P= IV , P is 240(not the power in an hour because the current is calculated in seconds), and V is 12V. Then you will get I, the current.

5 0

3 years ago