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3 years ago

Is water wet my friends have been arguing about this for about 3 months now but now it is time to hear other people

Physics

2 answers:

nirvana33 [79]3 years ago

8 0

Water is not wet, wet is a state of being. Water, however, can make things wet.

juin [17]3 years ago

4 0

I agree its not wet it only makes other objects wet its literally common sense
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What state of matter did the earth need to be in order for planetary differentiation to occur?

STatiana [176]

Gas. I know this because first you will need to do the equator so I will say gas

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2 years ago

If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i

kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from $15 cm$ to $21 cm$ will be , $E_{1}=27J$

The energy used for stretch the spring from $21 cm$ to $27 cm$ will be , $E_{2}=45J$

using the energy of spring formula ,we find that

$27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))$

$45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))$

Dividing both the equation will get,

$\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm$

Therefore, the natural length of the spring is, 9 cm.

4 0

2 years ago

Write the importance of international bureau of weights and measurement.

Taya2010 [7]

Answer:

The importance of international bureau of weights and measures are;

1. to bring unification of measurement system.

2. to establish and preserve fundamental international prototypes.

3. to verify national standards, etc.

8 0

3 years ago

A block of mass m=1kg sliding along a rough horizontal surface is traveling at a speed v0=2m/s when it strikes a massless spring

topjm [15]

Here we can use the work energy theorem

$W_f + W_s = K_f - K_i$

here we know that

$K_f = 0$

as it come to rest finally

$K_i = \frac{1}{2}mv_i^2$

$K_i = \frac{1}{2}\times 1\times 2^2$

$K_i = 2 J$

now work done by friction force will be given as

$W_f = - F_f \times d = -\mu mg d$

$W_f = - \mu(1)(9.8)(0.10) = - 0.98\mu$

Work done by spring force is given as

$W_s = \frac{1}{2}k(x_i^2 - x_f^2)$

$W_s = \frac{1}{2}(10)( 0 - 0.10^2)$

$W_s = -0.05 J$

so now plug in all data above

$- 0.05 - \mu(0.98) = 0 - 2$

$\mu = 1.99$

so above is the friction coefficient

4 0

3 years ago

A rocket is moving at 1/4 the speed of light relative to Earth. At the center of this rocket a light suddenly flashes. To an obs

Sedaia [141]

Answer:

B. the light will reach the front of the rocket at the same instant that it reaches the back of the rocket.

Explanation:

To an observer at rest in the rocket who can't see either sides of the rocket, the speed of the light is constant which means the distance to the front or the back is same and would appear to reach the rocket at the same time.

Although from the point of view of the person on the earth, the front of the rocket is travelling in opposite direction of the light while the back of the rocket is moving closer to the light. This means that the distance travelled by the light going forward will be longer going backwards. And since the speed of light is constant in both directions, the light will reach the back of the rocket before it reaches the front for the observer on the earth.

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3 years ago

Read 2 more answers