*Earth Science Question*

The solubility of most solids _______ as temperatures decreases.

Kitty [74]

Answer:

The solubility of most solids <u>decrease</u> as temperatures decreases.

Hope this helps :)

Have a great day !

5INGH

Explanation:

3 0

3 years ago

Collaborative learning activities:__________.

blagie [28]

Answer:

Options A, B as well as C are the correct choices.

Explanation:

Collaborative learning seems to be a circumstance where learning is an educational approach as well as make an effort to understand all this together.
Except for personal learning, people participating in this learning draw mostly on competencies and qualifications of each other.

This form of learning isn't always linked to the perhaps one given choice. And the response to the above will still be the appropriate one.

3 0

3 years ago

Which statement describes the relationship of resistance and current

slavikrds [6]

This is picture describe as

More current mean less resistance

Less resistance mean more current

6 0

3 years ago

Two blocks a and b ($m_a>m_b$) are pushed for a certain distance along a frictionless surface. how does the magnitude of the

Yuki888 [10]

Answer:

the magnitude of the work done by the two blocks is the same.

Explanation:

The work done by block a on block b is given by:

$W_a = F_a d$

where Fa is the force exerted by block a on block b, and d is the distance they cover.

The work done by block b on block a is given by:

$W_b = F_b d$

where Fb is the force exerted by block b on block a, and d is still the distance they cover.

For Newton's third law, the force exerted by block a on block b is equal to the force exerted by block b on block a, therefore

$F_a = F_b$

and so

$W_a=W_b$

3 0

3 years ago

Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as

Dmitriy789 [7]

Answer:

a) $v=5.6725\,m.s^{-1}$

b) $h= 1.6420\,m$

Explanation:

Given:

mass of the body, $M=20\,kg$

mass of the tyre, $m=10\,kg$

length of hanging of tyre, $l=3.5m$

distance run by the body, $d=10m$

acceleration of the body, $a=3.62m.s^{-2}$

(a)

Using the equation of motion :

$v^2=u^2+2a.d$ ..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

$u=0m.s^{-1}$

putting the values in eq. (1)

$v^2=0^2+2\times 3.62 \times 10$

$v=8.5088\,m.s^{-1}$

Now, the momentum of the body just before the jump onto the tyre will be:

$p=M.v$

$p=20\times 8.5088$

$p=170.1764\,kg.m.s^{-1}$

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

$(M+m)\times v'=p$

$(20+10)\times v'=170.1764$

$v'=5.6725\,m.s^{-1}$

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

$\frac{1}{2} (M+m).v'^2=(M+m).g.h$

$\frac{1}{2} (20+10)\times 5.6725^2=(20+10)\times 9.8\times h$

$h\approx 1.6420\,m$

above the normal hanging position.

3 0

3 years ago