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3 years ago

12

The capitary action of water is caused the cohesion of water molecules to different types of molecules and the adhesion of water

molecules with other water molecules true or false

1 answer:

s344n2d4d5 [400]3 years ago

5 0

Answer:

This is True

Explanation:

I just did this exact unit in bio last week I hope this helps ;)

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The amount of work done by two boys who apply 200 N of force in an

Aleksandr [31]

The amount of work done by two boys who apply 200 N of force in an unsuccessful attempt to move a stalled car is 0.

Answer: Option B

<u>Explanation: </u>

Work done is the measure of work done by someone to push an object from its present position. We can also define work done as the amount of forces needed to move an object from its present position to another position. So the amount of work done is directly proportionate to the product of forces acting on the object and the displacement of the object.

$\text { Work done }=\text { Force } \times \text { displacement }$

So in this present case, as the two boys have done an unsuccessful attempts to push a stalled car so that means the displacement of the car is zero as there is no change in the position of the car. But they have applied a force of 200 N each. So the amount of work done will be

$\text { Work done }=200 \mathrm{N} \times 0=0$

Thus, the amount of work done by two boys will be zero due to their unsuccessful attempt to move a stalled car.

8 0

2 years ago

Calculate the force of gravity on the 0.60-kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth r

KIM [24]

The force of gravity between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the mass of the object is $m_1=0.60 kg$ , while the Earth's mass is $m_2=5.97 \cdot 10^{24} kg$ . Their separation is $r=1.3 \cdot 10^7 m$ , therefore the gravitational force exerted on the object is
$F=(6.67 \cdot 10^{-11}m^3 kg^{-1} s^{-2}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}=1.4 N$

5 0

2 years ago

How long a star lives and what it becomes at the end of its life depends primarly on what

liraira [26]

I believe the answer would be mass. Low mass stars and medium mass stars often become white dwarfs when they die while high mass stars explode in violent explosions called supernovas and usually leave behind a black hole or a neutron star.

4 0

3 years ago

Which of your groups should you NOT change anything for? (in other

ella [17]

Answer:

B

Explanation:

The control is something that is meant to not be changed, the control is a comparison of the experimental.

7 0

3 years ago

You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?

Natali5045456 [20]

Equation of motion:

$y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}$

Since initial velocity is zero, the second term goes away:

$y_{f}=y_{o}+0+ \frac{1}{2} at^{2}$

$y_{f}=y_{o}+\frac{1}{2} at^{2}$

$y_{f}-y_{o}= \frac{1}{2} at^{2}$

$y_{f}-y_{o}=5.4m$

$5.4m= \frac{1}{2} at^{2}$

$\frac{2(5.4m) }{a} = t^{2}$

$a = g = 9.81 \frac{m}{ s^{2}}$

$\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}$

$1.1 s^{2} = t^{2}$ $\sqrt{1.1 s^{2}} = \sqrt{t^{2}}$

<u><em> $t = 1.05s$ </em></u>

4 0

3 years ago

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