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3 years ago

Patty decides to try another experiment. From the

Physics

1 answer:

Effectus [21]3 years ago

6 0

Answer: They will fall at the same time

Explanation:I had a google doc about this

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What is the weight on the surface of Earth of an object of mass 2.00 kilograms?

Anni [7]

Answer:

Explanation:

19.6 newtons

A 2.00-kilogram object weighs 19.6 newtons on Earth.

6 0

3 years ago

Read 2 more answers

Jobisdone [24]

Hey!!

here is your answer >>>

We know that, acids and strong bases are good conductors of electricity!. And gold conducts electricity , while the best is silver and niext comes gold!. And H20 which is water contains less ions , so it is not a good conductors , it is a conductors but not a good conductor!. Salt in it's solid for cannot produce ions, hence, it cannot conduct electricity , but when this salt is mixed with water or when it is in it's molten state it can conduct electricity!. So now,the answer for the question is ,

B) III and IV only!.

Hope my answer helps!

7 0

3 years ago

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a circular cylinder and isused to maintain a water depth of 4 m. That is, when the water depth exceeds 4 m, thegate opens slight

stiv31 [10]

Answer:

W / A = 39200 kg / m²

Explanation:

For this problem let's use the equilibrium equation of / newton

F = W

Where F is the force of the door and W the weight of water

W = mg

We use the concept of density

ρ = m / V

m = ρ V

The volume of the water column is

V = A h

We replace

W = ρ A h g

On the other side the cylinder cover has a pressure

P = F / A

F = P A

We match the two equations

P A = ρ A h g

P = ρ g h

P = 39200 Pa

The weight of the water column is

W = 1000 9.8 4 A

W / A = 39200 kg / m²

3 0

3 years ago

Children in a tree house lift a small dog in a basket 4.00 m up to their house. If it takes 187 J of work to do this, what is th

Vika [28.1K]

A=Fh
A - work
F - force
h - distance

F=mg
m - mass (god+basket)

so
A=mgh
187 = m*10*4
187=40m
m=187/40
m=4.675 kg
or 4kg and 675g

pretty small dog...

6 0

3 years ago

Two objects carry initial charges that are q1 and q2, respectively, where |q2| > |q1|. They are located 0.160 m apart and beh

mart [117]

Answer:

$\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges $\rm q_1$ and $\rm q_1$ , separated by a distance $\rm r$ , is given by

$\rm F = \dfrac{kq_1q_2}{r^2}.$

where k is the Coulomb's constant.

Initially,

$\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.$

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

$\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).$

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, $\rm F_f = +1.30\ N.$

The new charges on the two objects are

$\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.$

The new force is given by

$\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\$

Using (1),

$\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0$

$\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0$

$\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.$