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2 years ago

F = M x G Find the force of gravity acting upon a 1500Kg Hippopatamus.

Physics

1 answer:

expeople1 [14]2 years ago

6 0

Answer:........... .. .....

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2. What are the three phases of water?

Lorico [155]

Answer:

solid, liquid, gas

Explanation:

The three phases of water are the three states of matter water can be found in, and that is:

solid (as in ice)

liquid (as in water)

gas (as in water vapor)

3 0

2 years ago

5 uses of mechanical energy

Oksi-84 [34.3K]

Answer: The 5 uses of mechanical energy are: 1) Hammering a nail

2) Using Dart gun 3) Moon 4) Hydropower plant 5) Sharping a pencil.

8 0

2 years ago

When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to

Serjik [45]

Answer:

E) d/sqrt2

Explanation:

The initial electric force between the two charge is given by:

$F=k\frac{q_1 q_2}{d^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

d is the separation between the two charges

We can also rewrite it as

$d=\sqrt{k\frac{q_1 q_2}{F}}$

So if we want to make the force F twice as strong,

F' = 2F

the new distance between the charges would be

$d'=\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{1}{\sqrt{2}}\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{d}{\sqrt{2}}$

so the correct option is E.

8 0

2 years ago

The fumes of lighted 'incense stick' spread in the whole room due to-

Firlakuza [10]

Answer: burning of the incense and air drift in the room

Explanation: the smoke will automatically rise to the top of the room and then with air drift if would help spread further around the room and spread the smell

7 0

2 years ago

A small glass bead has been charged to 20 nC. What is the magnitude of acceleration in m/s^2 of an electron that is 1.0 cm from

MAVERICK [17]

Answer:

The acceleration is 3.16x10¹⁷ m/s².

Explanation:

First, we need to find the magnitude of the Coulombs force (F):

$|F| = \frac{Kq_{1}q_{2}}{d^{2}}$

<u>Where</u>:

K is the Coulomb constant = 9x10⁹ Nm²/C²

q₁ is the charge = 20x10⁻⁹ C

q₂ is the electron's charge = -1.6x10⁻¹⁹ C

d is the distance = 1.0 cm = 1.0x10⁻² m

$|F| = \frac{Kq_{1}q_{2}}{d^{2}} = \frac{9\cdot 10^{9}Nm^{2}/C^{2}*20 \cdot 10^{-9} C*(-1.6\cdot 10^{-19} C)}{(0.01 m)^{2}} = 2.88 \cdot 10^{-13} N$

Now, we can find the acceleration:

$a = \frac{F}{m} = \frac{2.88 \cdot 10^{-13} N}{9.1 \cdot 10^{-31} kg} = 3.16 \cdot 10^{17} m/s^{2}$

Therefore, the acceleration is 3.16x10¹⁷ m/s².

I hope it helps you!

7 0

2 years ago