All categories

3 years ago

State and explain the properties of static electricity charges

1 answer:

7 0

The phenomenon of static electricity requires a separation of positive and negative charges. When two materials are in contact, electrons may move from one material to the other, which leaves an excess of positive charge on one material, and an equal negative charge on the other.

You might be interested in

Which of the following does NOT use a turbine?

katovenus [111]

Answer:

laa a

Explanation:poque usairea ye aire tienfe luza preopia mkr

5 0

4 years ago

An 8.00-g sample of aluminum reacts with oxygen to form 15.20 g of aluminum oxide. how many grams of oxygen are used in the reac

konstantin123 [22]

I would say the the answer is 7.20 because if subtract the 8.00 grams from the 15.20 g you would get 7.20. Am I correct, please ley me know

7 0

3 years ago

What is the relation between centre of gravity and stability

maw [93]

Explanation:

tilting it will raise the height of its center of gravity.

8 0

3 years ago

A force of 6.0 Newton’s gives a 2.0kg block an acceleration of 3.0m/s to the right. The force of friction acting on the block is

DochEvi [55]

The force of friction is zero.
Using F = m a = the force needed is 6 N which means no friction can be acting.

3 0

4 years ago

- A cannon of 2000 kg fires a shell of 10 kg at

fgiga [73]

1) -0.5 m/s

We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.

Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:

$p=0$

After the shot, the momentum is:

$p=MV+mv$

where

M = 2000 kg is the mass of the cannon

m = 10 kg is the mass of the shell

v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)

V = ? is the velocity of the cannon

Since momentum is conserved, we can write

$0=MV+mv$

And solving for V, we find the velocity of the cannon:

$V=-\frac{mv}{M}=-\frac{(10)(100)}{2000}=-0.5 m/s$

where the negative sign indicates that the cannon moves in the direction opposite to the shell.

2) 0.5 m

The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:

$v^2-u^2=2as$

where

v is the final velocity of the cannon

u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)

$a=-0.25 m/s^2$ is the deceleration of the cannon

s is the distance travelled by the cannon

The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:

$s=\frac{v^2-u^2}{2a}=\frac{0-0.5^2}{2(-0.25)}=0.5 m$

7 0

3 years ago