<span>Which group in the periodic table is known as salt formers?
The correct option is the last one: Halogen family.
</span><span>
You can find the halogen or "</span>salt formers" in the group 17 of the periodic table. These are:
- Fluorine.
-Chlorine.
- Bromine.
- Iodine.
- Astatine.
All of them are non-metallic elements and they have 7 electrons.
Answer:
27.44 J
Explanation:
We can find the energy at the top of the slide by using the potential energy equation:
At the top of the slide, the swimmer has 0 kinetic energy and maximum potential energy.
The swimmer's mass is given as 7.00 kg.
The acceleration due to gravity is 9.8 m/s².
The (vertical) height of the water slide is 0.40 m.
Substitute these values into the potential energy equation:
- PE = (7.00)(9.8)(0.40)
- PE = 27.44
Since there is 0 kinetic energy at the top of the slide, the total energy present is the swimmer's potential energy.
Therefore, the answer is 27.44 J of energy when the swimmer is at the top of the slide.
Here is your answer:
First find the notations:
2×10^-3
=0002
And...
2.5×10^4=25000
Then divide:
0002÷25000=8E-9
Your answer:
=8 x 10-8
Answer:
17.1
Explanation:
The distance ahead, of the deer when it is sighted by the park ranger, d = 20 m
The initial speed with which the ranger was driving, u = 11.4 m/s
The acceleration rate with which the ranger slows down, a = (-)3.80 m/s² (For a vehicle slowing down, the acceleration is negative)
The distance required for the ranger to come to rest, s = Required
The kinematic equation of motion that can be used to find the distance the ranger's vehicle travels before coming to rest (the distance 's'), is given as follows;
v² = u² + 2·a·s
∴ s = (v² - u²)/(2·a)
Where;
v = The final velocity = 0 m/s (the vehicle comes to rest (stops))
Plugging in the values for 'v', 'u', and 'a', gives;
s = (0² - 11.4²)/(2 × -3.8) = 17.1
The distance the required for the ranger's vehicle to com to rest, s = 17.1 (meters).
mass of the box = 20 kg
force of friction on the box due to surface
similarly kinetic friction on it
now the weight of the suspended block will be
so here the weight of the suspended block is less than the limiting friction on it
So here we will say that friction will counter balance the weight of the suspended block and it will not move at all
So acceleration of the box will be zero
