a.
The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:
where F is the magnitude of the force, theta is the angle between them and d is the distance.
The problen gives the following data:
The magnitude of the force 750 N.
The angle between the force and the displacement which is 25°
The distance, 26 m.
Plugging this in the formula we have:
Therefore the work done is 17673 J.
b)
The power is given by:
the problem states that the time it takes is 6 s. Then:
Therefore the power is 2945.5 W
Answer:
O²⁻
Explanation:
Number of protons = 8
Number of neutrons = 9
Number of electrons = 10
What type of atom or ion is it = ?
Solution:
Protons are the positively charged particle in an atom
Neutrons do not carry any charges
Electrons are negatively charged particles
For this atom, the number of protons helps to identify what specie it is; so this is an oxygen atom.
Now,
Charge = Number of protons - Number of electrons
Charge = 8 - 10 = -2
The charge on the atom is -2 and so it is an oxygen ion with -2 charge
The ion is O²⁻
Suppose car A is moving with a velocity Va, and car b with a velocity Vb,
According the principle of conservation of momentum:
Va x Ma + Vb x Mb = (Ma + Mb) V
V = (Va x Ma + Vb x Mb)/(Ma +Mb)
V = speed of cars after coupling
V = (Va x 20 mg + Vb x 15 mg)/(20 mg + 15 mg)
Put in the values of Va and Vb, and get the V
<span>1078 kgm / s would be the answer I hope this helps!!!</span>
Answer:
<em>The answer is B</em>
Explanation:
<em>I got this from study island</em>
